Відео

understood

🙇‍♀🙇‍♀🙇‍♀🙇‍♀🙇‍♀🙇‍♀

Thank you for great video, we can avoid repeating the same code in both methods for finding the firstIndex and last using a simple boolean. flag: public int[] searchRange(int[] nums, int target) { int[] ans = {-1,-1}; if(nums.length == 0){ return ans; } int startIndex = search(nums, target, true); int endIndex = search(nums, target, false); ans[0] = startIndex; ans[1] = endIndex; return ans; } int search(int[] nums, int target, boolean findFirstIndex){ int ans = -1; int start = 0; int end = nums.length; while(start <= end){ int mid = (start + end) / 2; if (mid >= nums.length) return ans; if (target < nums[mid]) end = mid - 1; else if (target > nums[mid]) start = mid + 1; else { ans = mid; if (findFirstIndex) end = mid -1; else start = mid + 1; } } return ans; }

understood

ty sir

us

Understood

GOAT

understood

wtffffffffffffff

understood

Sir tussee great ho🫡❤️

ty sir

if anyone needs corrected code string minWindow(string s, string t) { int right=0, left=0, n = s.size(); map<char,int> mp; for(auto i : t){ mp[i]++; } int minLen = INT_MAX; int startInd = -1; int cnt = 0; int m = t.size(); while(right<n){ if(mp[s[right]]>0) cnt++; mp[s[right]]--; while(cnt==m){ if(minLen>right-left+1){ minLen = right-left+1; startInd = left; } mp[s[left]]++; if(mp[s[left]]>0) cnt--; left++; } right++; } if(startInd==-1) return ""; return s.substr(startInd,minLen); }

Understood! Thanks a lot Striver.

done 26/6

understood!

what when one of the value is not present in tree

you are amazing sir

thats amazing

For the base case, couldnt you just write , for(int i=1;i<=n;i++){ if(pattern[i-1]=='*' && dp[i-1][0]==1){ dp[i][0]=1; }else{ dp[i][0]=0; } }

Brilliant explanation and hats off to whoever though of this solution.

Understood!

understood

The way he explained made me to think how to really approach a problem with logical thinking...Great Striver!!!

Understood!

public static ArrayList<Integer> Intersection(int[] arr,int[] arr2,int n,int n2){ Set<Integer> set1=new HashSet<Integer>(); Set<Integer> intersect=new HashSet<Integer>(); for(int i=0;i<n;i++){ set1.add(arr[i]); } for(int i=0;i<n2;i++){ if(set1.contains(arr2[i])){ intersect.add(arr2[i]); } } ArrayList<Integer> a=new ArrayList<Integer>(intersect); return new ArrayList<Integer>(a); } Can we use intersection using this ???

class Solution { public: int lengthOfLongestSubstring(string s) { int n = s.length(), l = 0, r = 0, cnt = 0; unordered_set<char> seen; if (n == 0) return 0; while (r < n) { if (r < n && seen.find(s[r]) == seen.end()) { seen.insert(s[r]); cnt = max(cnt, r - l + 1); r++; } else { seen.erase(s[l]); l++; } } return cnt; } };

us

Understood

Understood

Striver I am from West Bengal. You are god of problem solving!! ❤❤❤❤❤❤

ud

Pivattt!

Pivaattt!

Understood

left + right kyu nhi kiya isme

thank you brother for this kinda of efforts !!! definitely it will help me a lot to get placed in reputed company🤗🤩

This is the first video that I have not understood of you. No matter how many times I watch I just can't understand. I'm just skipping this optimal approach for now. :)

thank you for expaination

you are amazing sir

alt approach: consider the stones as the nodes in dsu, instead of considering the rows/cols (can optimize the visited array size, i was just lazy) code-> (AC on LC) class DS { public: vector<int>parent,size; DS(int n) { size.resize(n,1); parent.resize(n); for(int i=0; i<n; i++) parent[i]=i; } int findUPar(int node) { if(parent[node]==node) return node; else return parent[node]=findUPar(parent[node]); } void unionBySize(int x, int y) { int ulp_x=findUPar(x); int ulp_y=findUPar(y); if(ulp_x==ulp_y) return; else if(size[ulp_x]>=size[ulp_y]) parent[ulp_y]=ulp_x, size[ulp_x]+=size[ulp_y]; else parent[ulp_x]=ulp_y, size[ulp_y]+=size[ulp_x]; } }; class Solution { public: int removeStones(vector<vector<int>>& stones) { //there are n stones, so I am making a dsu with the nodes as the n stones int n=stones.size(); //the coords can be anything from 0 to 1e4 //also the rows and cols need to be dealt separately as visited vector<int>visX(1e4+1,-1), visY(1e4+1,-1); DS dsu(n); for(int i=0; i<n; i++) { int x=stones[i][0], y=stones[i][1]; if(visX[x]==-1) visX[x]=i; else dsu.unionBySize(visX[x], i); if(visY[y]==-1) visY[y]=i; else dsu.unionBySize(visY[y], i); } //we can remove size-1 nodes from each connected component int ans=0; for(int i=0; i<n; i++) { if(dsu.parent[i]==i) ans+=dsu.size[i]-1; } return ans; } };

a very very thankyou for this playlist..from playlist to video and cheatsheet everything is just well managed...even paid courses doesn't provides what u r providing for free❤ and the biggest thank you to make this content as language independent and provide solutions in java too...❤

Presense of negative edge does not always result incorrect result ...that could be easily proved in the second example if there is a path say from 4 to 3 with -1 edge weight ....but if the negative edge cycle change the previous path length which we got after n-1 iteration then it cause problem.

nice explaination

Understood

us

Awesome

Understood

thankyou so much sir