title
DP 24. Rod Cutting Problem | 1D Array Space Optimised Approach

description
Lecture Notes/C++/Java Codes: https://takeuforward.org/dynamic-programming/striver-dp-series-dynamic-programming-problems/ Problem Link: https://bit.ly/3H10kYJ Pre-req for this Series: https://www.youtube.com/watch?v=yVdKa8dnKiE&list=PLgUwDviBIf0rGlzIn_7rsaR2FQ5e6ZOL9 a Make sure to join our telegram group for discussions: https://linktr.ee/takeUforward Full Playlist: https://www.youtube.com/watch?v=FfXoiwwnxFw&list=PLgUwDviBIf0qUlt5H_kiKYaNSqJ81PMMY In this video, we solve the problem of cutting rod. We start with memoization, then tabulation, then two-row space optimization, ending up with single row optimisation. This problem is the last problem on DP on Subsequences Pattern. Please watch DP 23 before watching this. If you have not yet checked our SDE sheet, you should definitely do it: https://takeuforward.org/interviews/strivers-sde-sheet-top-coding-interview-problems/ You can also get in touch with me at my social handles: https://linktr.ee/takeUforward

detail
{'title': 'DP 24. Rod Cutting Problem | 1D Array Space Optimised Approach', 'heatmap': [{'end': 757.38, 'start': 688.758, 'weight': 0.768}], 'summary': 'Discusses maximizing rod lengths to make n and ensuring it is maximized, with a focus on collecting rod lengths and maximizing the price while doing so, achieving the highest possible price of 12 in the rod cutting problem. it explores the similarity of the problem to the knapsack problem, the approach to pick rod lengths and sum them up to make the given n, and maximizing the price using unbounded knapsack. additionally, it delves into dynamic programming for rod cutting, algorithms for determining maximum price and cost of rods, writing base cases and recurrences in dynamic programming, and achieving space optimization into a 1d array.', 'chapters': [{'end': 244.983, 'segs': [{'end': 90.191, 'src': 'embed', 'start': 61.424, 'weight': 2, 'content': [{'end': 64.105, 'text': 'Now what you need to do is you need to cut the rod into pieces.', 'start': 61.424, 'duration': 2.681}, {'end': 66.788, 'text': 'Cutting, rod cutting, cut the rod into pieces.', 'start': 64.647, 'duration': 2.141}, {'end': 75.216, 'text': 'You can be like, hey, striver, I can cut the five rod into a length one, a length one, a length one, a length one, a length one.', 'start': 67.309, 'duration': 7.907}, {'end': 81.962, 'text': 'Possible? And for length 1, for length 2, length 3, length 4, length 5.', 'start': 76.417, 'duration': 5.545}, {'end': 86.267, 'text': 'For length 1, the price that you will get in the market is 2.', 'start': 81.962, 'duration': 4.305}, {'end': 90.191, 'text': 'So the pricing that you will get in the market is 2 for every length of rod.', 'start': 86.267, 'duration': 3.924}], 'summary': 'Cut the rod into pieces to get different prices in the market.', 'duration': 28.767, 'max_score': 61.424, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo61424.jpg'}, {'end': 244.983, 'src': 'embed', 'start': 181.817, 'weight': 0, 'content': [{'end': 183.958, 'text': 'try to think it in the other fashion.', 'start': 181.817, 'duration': 2.141}, {'end': 184.438, 'text': 'what fashion?', 'start': 183.958, 'duration': 0.48}, {'end': 186.195, 'text': 'Can I say this?', 'start': 185.555, 'duration': 0.64}, {'end': 189.357, 'text': 'You have a lot of lengths.', 'start': 187.436, 'duration': 1.921}, {'end': 202.223, 'text': 'Because if n is equal to 5, you know the rod lengths are, like the rod lengths can be 1, can be 2, can be 3, can be 4, can be 5.', 'start': 189.997, 'duration': 12.226}, {'end': 204.504, 'text': 'We can have 5 different rod lengths.', 'start': 202.223, 'duration': 2.281}, {'end': 211.407, 'text': '1 of a length 1, 1 of a length 2, 1 of a length 3, 1 of a length 4 and 1 of a length 5.', 'start': 204.524, 'duration': 6.883}, {'end': 226.123, 'text': 'So can I say, How do you collect rod lengths? Can I write this? Collect rod lengths to make N.', 'start': 211.407, 'duration': 14.716}, {'end': 230.106, 'text': 'And while collecting rod lengths, maximize the price.', 'start': 226.123, 'duration': 3.983}, {'end': 233.568, 'text': 'Can I say can I think the problem in the opposite fashion??', 'start': 230.386, 'duration': 3.182}, {'end': 242.754, 'text': 'Instead of saying break N into pieces of rod and sell, can I say collect rod lengths and make it N and make sure it is maximized?', 'start': 234.148, 