title
G-7. Number of Provinces | C++ | Java | Connected Components

description
GfG-Problem Link: https://bit.ly/3yR3dIB C++/Java/Codes and Notes Link: https://takeuforward.org/data-structure/number-of-provinces/ DP Series: https://www.youtube.com/watch?v=FfXoiwwnxFw&list=PLgUwDviBIf0qUlt5H_kiKYaNSqJ81PMMY SDE Sheet: https://takeuforward.org/interviews/strivers-sde-sheet-top-coding-interview-problems/ Check out our Website for curated resources: https://www.takeuforward.org/ Our Second Channel: https://www.youtube.com/channel/UCvEKHATlVq84hm1jduTYm8g In case you are thinking to buy courses, please check below: Code "takeuforward" for 15% off at GFG: https://practice.geeksforgeeks.org/courses Code "takeuforward" for 20% off on sys-design: https://get.interviewready.io?_aff=takeuforward Crypto, I use the Wazirx app: https://wazirx.com/invite/xexnpc4u Take 750 rs free Amazon Stock from me: https://indmoney.onelink.me/RmHC/idjex744 Earn 100 rs by making a Grow Account for investing: https://app.groww.in/v3cO/8hu879t0 Linkedin/Instagram/Telegram: https://linktr.ee/takeUforward ---------------------------------------------------------------------------------------------------------------------------

detail
{'title': 'G-7. Number of Provinces | C++ | Java | Connected Components', 'heatmap': [{'end': 784.558, 'start': 691.895, 'weight': 0.848}, {'end': 855.891, 'start': 795.63, 'weight': 0.721}], 'summary': 'Discusses finding the number of provinces in an undirected graph with v vertices, using dfs traversal and adjacency list conversion, emphasizing time complexity as near about v + 2e and space complexity as o(n), with a call for b go of n traversal.', 'chapters': [{'end': 88.677, 'segs': [{'end': 45.333, 'src': 'embed', 'start': 20.282, 'weight': 0, 'content': [{'end': 26.747, 'text': 'if there is a path from u to v or from v to u, your task is to find the number of provinces.', 'start': 20.282, 'duration': 6.465}, {'end': 28.408, 'text': 'so let me explain the question.', 'start': 26.747, 'duration': 1.661}, {'end': 37.374, 'text': 'so imagine a graph is given which is something like this like this is a particular piece, and then there is something like this.', 'start': 28.408, 'duration': 8.966}, {'end': 40.611, 'text': 'then imagine there is something like this okay.', 'start': 37.374, 'duration': 3.237}, {'end': 45.333, 'text': "so what they're saying is this is an entire graph that is given to you.", 'start': 40.611, 'duration': 4.722}], 'summary': 'Find the number of provinces in a given graph.', 'duration': 25.051, 'max_score': 20.282, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA20282.jpg'}], 'start': 3.271, 'title': 'Finding number of provinces in graph', 'summary': 'Discusses determining the number of provinces in an undirected graph with v vertices, where two vertices u and v belong to a single province if there is a path between them, providing a method to solve this task.', 'chapters': [{'end': 88.677, 'start': 3.271, 'title': 'Finding number of provinces in graph', 'summary': 'Discusses finding the number of provinces in a given undirected graph with v vertices, where two vertices u and v belong to a single province if there is a path from u to v or from v to u, and the task is to determine the number of provinces.', 'duration': 85.406, 'highlights': ['The problem involves finding the number of provinces in a given undirected graph with v vertices, where two vertices u and v belong to a single province if there is a path from u to v or from v to u.', 'A province is defined as a set of vertices in the graph where there is a path between every pair of vertices within the set.', 'It is essential to identify the connectivity between vertices to determine the provinces in the graph.']}], 'duration': 85.406, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA3271.jpg', 'highlights': ['The problem involves finding the number of provinces in a given undirected graph with v vertices.', 'A province is defined as a set of vertices in the graph where there is a path between every pair of vertices within the set.', 'It is essential to identify the connectivity between vertices to determine the provinces in the graph.']