L16. M-Coloring Problem | Backtracking

[@takeUforward]

✅ Instagram: instagram.com/striver_79/​ Please comment if you understand, a comments means a lot to me :)

[@fardeenshaikh6434]

Eid Mubarak

[@jayadubey_22]

bhaiyaa I was not able to understand the brute force approach solution of this at gfg please explain that also in some other problems

[@rosansenapati-pl5hr]

Striever why we don't use any of graph tarversal?

[@shivamdashore6864]

Why TC will be M^N ?? it should be (M^N) * N, Because we traverse adjacency list as well in isSafe method everytime and maximum cost of adjList will be of N So. TC will be (M^N) * N @takeUforward

[@lakhansingh_barod]

Tried solving without looking at any explanation for the first time in this playlist, took me almost three hours but did it on my own❤

[@arnobsaha7071]

Do u just solved the algo or full. Coding

[@lakhansingh_barod]

@@arnobsaha7071 Full code

[@umangsharma7623]

nice

[@manasanand3531]

I was looking for this problem explanation for the past few days but could'nt find a proper one. Now I can surely say, this one was the best amongst all. Thanks for this.🙌

[@takeUforward]

Thankyouu :)

[@human75788]

I solved the problem myself just with your explanation upto 13 minutes. Thanks Striver. The way you spoonfeed us nobody does, it just stays in my head.

[@tanayshah275]

Heads up: If you are practicing on gfg the graphs are stored in a different manner for different programing languages. For c++ : in matrix For Java: in an adjacency list For Python : in matrix For Javascript : in matrix Just posting because when I started implementing with python I was access the graph as if it is an adjacency matrix and it was resulting in wrong submission so, I thought if anyone else is trying to do the same can save some time by not repeating my mistake.

[@tanayshah275]

posting python version of same code : def is_safe(node, c, graph, color, n): for i in range(n): if i != node and graph[i][node] == 1 and color[i] == c: return False return True def solve(node, graph, color, m, n): if node == n: return True for c in range(1, m + 1): if is_safe(node, c, graph, color, n): color[node] = c if solve(node + 1, graph, color, m, n): return True color[node] = 0 return False color = [0] * V return solve(0, graph, color, k, V)

[@shubhambhardwaj8894]

Thank you brother! I'm preparing for my placements following your sde sheet and I'm getting so much confidence, please continue doing the great work 🙏👍

[@tekken1935]

how is the progress? Placed yet?

[@jaydeepkadam]

Sir placed? Please reply ✨

[@whocares7557]

Good explanation like always, thanks🙂❤️ Wanted to get clarified of two things here: 🤔0. Isn't the complexity M^N, because there are N places to fill with M choices for each, so wouldn't M be multiplied N times making it M^N? 🤔1. We haven't considered the time complexity for checking the possibility of filling the color(isSafe), which can be of order N at worse, but shouldn't we?

[@bharathkumar5870]

total time (M^N)*N(issafe)

[@whocares7557]

@@bharathkumar5870 yes, that's what I thought it should be.

[@parthsalat]

@@bharathkumar5870 Yes, That's what I believe is correct

[@shivamnegi7552]

@@bharathkumar5870 why not (M*N) ^ N ?

[@upamanyumukharji3157]

@@shivamnegi7552 yes checking will be taken for each color pick so (M*N)^N

[@kushalgupta2041]

hey guys this question didn't need even recursion solution for example we just need to check if it possible right ! We don't have to find which colour we use so it didn't create any difference if we use 1 for 0 or 2 for 0, So what I mean is if we can color a node means it is possible we can directly return the solve() function because it will not matter what colour we use, So basically what we will be doing is iterating from 0 to n nodes and checking if is possible to color them then assign them the color and if not possible then return false; Here is my code:- class Solution { private: bool solve(int node, int v, int m, vector &color, vector adj[]){ if(node == v) return true; for(int col = 0; col < m; col++){ if(isPossible(node, col, color, adj)){ color[node] = col; return solve(node+1, v, m, color, adj); } } return false; } bool isPossible(int node, int col, vector &color, vector adj[]){ for(auto it: adj[node]){ if(color[it] != -1 && color[it] == col) return false; } return 1; } public: bool graphColoring(int v, vector& edges, int m) { vector adj[v]; vector color(v, -1); for(auto ed: edges){ adj[ed.first].push_back(ed.second); adj[ed.second].push_back(ed.first); } return solve(0, v, m, color, adj); } }; TC:- O(m^2) + O(V) SC:- O(V) + O(V) + O(E) The reason tc is m^2 + v is because assume every node is connected to every other then you will traverse for first only 1 time then for second node 2 times then 3rd node 3 times and so on so overall the time it will take to iterate colour loop is m*2 and this is overall, and so overall vertices are V and that's it

[@sowndaryav6680]

U are doing a great job for students sir. Thank you so much for your efforts.