'duration': 8.606}, {'end': 244.983, 'text': "They'll be like, yes, Trevor, makes sense.", 'start': 243.482, 'duration': 1.501}], 'summary': 'Maximize price by collecting rod lengths to make n.', 'duration': 63.166, 'max_score': 181.817, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo181817.jpg'}], 'start': 0.189, 'title': 'Maximizing rod lengths', 'summary': 'Discusses the concept of maximizing rod lengths to make n and ensuring it is maximized, with a focus on collecting rod lengths and maximizing the price while doing so. the highest possible price being 12 is considered in the rod cutting problem.', 'chapters': [{'end': 181.817, 'start': 0.189, 'title': 'Rod cutting problem', 'summary': 'Covers the rod cutting problem, where given a rod of length n and prices for each length, the task is to maximize the total price obtained after cutting and selling the rod, with the highest possible price being 12.', 'duration': 181.628, 'highlights': ['The problem focuses on the rod cutting scenario, where a rod of length 5 with corresponding prices 2, 5, 7, 8, and 10 is given, and the goal is to maximize the total price obtained after cutting and selling the rod, with the highest possible price being 12.', 'It emphasizes the need to cut the rod into pieces and sell them in the market to maximize the total price, showcasing various cutting combinations to achieve the maximum cost, such as cutting it into lengths of 1, 2, and 2, yielding a total price of 12.', 'The problem requires trying every possible way to determine the maximal cost, with an example demonstrating that cutting the rod into lengths of 2 and 3 also yields the maximum cost of 12.', 'The task ultimately requires printing the maximum cost obtained, which in this case is 12, emphasizing the goal of maximizing the price obtained from selling the cut rod pieces in the market.']}, {'end': 244.983, 'start': 181.817, 'title': 'Maximizing rod lengths', 'summary': 'Discusses the concept of maximizing rod lengths to make n and ensuring it is maximized, with a focus on collecting rod lengths and maximizing the price while doing so.', 'duration': 63.166, 'highlights': ['Collecting rod lengths to make N and maximizing the price while doing so.', 'Thinking about the problem in the opposite fashion of breaking N into pieces of rod and selling.']}], 'duration': 244.794, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo189.jpg', 'highlights': ['The problem focuses on maximizing the total price obtained after cutting and selling the rod, with the highest possible price being 12.', 'The task requires trying every possible way to determine the maximal cost, with an example demonstrating that cutting the rod into lengths of 2 and 3 also yields the maximum cost of 12.', 'It emphasizes the need to cut the rod into pieces and sell them in the market to maximize the total price, showcasing various cutting combinations to achieve the maximum cost.', 'Collecting rod lengths to make N and maximizing the price while doing so.', 'Thinking about the problem in the opposite fashion of breaking N into pieces of rod and selling.']}, {'end': 460.451, 'segs': [{'end': 303.984, 'src': 'embed', 'start': 274.602, 'weight': 0, 'content': [{'end': 279.565, 'text': 'i say this if i write down the index number 0, 1, 2, 3, 4.', 'start': 274.602, 'duration': 4.963}, {'end': 283.308, 'text': 'i know index 4 means a length 5.', 'start': 279.565, 'duration': 3.743}, {'end': 285.689, 'text': 'i know index 3 means a length 4.', 'start': 283.308, 'duration': 2.381}, {'end': 287.931, 'text': 'i know index 2 means a length 3.', 'start': 285.689, 'duration': 2.242}, {'end': 290.032, 'text': 'i know index 1 means a length 2.', 'start': 287.931, 'duration': 2.101}, {'end': 292.573, 'text': 'i know index 0 means a length 1.', 'start': 290.032, 'duration': 2.541}, {'end': 293.854, 'text': 'so can i do?', 'start': 292.573, 'duration': 1.281}, {'end': 297.356, 'text': 'can i solve the problem in a similar way to knapsack?', 'start': 293.854, 'duration': 3.502}, {'end': 297.897, 'text': 'i think i can.', 'start': 297.356, 'duration': 0.541}, {'end': 303.984, 'text': "So what I'll do is I will try to pick lengths.", 'start': 299.761, 'duration': 4.223}], 'summary': 'The speaker discusses picking lengths with index numbers and solving a problem similar to knapsack.', 'duration': 29.382, 'max_score': 274.602, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo274602.jpg'}], 