}, {'end': 451.378, 'segs': [{'end': 220.511, 'src': 'embed', 'start': 194.227, 'weight': 0, 'content': [{'end': 203.395, 'text': 'right, this can be like i can call the dfs traversals starting from three different starting nodes and thereby i can get three provinces.', 'start': 194.227, 'duration': 9.168}, {'end': 205.497, 'text': 'right, we know how to do a dfs traversal.', 'start': 203.395, 'duration': 2.102}, {'end': 207.358, 'text': 'we have learned about that.', 'start': 205.497, 'duration': 1.861}, {'end': 210.141, 'text': 'but in that case there was a single starting node.', 'start': 207.358, 'duration': 2.783}, {'end': 212.183, 'text': 'over here the starting node can be multiple.', 'start': 210.141, 'duration': 2.042}, {'end': 220.511, 'text': 'so we can see if somehow we can figure out the starting points, then we will be able to figure out three provinces.', 'start': 212.643, 'duration': 7.868}], 'summary': 'Using dfs traversals from multiple starting nodes to find three provinces.', 'duration': 26.284, 'max_score': 194.227, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA194227.jpg'}, {'end': 273.924, 'src': 'embed', 'start': 238.226, 'weight': 1, 'content': [{'end': 241.308, 'text': 'it will make sure it visits everyone right.', 'start': 238.226, 'duration': 3.082}, {'end': 246.571, 'text': 'so somewhere down the line, if i can figure out the starting points, my job will be done.', 'start': 241.308, 'duration': 5.263}, {'end': 251.214, 'text': 'this is where what we do is we use something as a visited array.', 'start': 246.571, 'duration': 4.643}, {'end': 253.236, 'text': 'now we see there are eight nodes.', 'start': 251.214, 'duration': 2.022}, {'end': 260.32, 'text': 'so we will create a nine size visited array one, two, three, four, five, six, seven, eight.', 'start': 253.236, 'duration': 7.084}, {'end': 265.142, 'text': "so it's like zero one, two, three, four, five, six, seven, eight.", 'start': 260.32, 'duration': 4.822}, {'end': 273.924, 'text': 'perfect, we have created a nine size visited array and we mark everyone as false, and this is known as the visited array done.', 'start': 265.142, 'duration': 8.782}], 'summary': 'Creating a visited array with nine elements to mark everyone as false in a task involving visiting eight nodes.', 'duration': 35.698, 'max_score': 238.226, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA238226.jpg'}, {'end': 363.559, 'src': 'embed', 'start': 333.675, 'weight': 3, 'content': [{'end': 338.137, 'text': 'because a DFS algorithm makes sure that you visit all the nodes in a graph.', 'start': 333.675, 'duration': 4.462}, {'end': 340.618, 'text': 'So it will make sure that it visits right?', 'start': 338.457, 'duration': 2.161}, {'end': 343.06, 'text': 'and then the call will be over.', 'start': 341.238, 'duration': 1.822}, {'end': 349.866, 'text': 'the call came from dfs1, it will go to dfs2, it will go to dfs3 and then it will come back.', 'start': 343.06, 'duration': 6.806}, {'end': 352.228, 'text': 'then it will come back and then it will come back.', 'start': 349.866, 'duration': 2.362}, {'end': 357.313, 'text': 'now in the process, when it goes from one, it marks one as visited, if you remember.', 'start': 352.228, 'duration': 5.085}, {'end': 363.559, 'text': 'then it marks two as visited, then it marks three as visited and then it goes back ultimately.', 'start': 357.313, 'duration': 6.246}], 'summary': 'Dfs algorithm ensures visiting all nodes in a graph, marking each as visited.', 'duration': 29.884, 'max_score': 333.675, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA333675.jpg'}, {'end': 411.073, 'src': 'embed', 'start': 385.962, 'weight': 4, 'content': [{'end': 391.629, 'text': 'so what happens is it again calls the dfs for four, but this time the dfs is called for four.', 'start': 385.962, 'duration': 5.667}, {'end': 395.173, 'text': 'so if you see, the first time it is called for one, now it is called for four.', 'start': 391.629, 'duration': 3.544}, {'end': 397.996, 'text': 'now when it is called for four, it marks four as visited.', 'start': 395.173, 'duration': 2.823}, {'end': 409.652, 'text': 'then it goes to five, then it goes to six and makes sure everyone is marked as visited in that call, and then it comes back, comes back, comes back,', 'start': 397.996, 'duration': 11.656}, {'end': 411.073, 'text': 'so apparently comes back.', 'start': 409.652, 'duration': 1.421}], 'summary': 'Dfs algorithm visits nodes 4, 5, and 6, marking them as visited.', 'duration': 25.111, 'max_score': 385.962, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA385962.jpg'}, {'end': 460.71, 'src': 'embed', 'start': 433.765, 'weight': 5, 'content': [{'end': 441.53, 'text': 'so this is how we can say the