[@vegitogamingpubg3364]

Very detailed and easy to understand explanation.. 10 times better than the so called paid courses.. Thank you so much bhaiya ❤

[@devanshikapla7491]

Amazing explanation as well as the code. I haven't seen so much clear explanation on any other channel. Thankyou for this❤️

[@Tomharry910]

Beautifully explained a tough problem very simply and in an easy to understand way. Thank you so much!

[@kamalkantrajput4431]

time complexity = O(m ^ n) not o(n^m) thanks bhaiya . as we have m choice for each node .

[@SJ_46]

yes right

[@ROSHANKUMAR-rl6bf]

If the graph is connected simple dfs based recursion also works but one can only appreciate this if he wrote code for connected and realises if there are more than one components what should be done

[@vankshubansal6495]

True, I skipped this part. Attaching the DFS solution which handles all cases bool solve(int sv, bool graph[101][101], int V, vector& visited, int colors) { unordered_map visitedColors; for(int i = 0; i < V; i++) { if(graph[sv][i] == true && visited[i] != -1) visitedColors[visited[i]] = 1; } if(visitedColors.size() == colors) return false; for(int i = 0; i < colors; i++) { if(visitedColors[i] > 0) continue; visited[sv] = i; bool isAll = true; int j = 0; for(j = 0; j < V; j++) { if(graph[sv][j] == true && visited[j] == -1) { if(solve(j, graph, V, visited, colors) == false) break; } } if(j == V) return true; } visited[sv] = -1; return false; } bool graphColoring(bool graph[101][101], int m, int V) { vector visited(V, -1); for(int i = 0; i < V; i++) { if(visited[i] == -1) { if(solve(i, graph, V, visited, m) == false) return false; } } return true; }

[@rishabhgupta9846]

@@vankshubansal6495 can you pls explain how your code is working?

[@sairamsudireddy6191]

@@vankshubansal6495 is this a working solution ?? for(j = 0; j < V; j++) { if(graph[sv][j] == true && visited[j] == -1) { if(solve(j, graph, V, visited, colors) == false) break; } } In this part of the soution, if you are able to color some of the child nodes, and not able to color a child node, where are you backtracking (un coloring )all the child nodes that are colored ?

[@pranavdharkar]

i was not able to solve questions on my own before but i solved this question completely by my own today the way he taught intutions is amazing 😍

[@sasirekhamsvl9504]

The best explanation I have found on youtube. Thank you so much.

[@AdityaRajVerma-io3pr]

i was not able to draw the recursion tree, now as I understood the approach, I just coded it by myself, thanks bhaiya

[@mohitsingh7793]

Correction in Time-Complexity discussed in striver's vedio T.C = O(M^N)*N(isSafe-fxn) For one node ,we have m different colors. For n node we have m*m*m.....n times =M^N Also isSafe() takes O(N) in worst case. Let me know if I was wrong.

[@tushar7305]

It N^m* N

[@snehagoyal4978]

Thank you striver, after watching your previous videos of this playlist, I could do this on my own;

[@nitinvadadoriya8280]

Small Correction In Time Complexity T.C = ~M^N not N^M Because we have M-Color choice for every nodes. Tell me am i correct or not..

[@98ujaal]

Correction: Time complexity is O(M^N) not O(N^M) (which solves the P vs NP problem and he would have been a millionaire by now)

[@priyanshudwivedi7535]

What do you mean by him being a millionaire?

[@shivammishra6381]

@@priyanshudwivedi7535 P vs NP problems are problems can truly revolutionize the world of computing and maths. and have reward of a million dollars for the person who solve them.

[@Rohan-hj9gg]

The time complexity has to be M^N ? Because the height of tree will go till N and for each node we have M choices.