'start': 245.583, 'title': 'Similarity to knapsack problem, rod cutting problem, and maximizing price with unbounded knapsack', 'summary': 'Explores the similarity of a given problem to the knapsack problem, the approach to pick rod lengths and sum them up to make the given n, and maximizing the price using unbounded knapsack, providing insights on various methods and similarities to known problems.', 'chapters': [{'end': 303.984, 'start': 245.583, 'title': 'Similarity to knapsack problem', 'summary': 'Discusses the similarity of a given problem to the knapsack problem by using index numbers to represent lengths and formulating a solution in a similar way to the knapsack problem.', 'duration': 58.401, 'highlights': ['The problem is similar to the knapsack problem where we are given an array with values and weights and try to form the n.', 'Using index numbers to represent lengths (0, 1, 2, 3, 4) and solving the problem in a similar way to the knapsack problem.', 'Attempting to pick lengths to solve the problem in a manner similar to the knapsack problem.']}, {'end': 368.49, 'start': 305.324, 'title': 'Rod cutting problem', 'summary': 'Discusses the approach to pick rod lengths and sum them up to make the given n, using various ways and recursion, with a focus on expressing in terms of index when given an array of rod lengths and prices.', 'duration': 63.166, 'highlights': ['The key focus is on picking rod lengths and summing them up to form the given n, using various ways and recursion.', 'The approach emphasizes expressing in terms of index when given an array of rod lengths and prices.']}, {'end': 460.451, 'start': 368.49, 'title': 'Maximizing price with unbounded knapsack', 'summary': 'Discusses maximizing the price using unbounded knapsack by exploring all possibilities and finding the maximal cost of formatting a given input, similar to the pricing of rodlets in the example provided.', 'duration': 91.961, 'highlights': ['The question involves maximizing the price using unbounded knapsack, similar to the pricing of rodlets.', 'Exploring all possibilities to maximize the price is emphasized in the discussion.', 'The concept of maximal cost of formatting a given input is highlighted as a key point in the chapter.', 'The specific values provided (2, 5, 7, 8, 10) showcase the pricing of rodlets, demonstrating the practical application of the discussed concepts.', 'The function f(n-1, N) is used to determine the maximal cost that can be formatted, emphasizing the practical application of the concepts in the question.']}], 'duration': 214.868, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo245583.jpg', 'highlights': ['The problem is similar to the knapsack problem where we are given an array with values and weights and try to form the n.', 'The question involves maximizing the price using unbounded knapsack, similar to the pricing of rodlets.', 'The approach emphasizes expressing in terms of index when given an array of rod lengths and prices.', 'Exploring all possibilities to maximize the price is emphasized in the discussion.', 'Using index numbers to represent lengths (0, 1, 2, 3, 4) and solving the problem in a similar way to the knapsack problem.']}, {'end': 592.604, 'segs': [{'end': 600.768, 'src': 'embed', 'start': 565.413, 'weight': 0, 'content': [{'end': 567.863, 'text': 'yes, baba, yes, you stand at the same index.', 'start': 565.413, 'duration': 2.45}, {'end': 574.897, 'text': 'why? because because, if you carefully remember, You can actually pick up one as many times as you want.', 'start': 567.863, 'duration': 7.034}, {'end': 579.279, 'text': 'Which case, infinite supply of rods.', 'start': 575.577, 'duration': 3.702}, {'end': 582, 'text': 'Infinite supply of rods.', 'start': 580.039, 'duration': 1.961}, {'end': 588.703, 'text': 'And whenever there is an infinite supply of rods, what do you do? I have already taught this in the previous lectures.', 'start': 582.58, 'duration': 6.123}, {'end': 590.784, 'text': 'You stand at the same index.', 'start': 589.183, 'duration': 1.601}, {'end': 592.604, 'text': 'You stand at the same index.', 'start': 591.224, 'duration': 1.38}, {'end': 600.768, 'text': 'And right at the end of the day, you can return the max of tick and not tick.', 'start': 593.104, 'duration': 7.664}], 'summary': 'Infinite supply of rods allows standing at the same index and returning the max of tick and not tick.', 'duration': 35.355, 'max_score': 