visited array performs and this after this term will be 8 and it will not find it.', 'start': 433.765, 'duration': 7.765}, {'end': 444.411, 'text': 'then it will be 9 and the for loop will be over.', 'start': 441.53, 'duration': 2.881}, {'end': 451.378, 'text': 'so can i say this particular if statement was executed twice, can i say this particular statement was executed twice.', 'start': 444.411, 'duration': 6.967}, {'end': 460.71, 'text': 'thereby can i just put a counter plus plus over here, which will make sure, make sure, that the provinces are counted right separately.', 'start': 451.378, 'duration': 9.332}], 'summary': 'The if statement was executed twice, and adding a counter will ensure provinces are counted correctly.', 'duration': 26.945, 'max_score': 433.765, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA433765.jpg'}], 'start': 88.677, 'title': 'Graph traversal techniques', 'summary': 'Discusses using dfs traversal to count provinces in a graph with multiple starting nodes and illustrates the use of a visited array. it also explains performing dfs and bfs on a graph, emphasizing marking nodes as visited and discussing the execution of a specific statement.', 'chapters': [{'end': 292.693, 'start': 88.677, 'title': 'Counting provinces with dfs traversal', 'summary': 'Explains how to use dfs traversal to count provinces in a graph, where the starting node can be multiple, and illustrates the use of a visited array to track visited nodes.', 'duration': 204.016, 'highlights': ['DFS traversal can be used to count provinces in a graph by starting from multiple nodes, ensuring the visitation of all nodes. This approach can identify starting points and ultimately determine the number of provinces.', 'The use of a visited array with a size equal to the number of nodes allows for tracking the visited status of each node, facilitating the identification of starting points and the determination of provinces.', 'The starting node for DFS traversal to count provinces can be any node in the graph, as long as it ensures the visitation of all nodes.']}, {'end': 451.378, 'start': 292.693, 'title': 'Dfs and bfs in graphs', 'summary': 'Explains how to perform depth first search (dfs) and breadth first search (bfs) on a graph, highlighting the process of marking nodes as visited and discussing the execution of a specific statement.', 'duration': 158.685, 'highlights': ['The DFS algorithm ensures that it visits all the nodes in a graph and marks them as visited, as demonstrated by the process of marking nodes 1, 2, and 3 as visited during the DFS call for node 1.', 'The process of marking nodes as visited and performing DFS calls for unvisited nodes is detailed, showcasing the marking of nodes 4, 5, and 6 as visited during the DFS call for node 4.', 'The execution of a specific if statement is discussed, highlighting its occurrence twice within the context of the algorithm.']}], 'duration': 362.701, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA88677.jpg', 'highlights': ['DFS traversal can be used to count provinces in a graph by starting from multiple nodes, ensuring the visitation of all nodes. This approach can identify starting points and ultimately determine the number of provinces.', 'The use of a visited array with a size equal to the number of nodes allows for tracking the visited status of each node, facilitating the identification of starting points and the determination of provinces.', 'The starting node for DFS traversal to count provinces can be any node in the graph, as long as it ensures the visitation of all nodes.', 'The DFS algorithm ensures that it visits all the nodes in a graph and marks them as visited, as demonstrated by the process of marking nodes 1, 2, and 3 as visited during the DFS call for node 1.', 'The process of marking nodes as visited and performing DFS calls for unvisited nodes is detailed, showcasing the marking of nodes 4, 5, and 6 as visited during the DFS call for node 4.', 'The execution of a specific if statement is discussed, highlighting its occurrence twice within the context of the algorithm.']