[@soumavanag5025]

correct, it confused me

[@rncsMahimaMahendru]

even i think so

[@abhisheksa6635]

Badhiya bhai, woh tumney strategy same rakha h parallel recursion wala and saarey backtracking solve kar liye.

[@stith_pragya]

Understood............Thanks a ton for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@aditya-bl5xh]

Shuru majburi mei kiye the ab mazza aa rha hai...

[@anabildebnath2590]

Such great explaination, kudos

[@ashutoshkumawat7854]

I Think it's Complexity is M^n because m is no.of choice like it happen in printing all sub sets 2^n.......please correct me if i'm wrong

[@janaSdj]

One more optimization,if m>=4 then it is always possible to colour a graph by 4 colour theorem. So if m>=4 return true.

[@pranavpandey4331]

I think the time complexity should be O(m^N)

[@sharmaji490]

Yupp it will be

[@narayanbung7550]

Your videos make problem look very simple.Thanks

[@animearena8443]

python code for anyone struggling with it: def graphColoring(graph, k, V): color=[0]*V def mcolor(vertex,graph,k,V): if vertex==V: return True for i in range(1,k+1): flag=1 for j in range(V): if graph[vertex][j]==1 and color[j]==i: flag=0 break if flag==1: color[vertex]=i if mcolor(vertex+1,graph,k,V): return True color[vertex]=0 return False return mcolor(0,graph,k,V)

[@riyakumari8377]

hey can u tell me why are we doing => graph[vertex][j]==1 ?

[@nirupamsuraj1634]

​@@riyakumari8377to check if the node j is connected to vertex node

[@PrashantSingh-jy6zp]

for skip ads go to 4:01

[@omkarsawant9267]

C++ Code for M coloring Problem--> TC-- M^V v is no of vertices and M is max no if colors that can be used. Exponential TC due to Recursive nature of algorithm SC-- O(V) as color vector stores color assigned to each vertex. Size is proportional to no of vertices in the graph.But it is at most O(V) due to depth of recursion. ----- Code Follows ------ #include #include using namespace std; class Graph { private: int V; // No of Vertices vector adj; // adjacent list; public: // constructor Graph(int Vertices) : V(Vertices), adj(Vertices) {} // Funciton to add an edge to the graph void addEdge(int u, int v) { adj[u].push_back(v); adj[v].push_back(u); } // func to check if it is possible to color the graph with at most M colors bool graphColoring(int M); private: // utility function for graph coloring; bool graphColoringUtil(int vertex, vector &color, int M); }; bool Graph :: graphColoringUtil(int vertex, vector &color, int M) { // check if we have assigned colors to all vertices if (vertex == V) return true; // try assigning different colors to current vertex for (int c = 1; c

[@factfactorial632]

I have doubt , time complexity should be N*(N^m) because we have that IsSafe function also there and at max a node can be connected to all the others node So O(N) for chacking btw Great explanation

[@tushar7305]

Yes it should be there

[@mehershrishtinigam5449]

@@tushar7305 that will just be O(N^(m+1)) which will be O(N^m) only, i think?

[@shwetabhagat8920]

you make all the problems easy⭐

[@mohammedwaseem8599]

Hello bhaiya i hope you are doing extremely well

[@kaichang8186]

understood, thanks for the perfect explanation

[@reshubnigam8375]

How do we get the intuition that here we had to traverse serially on the nodes and not initiate dfs for every connected component? Had been doing that and got stuck at what the base case would be for dis-connected components as backtracking was erasing everything :|

[@RahulKashyap-yv5ox]

Yes I was also stucked at the same point

[@parthsalat]

C++ code starts at 20:53

[@bitturanjan9539]

Great explanation man❤️

[@rishukumarsingh3926]

In this problem, we are trying every color, if and only if the color is possible to take, i.e. we are filtering and taking, we are not waiting to mismatch!

[@sohamsonwane1534]

The best what you get i have done the recursion and backtracking from another paid courser but it is not worth it but the striver recursion playlist is bast of class i have done all recursion from 1 to last and now the recursion feel like game not very easy but so far so good

[@supratimbhattacharjee5324]

Best explanation....learning for my practical exams

[@siddiqabr7110]

Why it is on the recursion list instead of graph I don't even know what a graph is

[@vaishnavi5070]

shouldn't the graph be List where every index is considered as node and the list that is there in that index are the adjacent nodes??