565.413, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo565413.jpg'}], 'start': 460.451, 'title': 'Dynamic programming for rod cutting', 'summary': 'Discusses dynamic programming for rod cutting, explaining the process of determining the maximum price for cutting a rod of varying lengths while iteratively considering different cases and indices.', 'chapters': [{'end': 592.604, 'start': 460.451, 'title': 'Dynamic programming for rod cutting', 'summary': 'Discusses dynamic programming for rod cutting, explaining the process of determining the maximum price for cutting a rod of varying lengths while iteratively considering different cases and indices.', 'duration': 132.153, 'highlights': ['The rod length is index plus 1, and it must be less than or equal to the length being sought, with the price added determined by the function of index n minus the rod length.', "When looking for maximum pricing, it's implied that the rod can only be cut if the rod length corresponds to the value being sought, and in the case of an infinite supply of rods, one can stand at the same index.", 'The speaker emphasizes the concept of standing at the same index when there is an infinite supply of rods, highlighting its significance in the rod cutting process.']}], 'duration': 132.153, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo460451.jpg', 'highlights': ['The rod length is index plus 1, and it must be less than or equal to the length being sought, with the price added determined by the function of index n minus the rod length.', "When looking for maximum pricing, it's implied that the rod can only be cut if the rod length corresponds to the value being sought, and in the case of an infinite supply of rods, one can stand at the same index.", 'The speaker emphasizes the concept of standing at the same index when there is an infinite supply of rods, highlighting its significance in the rod cutting process.']}, {'end': 830.708, 'segs': [{'end': 757.38, 'src': 'heatmap', 'start': 620.456, 'weight': 0, 'content': [{'end': 621.557, 'text': 'So zero is the base case.', 'start': 620.456, 'duration': 1.101}, {'end': 624.759, 'text': "Let's just think it as a single element.", 'start': 622.197, 'duration': 2.562}, {'end': 626.88, 'text': 'So we have the zeroth index.', 'start': 625.499, 'duration': 1.381}, {'end': 629.982, 'text': 'We know the rod length for zeroth index is one.', 'start': 627.6, 'duration': 2.382}, {'end': 631.963, 'text': 'And it will have a price.', 'start': 630.922, 'duration': 1.041}, {'end': 634.125, 'text': 'Assume it has a price of 6.', 'start': 632.163, 'duration': 1.962}, {'end': 636.367, 'text': "And I'm coming up with a rod length.", 'start': 634.125, 'duration': 2.242}, {'end': 639.689, 'text': "Yes, I'm coming up with a rod length 12.", 'start': 636.487, 'duration': 3.202}, {'end': 643.392, 'text': "This is the remaining rod length that I'm looking to make.", 'start': 639.689, 'duration': 3.703}, {'end': 648.096, 'text': "Initially, whatever it is, this is the remaining rod length I'm looking to make.", 'start': 644.093, 'duration': 4.003}, {'end': 649.597, 'text': 'Looking to make this.', 'start': 648.717, 'duration': 0.88}, {'end': 654.041, 'text': 'Because that is what the recurrence was stating, right? Because that is what the recurrence was stating.', 'start': 650.238, 'duration': 3.803}, {'end': 660.473, 'text': 'Till index 4, what is the maximum price you can obtain in order to make n right?', 'start': 655.452, 'duration': 5.021}, {'end': 666.334, 'text': 'So whatever n comes, if at the last index n is 12, how can you make this?', 'start': 660.733, 'duration': 5.601}, {'end': 667.434, 'text': '12 is the question.', 'start': 666.334, 'duration': 1.1}, {'end': 674.636, 'text': "So the question is, how can you make this rod length 12? Because over here, it's a rod length 1.", 'start': 667.754, 'duration': 6.882}, {'end': 686.317, 'text': 'So if you can take 1, 1, 1, 1, dot dot dot dot 1, 12 times 12, i say that will make this remaining length 12 black.', 'start': 674.636, 'duration': 11.681}, {'end': 688.758, 'text': 'yes, yes, it will, it will.', 'start': 686.317, 'duration': 2.441}, {'end': 700.845, 'text': 'so whatever n is left at the end of the day at index 0, can i say the pricing will be because the rod length is 1.', 'start': 688.758, 'duration': 12.087}, {'end': 710.347, 'text': 'i can make that rod length because if the remaining length is n, it will require n rods.', 