}, {'end': 919.973, 'segs': [{'end': 511.99, 'src': 'embed', 'start': 475.224, 'weight': 2, 'content': [{'end': 478.526, 'text': "so the input is slightly tricky and it's slightly different.", 'start': 475.224, 'duration': 3.302}, {'end': 482.269, 'text': "so you're given an uh, adjacency matrix.", 'start': 478.526, 'duration': 3.743}, {'end': 486.272, 'text': "very careful, it's not a list, it's an adjacency matrix.", 'start': 482.269, 'duration': 4.003}, {'end': 494.539, 'text': 'okay, so one of the ways is either you create your adjacency list and then just go across.', 'start': 486.272, 'duration': 8.267}, {'end': 495.94, 'text': 'that is one of the ways that you can do it.', 'start': 494.539, 'duration': 1.401}, {'end': 501.403, 'text': 'or what you can do is you can use this adjacency matrix and solve this particular problem.', 'start': 496.6, 'duration': 4.803}, {'end': 511.99, 'text': "now what i will do is, since we have been doing everything in adjacency list, what i'll do is i'll just quickly, uh, create our adjacency list.", 'start': 501.403, 'duration': 10.587}], 'summary': 'Given an adjacency matrix, create an adjacency list for problem-solving.', 'duration': 36.766, 'max_score': 475.224, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA475224.jpg'}, {'end': 688.432, 'src': 'embed', 'start': 655.667, 'weight': 3, 'content': [{'end': 659.088, 'text': 'and you say hey listen, if not visited it.', 'start': 655.667, 'duration': 3.421}, {'end': 666.991, 'text': "what i'll do is i'll go and call the dfs for it adjacency list and visited, and this will be my dfs calls and at the end of the day,", 'start': 659.088, 'duration': 7.903}, {'end': 669.212, 'text': "since i've counted the number of provinces, i'll return it.", 'start': 666.991, 'duration': 2.221}, {'end': 672.233, 'text': "now let's super quickly compile this and check it out.", 'start': 669.212, 'duration': 3.021}, {'end': 673.754, 'text': 'okay, so okay.', 'start': 672.233, 'duration': 1.521}, {'end': 677.415, 'text': 'so, since the adjacency list was created like this, uh, this will be an array.', 'start': 673.754, 'duration': 3.661}, {'end': 680.985, 'text': 'my bad, Perfect.', 'start': 677.415, 'duration': 3.57}, {'end': 681.826, 'text': 'I think we should be fine.', 'start': 681.025, 'duration': 0.801}, {'end': 688.432, 'text': 'So usually in all the problems you will not find the adjacency matrix, just because, if you see, the constraints are very small.', 'start': 682.106, 'duration': 6.326}], 'summary': 'The speaker discusses calling dfs for adjacency list, counting provinces, and constraints.', 'duration': 32.765, 'max_score': 655.667, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA655667.jpg'}, {'end': 784.558, 'src': 'heatmap', 'start': 691.895, 'weight': 0.848, 'content': [{'end': 697.701, 'text': 'You will mostly find adjacency list given to you and then you can perform this particular algorithm and then this DFS call.', 'start': 691.895, 'duration': 5.806}, {'end': 701.844, 'text': "So let's discuss about the time complexity.", 'start': 698.041, 'duration': 3.803}, {'end': 705.187, 'text': 'because that is something which is very important to understand,', 'start': 702.685, 'duration': 2.502}, {'end': 713.514, 'text': 'because going forward we will be solving problems where we will be running this particular for loop and then we will be calling these dfs or bfs calls.', 'start': 705.187, 'duration': 8.327}, {'end': 716.336, 'text': "so at first let's discuss the space complexity.", 'start': 713.514, 'duration': 2.822}, {'end': 717.256, 'text': 'are we using something?', 'start': 716.336, 'duration': 0.92}, {'end': 719.338, 'text': 'yes, we created an adjacency list.', 'start': 717.256, 'duration': 2.082}, {'end': 723.882, 'text': "let's not take that into account, because most of the problems, adjacency list, will be given to you.", 'start': 719.338, 'duration': 4.544}, {'end': 726.484, 'text': "so let's not discuss about the adjacency list.", 'start': 723.882, 'duration': 2.602}, {'end': 731.508, 'text': "apart from that, we're using a visited array which takes, we go often and we're using a recursion stack.", 'start': 726.484, 'duration': 5.024}, {'end': 734.79, 'text': 'space, which are the worst case, can go up to big often.', 'start': 731.508, 'duration': 3.282}, {'end': 738.593, 'text': "imagine you're given a graph, a skewed graph, something like this.", 'start': 734.79, 'duration': 3.803}, {'end': 741.476, 'text': 'then the dfs call will go from here to here to here.', 'start': 738.593, 'duration': 2.883}, {'end': 748.001, 'text': 'so it will go the entire way and thereby it will take a recursion stack, space of bico, of n.', 'start': 741.476, 'duration': 6.525}, {'end': 750.283, 'text': "okay. so you've understood the space complexity.", 'start': 748.001, 