[@sahelidebnath9972]

For adjacency list, shouldn't we check: public boolean isColorPossible(int node, int length, int colorTochoose,boolean graph[][], int[] color ) { for(int i=0;i

[@SeloniSinha]

Wonderful explanation sir!! Thank you !!!

[@sanjana-singla]

Can't we just count the maximum number of adjacent nodes present in the graph? if the maximum nodes is greater than the number of colors, return false, else return true?

[@akshitkathuria3516]

I had the same thought and tried applying it on the bipartite graph problem but one test case failed which is this one: [[4,1],[0,2],[1,3],[2,4],[3,0]] Here every node has 2 adjacent but we still cannot color the graph using 2 colors. So i think it wont work for this problem as well

[@kannupriyarana4971]

clear and straight-forward explanation. Thanks bro :)

[@adrikagupta5573]

Great video! I have one doubt though. Shouldn't the time complexity be O(m^N)?

[@abhinavpandey3356]

Can u explain why n^m seems correct as for every node there are m possibility

[@sparshsharma6068]

​@@abhinavpandey3356 Here's how i devised the TC for this: At each node, there will be at max there will be m operations and for each operation on a node, their childs will have their own respective m operations. i.e if a graph was: 1 / \ 2 3 then at node 1 there will be m operations but for each operation on node 1, there will be m operations on its successive nodes(here on the node 2 and then on node 3). i.e m*m*m = m^3 == m^n thus on graphs having n nodes, starting from source node, there will be: m*m*m*...........*m(n terms) == m^n. Thus, TC will be O(m^n). I hope it helps!

[@PiyushSharma-bo6pp]

@@abhinavpandey3356 easy o(n) approach is there

[@Pawan_Kumar_Mehta]

Yes yr right cause the height of the tree will be N and at each level of rec we will have m nodes.

[@parthsalat]

Yes, imo that should be the correct time complexity

[@prasannapm3220]

Amazing thought process sir !!

[@RitikKumar-lv8cm]

hi in this problem we can not simply check for each node. instead we must create adjacent node for each node and then check for possibility of coloring

[@siddharthvs1770]

Can we not use a greedy approach for this problem? Would it fail, if yes why?

[@krishnakanttiwari517]

Thank You So Much Sir

[@madhurgupta4220]

A Detailed And A Great Explanation .

[@rushidesai2836]

Very classic recusion problem.

[@paragroy5359]

Thanks for the playlist sir......it's really helpful

[@abhinavpandey3356]

What if I don't put colour [node]=0 Does it affect anything ,as I'm looking colr of negbour node to match with the colr I'm am trying to color for a particular node for deciding it's color

[@fardeenshaikh6434]

Eid Mubarak striver Bhai

[@takeUforward]

Eid mubarak bhai ❤️

[@Ace-ex9gg]

explanation is good but time complexity is (n-1)*m^n by the way.

[@ganeshkamath89]

I have 1questions: 1) For Java you are just checking 1 condition in the loop if (color[it] == col) return false; 2) Whereas in C++ you are checking multiple conditions: if (k != node && graph[k][node] == 1 && color[k] == col) Is this because you have done some preprocessing in Java to have graph as an adjacency list but are passing the graph itself in C++?

[@SanthoshSunny21]

In cpp he used adjacency Matrix nd in Java adjacency list In Matrix you will check if there is an edge with every other vertex (g[i][j]==1) but in adjacency list only edges are present so no need to check if there's an edge

[@ganeshkamath89]

@@SanthoshSunny21 Thanks

[@hackstreet781]

Time complexity should be M^N and not N^M

[@anshulgoel1940]

Will the time complexity be N^M or M^N ?

[@chase.2595]

i think m^n m choices in n rec calls right?

[@skyy-v5o]

Hello striver , I see there are no articles linked to bit manipulation , linked list problems . Plus there are no javascript code on striver website .

[@asishcodes]

Do i need to learn graphs for this problem?

[@lettry5297]

Thanks you sir for this video. Sir can you please clarify whether for SDE profile it is mandatory to know C++ or Java ? Python is not sufficient for cracking test , I am practicing with Python only? Sir please guide..... 🤷‍♂️

[@shivanshkhare2724]

After watching this series , I understood why striver is god of coding.