'start': 700.845, 'duration': 9.502}, {'end': 714.288, 'text': 'n rods of length one, n rods of length one is what it will require.', 'start': 710.347, 'duration': 3.941}, {'end': 719.65, 'text': 'and if it requires n rod of length one, what will be the price?', 'start': 714.288, 'duration': 5.362}, {'end': 721.45, 'text': 'n into price of zero?', 'start': 719.65, 'duration': 1.8}, {'end': 723.671, 'text': 'perfect, perfect.', 'start': 721.45, 'duration': 2.221}, {'end': 725.772, 'text': "so that's what you'll return.", 'start': 723.671, 'duration': 2.101}, {'end': 735.612, 'text': "you will see n into price of zero is what i'll return because because the last guy is of length one and if it is of length one,", 'start': 725.772, 'duration': 9.84}, {'end': 741.674, 'text': "i know i'll require n rods of length one and the pricing will be into price of zero.", 'start': 735.612, 'duration': 6.062}, {'end': 743.955, 'text': "that's the base case, super simple.", 'start': 741.674, 'duration': 2.281}, {'end': 746.996, 'text': "so once i've written the base case, what's the next step?", 'start': 743.955, 'duration': 3.041}, {'end': 749.017, 'text': "it's a recursion code, right.", 'start': 746.996, 'duration': 2.021}, {'end': 753.599, 'text': 'so you have to convert the recursion into memo i asian.', 'start': 749.017, 'duration': 4.582}, {'end': 757.38, 'text': 'So overlapping subproblems will be there.', 'start': 755.259, 'duration': 2.121}], 'summary': 'The transcript discusses the process of determining rod length and pricing, with a focus on base cases and recursion.', 'duration': 80.389, 'max_score': 620.456, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo620456.jpg'}, {'end': 800.645, 'src': 'embed', 'start': 774.707, 'weight': 1, 'content': [{'end': 783.571, 'text': 'What is the space complexity? The auxiliary stack space is the recursion depth and it can go on till we go off target.', 'start': 774.707, 'duration': 8.864}, {'end': 789.438, 'text': "whatever target is, because you're taking the target, you're taking the target again and again.", 'start': 785.356, 'duration': 4.082}, {'end': 791.4, 'text': 'so we go off target.', 'start': 789.438, 'duration': 1.962}, {'end': 797.783, 'text': 'it can go as if you do memoize index and then couple of changing parameters.', 'start': 791.4, 'duration': 6.383}, {'end': 800.645, 'text': 'index is n, n is also n.', 'start': 797.783, 'duration': 2.862}], 'summary': 'The space complexity is related to recursion depth and memoization, with index n and target repetitions.', 'duration': 25.938, 'max_score': 774.707, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo774707.jpg'}], 'start': 593.104, 'title': 'Rod length algorithms', 'summary': 'Discusses algorithms for determining maximum price and cost of rods using recursion and memoization, emphasizing base case, time complexity of o(n^2), space complexity of o(n^2), and the need for optimization.', 'chapters': [{'end': 700.845, 'start': 593.104, 'title': 'Rod length pricing algorithm', 'summary': 'Discusses the process of determining the maximum price for a given rod length using a recursive algorithm, with specific emphasis on the base case and the calculation for a rod length of 12.', 'duration': 107.741, 'highlights': ['The base case is when the index is equal to zero, representing a rod length of 1 with a price of 6.', 'The algorithm aims to determine the maximum price for a rod length of 12, with the recursive process involving breaking down the rod into smaller units and calculating the pricing accordingly.', 'The initial starting point for the recursive algorithm is at n minus one, leading to the exploration of various rod length possibilities and their corresponding prices.']}, {'end': 753.599, 'start': 700.845, 'title': 'Rod length calculation', 'summary': 'Discusses a recursive algorithm for calculating the cost of rods of varying lengths and converting it into memoization, emphasizing the base case and the need for optimization.', 'duration': 52.754, 'highlights': ['The algorithm involves calculating the cost of n rods of length one, resulting in a pricing of n * price of one, serving as the base case.', 'The next step involves converting the recursive algorithm into memoization for optimization.']