'duration': 2.282}, {'end': 751.664, 'text': 'what about the time complexity?', 'start': 750.283, 'duration': 1.381}, {'end': 753.205, 'text': 'how much time does it take?', 'start': 751.664, 'duration': 1.541}, {'end': 757.669, 'text': 'so you can argue that striver, there is a for loop that runs and then there is a dfs call.', 'start': 753.205, 'duration': 4.464}, {'end': 761.671, 'text': 'right. so you might see that we go of n into the dfs.', 'start': 758.089, 'duration': 3.582}, {'end': 763.833, 'text': 'we now takes v plus 2e.', 'start': 761.671, 'duration': 2.162}, {'end': 768.196, 'text': "so, but this is wrong, it's not this.", 'start': 763.833, 'duration': 4.363}, {'end': 772.679, 'text': 'instead of that, it is n plus v plus 2e.', 'start': 768.196, 'duration': 4.483}, {'end': 775.16, 'text': "why let's go back?", 'start': 772.679, 'duration': 2.481}, {'end': 778.482, 'text': 'so how many times is this dfs executed?', 'start': 775.16, 'duration': 3.322}, {'end': 784.558, 'text': 'three times for this particular graph once for here, Once for here, once for here.', 'start': 778.482, 'duration': 6.076}], 'summary': 'The time complexity is o(n + v + 2e), and the space complexity can go up to o(n) in the worst case.', 'duration': 92.663, 'max_score': 691.895, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA691895.jpg'}, {'end': 768.196, 'src': 'embed', 'start': 738.593, 'weight': 1, 'content': [{'end': 741.476, 'text': 'then the dfs call will go from here to here to here.', 'start': 738.593, 'duration': 2.883}, {'end': 748.001, 'text': 'so it will go the entire way and thereby it will take a recursion stack, space of bico, of n.', 'start': 741.476, 'duration': 6.525}, {'end': 750.283, 'text': "okay. so you've understood the space complexity.", 'start': 748.001, 'duration': 2.282}, {'end': 751.664, 'text': 'what about the time complexity?', 'start': 750.283, 'duration': 1.381}, {'end': 753.205, 'text': 'how much time does it take?', 'start': 751.664, 'duration': 1.541}, {'end': 757.669, 'text': 'so you can argue that striver, there is a for loop that runs and then there is a dfs call.', 'start': 753.205, 'duration': 4.464}, {'end': 761.671, 'text': 'right. so you might see that we go of n into the dfs.', 'start': 758.089, 'duration': 3.582}, {'end': 763.833, 'text': 'we now takes v plus 2e.', 'start': 761.671, 'duration': 2.162}, {'end': 768.196, 'text': "so, but this is wrong, it's not this.", 'start': 763.833, 'duration': 4.363}], 'summary': 'Dfs call takes o(bico^n) space, time complexity is not n*v + 2e', 'duration': 29.603, 'max_score': 738.593, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA738593.jpg'}, {'end': 855.891, 'src': 'heatmap', 'start': 795.63, 'weight': 0.721, 'content': [{'end': 799.074, 'text': 'Whenever it is executed for 4, in total, 3 DFS calls are made.', 'start': 795.63, 'duration': 3.444}, {'end': 802.558, 'text': 'Whenever it is executed for 7, in total, 2 DFS calls are made.', 'start': 799.474, 'duration': 3.084}, {'end': 807.927, 'text': "so overall, can i see i'm making a dfs call for all the end nodes.", 'start': 802.998, 'duration': 4.929}, {'end': 810.912, 'text': "i'm making a dfs call for all the end nodes.", 'start': 807.927, 'duration': 2.985}, {'end': 820.508, 'text': 'can i say that i can, so thereby i can say this is the outer loop and the inner loop overall throughout the loop journey.', 'start': 810.912, 'duration': 9.596}, {'end': 824.571, 'text': 'throughout the loop journey runs for the dfs times for the entire graph.', 'start': 820.508, 'duration': 4.063}, {'end': 831.536, 'text': 'it runs throughout because partially it will run for when the starting node is one, then the starting node is four, then the starting node is seven.', 'start': 824.571, 'duration': 6.965}, {'end': 833.017, 'text': 'so partial dfs, partial dfs,', 'start': 831.536, 'duration': 1.481}, {'end': 843.003, 'text': 'partial dfs sums up to overall dfs of the graph and the overall dfs of the graph is having a complexity of near about v plus 2e, near about v plus 2e,', 'start': 833.017, 'duration': 9.986}, {'end': 853.17, 'text': "and this can be like equivalently see it as near about a linear stuff, near about again it's not exactly because we v plus 2e is there.", 'start': 843.003, 'duration': 10.167}, {'end': 855.891, 'text': 'then depends on the number of neighbor nodes as well.', 'start': 853.17, 'duration': 2.721}], 'summary': 'Dfs calls made for 4 and 7, with