[@viswanathank2551]

thanks for the best explanation in yt

[@sahelidebnath5085]

Time complexity will be O(M^N) isn't it? (worst case)M colors for every node(N).

[@vishnuthulasidosss]

I wonder why the BFS solution is not working! Could you tell me what's wrong with BFS?

[@rishurana9655]

This question is exactly similar to n queen problem for those who have problem can first try that one and then come to this question

[@nakuljindal1421]

Bro why we have backtracked in this as for loop itself change its colour suppose if statement gives u false then ultimately it will go to another colour why does it makes difference whether we have done 0 previously or not

[@sahiljain2524]

Bhaiya Time complexity M^N ayegi na ?

[@takeUforward]

Hnn

[@sahiljain2524]

@@takeUforward Ok bhaiya

[@factfactorial632]

@@takeUforward Bhaiya it should be N*(N^m) because we have that IsSafe function also there and at max a node can be connected to all the others node So O(N) for chacking

[@YogaJournalWithMimansa]

Hi All, I was really confused between M coloring problem and bi partite graph problem. The Bi partite graph problem uses DFS/BFS for checking if no 2 adjacent nodes are filled with the same color(2 colors). So I was wondering if we could use DFS/BFS here as well. Turns out no, as in the GFG practice problem, in one TC. we have disconnected nodes as well in the graph and this cant be covered with DFS/BFS. Please let me know if my understanding is correct.

[@Yash_Parashar]

But we can us the dfs and bfs for disconnected nodes also.

[@K_EN_VisheshSaini]

I havent done Graphs yet, do i need to know graphs for this question or Recursion is sufficient alone?

[@gauravagrawal7988]

You should have very basic knowledge of graphs for this

[@raviashwin1157]

God level explanation🔥....really appreciate your efforts❤️....what was that song in the end??

[@techfreak416]

why do we have to check all possible safe colors for every node? why cant we just assign any 1 of the colors which are not adjacent to current node.

[@083_h_nitishkumarjha3]

why we are calling the function for next serial node, shouldn't we call for the nodes attached to this node ?

[@riteshadwani9084]

Amazing explanation!

[@rishukumarsingh3926]

Why in this problem, we didn't follow standard dfs and used as node numbers as a part of traversal?

[@nipunrawat7137]

same doubt

[@aasthajha1051]

explanation ek no.!!!!!!!!!!!!!!!!!

[@sathishkumar-dc9ce]

Great job anna.. luv from TN 😍

[@VikashKumar-tr9cq]

will this algorithm work if there is more than one connected components?

[@lakshmiprasanna7058]

Understood 💯💯💯

[@chase.2595]

isnt time complexity m^n? as m choices of colour available in n recursive calls?

[@dipjoybasak3156]

Brother cant we just check for bipartiteness of the graph? Like if it is bipartite, we can always color using m>=2 and if its not then anything m>=3... Wont this work?

[@nameetjain2251]

Exactly what i was thinking.

[@sharmaji490]

Consider when you are asked to tell if a graph can be colored using 5 colours and suppose it is not biparatite. Then what will you return. You are not sute about 3,4,5 colors. So only checking biparatiye won't work every time. Hope it is helpful in some way

[@sharmaji490]

If a graph is not paratite m>=3 might work

[@parthsalat]

Thanks

[@alishashaikh3859]

I do not know graph , can I solve this problem by using only recursion

[@sachin_yt]

You are the best! 🙌

[@aloklaha8694]

Thanks brother. Understood

[@sahilnegi2789]

Thanks bro for amazing content .

[@ravindermaan1887]

Time Complexity: O(M^N).

[@amanupadhyay1275]

"Definitely" some great stuff Striver. Thanks a lot.

[@UECAshutoshKumar]

Thank you sir

[@PrashantSingh-jy6zp]

c++ code at 20:55

[@ritugoyak7238]

Thank you so much sirr

[@ratankumar1399]

can you make some videos on BFS approach of this ques ,its bit confusing for me

[@josephstark5810]

i think we cant do it by graph traversal beacause it give tle in case of cycle and also hard to manage backtracking right?

[@mohitsingh13]

Understood ❤

[@bhaveshkumar6842]

Thank you!

[@gunanka6860]

again solved on my own even without the explanation... i think im getting better : ")