}, {'end': 830.708, 'start': 755.259, 'title': 'Recursion time and space complexity', 'summary': 'Discusses the time and space complexity of recursion, emphasizing that the time complexity can be o(n^2) and the space complexity can be o(n^2) as well, and suggests converting recursion to tabulation to avoid auxiliary stack space.', 'duration': 75.449, 'highlights': ['The time complexity can be O(n^2) and the space complexity can be O(n^2) as well, due to staying at the same index and memoization with changing parameters.', 'The auxiliary stack space in recursion can go as high as the target, leading to a space complexity of O(n^2) plus O(target).', 'Suggests converting recursion to tabulation code to avoid the use of auxiliary stack space and mentions following certain rules for the conversion.']}], 'duration': 237.604, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo593104.jpg', 'highlights': ['The time complexity can be O(n^2) and the space complexity can be O(n^2) as well, due to staying at the same index and memoization with changing parameters.', 'The auxiliary stack space in recursion can go as high as the target, leading to a space complexity of O(n^2) plus O(target).', 'The algorithm involves calculating the cost of n rods of length one, resulting in a pricing of n * price of one, serving as the base case.', 'The next step involves converting the recursive algorithm into memoization for optimization.', 'The initial starting point for the recursive algorithm is at n minus one, leading to the exploration of various rod length possibilities and their corresponding prices.']}, {'end': 948.48, 'segs': [{'end': 860.835, 'src': 'embed', 'start': 830.708, 'weight': 1, 'content': [{'end': 834.681, 'text': 'write the base case, Write the two changing parameters.', 'start': 830.708, 'duration': 3.973}, {'end': 836.662, 'text': 'Copy the recurrence.', 'start': 835.681, 'duration': 0.981}, {'end': 838.523, 'text': 'As simple as that.', 'start': 837.562, 'duration': 0.961}, {'end': 840.344, 'text': 'As simple as it can get.', 'start': 839.203, 'duration': 1.141}, {'end': 842.305, 'text': "Let's write the base case at first.", 'start': 840.884, 'duration': 1.421}, {'end': 845.707, 'text': "Let's go back to the recurrence.", 'start': 843.566, 'duration': 2.141}, {'end': 854.332, 'text': 'By the way, in order to convert it into a memorization, just make sure you write dp of index n equal to, and over here,', 'start': 846.507, 'duration': 7.825}, {'end': 860.015, 'text': 'dp of index n not equal to minus 1, and you try to return.', 'start': 854.332, 'duration': 5.683}, {'end': 860.835, 'text': 'Simple as that.', 'start': 860.355, 'duration': 0.48}], 'summary': 'Explanation of writing base case, changing parameters, and converting to memorization.', 'duration': 30.127, 'max_score': 830.708, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo830708.jpg'}, {'end': 954.725, 'src': 'embed', 'start': 919.949, 'weight': 0, 'content': [{'end': 921.27, 'text': 'superb. what about this?', 'start': 919.949, 'duration': 1.321}, {'end': 923.911, 'text': 'what did you do in the recursion?', 'start': 922.21, 'duration': 1.701}, {'end': 930.093, 'text': 'in the recursion it went from n minus 1 till 0, where 0 being the base case.', 'start': 923.911, 'duration': 6.182}, {'end': 933.034, 'text': 'what is tabulation bottom up?', 'start': 930.093, 'duration': 2.941}, {'end': 935.375, 'text': 'what is tabulation bottom up?', 'start': 933.034, 'duration': 2.341}, {'end': 938.916, 'text': 'so from base case you go till the ultimate.', 'start': 935.375, 'duration': 3.541}, {'end': 940.656, 'text': 'so base case is done.', 'start': 938.916, 'duration': 1.74}, {'end': 943.137, 'text': 'so you start from 1 till n minus 1.', 'start': 940.656, 'duration': 2.481}, {'end': 944.118, 'text': 'what about n?', 'start': 943.137, 'duration': 0.981}, {'end': 946.238, 'text': 'you do it in the opposite Done.', 'start': 944.118, 'duration': 2.12}, {'end': 948.48, 'text': 'And now you just need to copy the records.', 'start': 946.518, 'duration': 1.962}, {'end': 954.725, 'text': "So without actually wasting any time, let's super quickly code this up.", 'start': 949.06, 'duration': 5.665}], 'summary': 'Discussion covers recursion, tabulation, and coding implementation.', 'duration': 34.776, 'max_score': 919.949, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo919949.jpg'}], 'start': 830.708, 'title': 'Dynamic programming base case and recurrence', 'summary': "Discusses writing base cases and recurrences in dynamic programming, emphasizing the flexibility of parameter 'n' and the iterative approach of