overall complexity near v + 2e.', 'duration': 60.261, 'max_score': 795.63, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA795630.jpg'}, {'end': 855.891, 'src': 'embed', 'start': 833.017, 'weight': 0, 'content': [{'end': 843.003, 'text': 'partial dfs sums up to overall dfs of the graph and the overall dfs of the graph is having a complexity of near about v plus 2e, near about v plus 2e,', 'start': 833.017, 'duration': 9.986}, {'end': 853.17, 'text': "and this can be like equivalently see it as near about a linear stuff, near about again it's not exactly because we v plus 2e is there.", 'start': 843.003, 'duration': 10.167}, {'end': 855.891, 'text': 'then depends on the number of neighbor nodes as well.', 'start': 853.17, 'duration': 2.721}], 'summary': 'Overall depth-first search complexity is near v + 2e, approaching linearity.', 'duration': 22.874, 'max_score': 833.017, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA833017.jpg'}], 'start': 451.378, 'title': 'Dfs traversal and time complexity', 'summary': 'Explains the process of converting an adjacency matrix to an adjacency list, implementing dfs traversal, and analyzing space complexity, along with insights on the time complexity of dfs as near about v + 2e and the space complexity as o(n), with a call for b go of n traversal.', 'chapters': [{'end': 738.593, 'start': 451.378, 'title': 'Dfs traversal and adjacency matrix to list conversion', 'summary': 'Explains the process of converting an adjacency matrix to an adjacency list, implementing dfs traversal and analyzing the space complexity, with a demonstration of c++ and java code.', 'duration': 287.215, 'highlights': ['The process of converting an adjacency matrix to an adjacency list is explained, including the creation of the adjacency list and the use of loops to populate it.', 'The implementation of DFS traversal is outlined, covering the creation of a visited matrix, the iterative process using DFS calls, and the counting of provinces.', 'The discussion of space complexity emphasizes the usage of a visited array and a recursion stack, providing insight into the potential space requirements for skewed graphs.']}, {'end': 919.973, 'start': 738.593, 'title': 'Dfs time and space complexity', 'summary': 'Explains the time complexity of dfs as near about v + 2e, and the space complexity as o(n), with insights on the impact of neighbor nodes on time complexity and a call for b go of n traversal.', 'duration': 181.38, 'highlights': ['The time complexity of the DFS is near about v + 2e, impacted by the number of neighbor nodes, and can be summarized as B go of n traversal.', 'The space complexity is O(n) due to the recursion stack space of B go of n.', 'The DFS is executed for all the end nodes, resulting in a total of 3 DFS calls for some nodes and 2 DFS calls for others.', 'The chapter emphasizes the impact of neighbor nodes on time complexity, stating that it will vary depending on the number of neighbor nodes, and provides an example of a graph with single nodes to illustrate the variation.']}], 'duration': 468.595, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/ACzkVtewUYA/pics/ACzkVtewUYA451378.jpg', 'highlights': ['The time complexity of the DFS is near about v + 2e, impacted by the number of neighbor nodes, and can be summarized as B go of n traversal.', 'The space complexity is O(n) due to the recursion stack space of B go of n.', 'The process of converting an adjacency matrix to an adjacency list is explained, including the creation of the adjacency list and the use of loops to populate it.', 'The implementation of DFS traversal is outlined, covering the creation of a visited matrix, the iterative process using DFS calls, and the counting of provinces.']}], 'highlights': ['The time complexity of the DFS is near about v + 2e, impacted by the number of neighbor nodes, and can be summarized as B go of n traversal.', 'The space complexity is O(n) due to the recursion stack space of B go of n.', 'DFS traversal can be used to count provinces in a graph by starting from multiple nodes, ensuring the visitation of all nodes. This approach can identify starting points and ultimately determine the number of provinces.', 'The process of converting an adjacency matrix to an adjacency list is explained, including the creation of the adjacency list and the use of loops to populate it.', 'The problem involves finding the number of provinces in a given undirected graph with v vertices.']}