tabulation. it highlights the importance of declaring dp values for each 'n' and the bottom-up approach in tabulation.", 'chapters': [{'end': 948.48, 'start': 830.708, 'title': 'Dynamic programming base case and recurrence', 'summary': "Discusses the process of writing base cases and recurrences in dynamic programming, emphasizing the flexibility of parameter 'n' and the iterative approach of tabulation. it highlights the importance of declaring dp values for each 'n' and the bottom-up approach in tabulation.", 'duration': 117.772, 'highlights': ["The chapter emphasizes the flexibility of parameter 'n' and the need to declare dp values for each 'n', indicating that 'n' can vary from 0 to the original 'n'.", "It discusses the iterative approach of tabulation, explaining the process from base case to the ultimate value, highlighting the opposite direction from 'n' to 1 till 'n-1'.", "The chapter stresses the importance of writing base cases and recurrences in a simple and straightforward manner, emphasizing the process of copying the same approach for every 'n'."]}], 'duration': 117.772, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo830708.jpg', 'highlights': ["The chapter emphasizes the flexibility of parameter 'n' and the need to declare dp values for each 'n', indicating that 'n' can vary from 0 to the original 'n'.", "It discusses the iterative approach of tabulation, explaining the process from base case to the ultimate value, highlighting the opposite direction from 'n' to 1 till 'n-1'.", "The chapter stresses the importance of writing base cases and recurrences in a simple and straightforward manner, emphasizing the process of copying the same approach for every 'n'."]}, {'end': 1364.514, 'segs': [{'end': 978.052, 'src': 'embed', 'start': 949.06, 'weight': 1, 'content': [{'end': 954.725, 'text': "So without actually wasting any time, let's super quickly code this up.", 'start': 949.06, 'duration': 5.665}, {'end': 957.547, 'text': "So you're given the price and you're given the n right?", 'start': 954.745, 'duration': 2.802}, {'end': 969.417, 'text': 'So what I can do is I can write the function with index and the capital N and we can take the vector int and the price correct?', 'start': 957.888, 'duration': 11.529}, {'end': 970.798, 'text': "So that's what I will be taking.", 'start': 969.437, 'duration': 1.361}, {'end': 978.052, 'text': 'Can I say if index is equal to equal to zero, I need to return the n into price of zero.', 'start': 973.51, 'duration': 4.542}], 'summary': 'Quickly code function to calculate price based on index and n.', 'duration': 28.992, 'max_score': 949.06, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo949060.jpg'}, {'end': 1036.171, 'src': 'embed', 'start': 1006.316, 'weight': 0, 'content': [{'end': 1011.599, 'text': 'so i need to check if the rod length is lesser than the required end.', 'start': 1006.316, 'duration': 5.283}, {'end': 1014.241, 'text': "if that's the case, i can definitely take it.", 'start': 1011.599, 'duration': 2.642}, {'end': 1016.962, 'text': 'and if i take it it will be price of index plus function of.', 'start': 1014.241, 'duration': 2.721}, {'end': 1022.265, 'text': 'you stay at the same index and the rod length that you are looking for will reduce by the rod length.', 'start': 1016.962, 'duration': 5.303}, {'end': 1030.469, 'text': 'perfect, and you just pass on the price and over here you can return the max of take and not tick.', 'start': 1022.265, 'duration': 8.204}, {'end': 1036.171, 'text': 'and right here you can say return f of you start from the last index, which is n minus 1.', 'start': 1030.469, 'duration': 5.702}], 'summary': 'Check rod length, take at price of index plus function, return max of take and not tick.', 'duration': 29.855, 'max_score': 1006.316, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo1006316.jpg'}, {'end': 1082.103, 'src': 'embed', 'start': 1048.636, 'weight': 3, 'content': [{'end': 1056.442, 'text': 'So create a dp array and try to memorize this into n, into vector of int, n plus 1 comma minus 1..', 'start': 1048.636, 'duration': 7.806}, {'end': 1058.344, 'text': 'Once you memorize this, this will go as dp.', 'start': 1056.442, 'duration': 1.902}, {'end': 1064.469, 'text': 'Obviously this will go as vector vector int dp.', 'start': 1059.104, 'duration': 5.365}, {'end': 1069.298, 'text': 'will be dp of index n.', 'start': 1065.916, 'duration': 3.382}, {'end': 1077.321, 'text': 'this you can say if dp of index n not equal to minus one, you can say return dp of index n.', 'start': 1069.298, 'duration': 8.023}, {'end': 1080.563, 'text': 'perfect, and just make sure you pass a dp here.', 'start': 1077.321, 'duration': 3.242}, {'end': 1082.103, 'text': 'make sure you pass a dp here.', 'start': 1080.563, 'duration': 1.54}], 'summary': 'Create a dp array to memorize into n, vector of int, n + 1, -1. use dp of index n and return if not equal to -1.', 'duration': 33.467, 'max_score': 1048.636, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo1048636.jpg'}], 'start': 949.06, 'title': 'Dynamic programming applications', 'summary': 'Covers dynamic programming for the rod cutting problem and subsequences, achieving 100 marks in submissions and successfully space optimizing into a 1d array.', 'chapters': [{'end': 1030.469, 'start': 949.06, 'title': 'Dynamic programming for rod cutting problem', 'summary': 'Explains a dynamic programming approach for the rod cutting problem, demonstrating the function and logic for finding the maximum revenue from cutting a rod, using index, price, and n as inputs.', 'duration': 81.409, 'highlights': ['The function for the rod cutting problem is defined using the inputs index, capital N, and the vector int price, to efficiently calculate the maximum revenue from cutting the rod.', 'The logic involves determining the not take and take cases, with the take case aiming to find the maximum revenue by considering the rod length and price at each index.', 'The process includes checking the rod length against the required end, reducing the rod length, and returning the maximum value between take and not tick.']}, {'end': 1364.514, 'start': 1030.469, 'title': 'Dp on subsequences: memoization, tabulation, and space optimization', 'summary': 'Covers the implementation of dp on subsequences using memoization, tabulation, and space optimization techniques, achieving 100 marks in submissions and successfully space optimizing into a 1d array.', 'duration': 334.045, 'highlights': ['The chapter covers the implementation of DP on subsequences using memoization, tabulation, and space optimization techniques, achieving 100 marks in submissions.', 'The process involves creating a dp array and memorizing it into n, into vector of int, n plus 1 comma minus 1, which then goes as dp, and implementing tabulation by writing base cases and iterating through the array.', 'Space optimization is achieved by using a single array and rewriting the current values over the previous ones, eliminating the need for previous and current arrays.']}], 'duration': 415.454, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/mO8XpGoJwuo/pics/mO8XpGoJwuo949060.jpg', 'highlights': ['Covers dynamic programming for the rod cutting problem with efficient revenue calculation', 'Covers DP on subsequences using memoization, tabulation, and space optimization techniques', 'Achieved 100 marks in submissions for the implementation of DP on subsequences', 'Successfully space optimized into a 1d array for the rod cutting problem']}], 'highlights': ['The problem focuses on maximizing the total price obtained after cutting and selling the rod, with the highest possible price being 12.', 'The task requires trying every possible way to determine the maximal cost, with an example demonstrating that cutting the rod into lengths of 2 and 3 also yields the maximum cost of 12.', 'The problem is similar to the knapsack problem where we are given an array with values and weights and try to form the n.', 'The rod length is index plus 1, and it must be less than or equal to the length being sought, with the price added determined by the function of index n minus the rod length.', 'The time complexity can be O(n^2) and the space complexity can be O(n^2) as well, due to staying at the same index and memoization with changing parameters.', 'Covers dynamic programming for the rod cutting problem with efficient revenue calculation', 'Collecting rod lengths to make N and maximizing the price while doing so.', "The chapter emphasizes the flexibility of parameter 'n' and the need to declare dp values for each 'n', indicating that 'n' can vary from 0 to the original 'n'.", 'The approach emphasizes expressing in terms of index when given an array of rod lengths and prices.', 'The algorithm involves calculating the cost of n rods of length one, resulting in a pricing of n * price of one, serving as the base case.']}