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Codeforces|Leetcode|Codechef free solutions

Codeforces|Leetcode|Codechef free solutions

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Free codeforces, Codechef, Leetcode solutions are available 😍😍😍😍😍😍 Helped More than 200+ students to crack coding round in 2022 and helped placed them in Good companies. 🥳🥳🥳🤩🤩🤩 Dm @Cpsoln if you want help in coding round.

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If anyone want help in codekaze coding round on 17 then Dm @Cpsoln.

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Since your placement and intern session is coming 🤓🤓 So don't miss any opportunity in this recession😇😇😇😇 📌📌📌📌📌📌📌📌📌📌📌📌 No scam No fraud because we are not like other telegram channels. If you want help in coding round of any company then Dm @Cpsoln and book your slot ✅✅✅✅

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class Solution:     def goodSubsetofBinaryMatrix(self, grid: List[List[int]]) -> List[int]:         for i, e in enumerate(grid):             if sum(e) == 0:                 return [i]         for i in range(len(grid)):             for j in range(i+1, len(grid)):                 if sum([a and b for a, b in zip(grid[i], grid[j])]) == 0:                     return [i, j]                         return [] Solution to Leetcode D

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Leetcode D guys in Cpp🔥🔥🔥❤❤❤❤😁😁

class Solution { public: vector goodSubsetofBinaryMatrix(vector>& grid) { vector result; for (int i = 0; i < grid.size(); i++) { int rowSum = 0; for (int j = 0; j < grid[i].size(); j++) { rowSum += grid[i][j]; } if (rowSum == 0) { result.push_back(i); return result; } } for (int i = 0; i < grid.size(); i++) { for (int j = i + 1; j < grid.size(); j++) { int colSum = 0; for (int k = 0; k < grid[i].size(); k++) { colSum += (grid[i][k] && grid[j][k]); } if (colSum == 0) { result.push_back(i); result.push_back(j); return result; } } } return result; } };

D is coming soon🔥🔥🔥🔥🔥

class Solution { public:     int mod=(1e9 + 7);     int sumDistance(vector& nums, string s, int d) {         for (int i = 0; i < s.size(); i++) {             if (s[i] == 'L') {                 nums[i] -= d;             } else if (s[i] == 'R') {                 nums[i] += d;             }         }         sort(nums.begin(), nums.end());         long long sum = 0;         long long ans = 0;         for (int i = 0; i < nums.size(); i++) {             ans += 1LL*nums[i] * i - sum;             ans%=mod;             sum += nums[i];             sum%=mod;         }         return ans % (int)(1e9 + 7);     } };

class Solution { public:     int longestSemiRepetitiveSubstring(string s) {                 int ans=0;                 for(int i=0;i=2) flag = 1;                     }                 }                 if(!flag) ans=max(ans, (int)t.size());             }         }         return ans;             } };

class Solution { public:     int mod=(1e9 + 7);     int sumDistance(vector& nums, string s, int d) {         for (int i = 0; i < s.size(); i++) {             if (s[i] == 'L') {                 nums[i] -= d;             } else if (s[i] == 'R') {                 nums[i] += d;             }         }         sort(nums.begin(), nums.end());         long long sum = 0;         long long ans = 0;         for (int i = 0; i < nums.size(); i++) {             ans += 1LL*nums[i] * i - sum;             ans%=mod;             sum += nums[i];             sum%=mod;         }         return ans % (int)(1e9 + 7);     } };

Codeforces Soultion: #include <iostream> #include<bits/stdc++.h> using namespace std; #define int long long void sol(){           string a,b;   cin>>a>>b;     int t,q;   cin>>t>>q;         vector<char> aa,bb;     for(auto val:a){       aa.push_back(val);   }       for(auto val:b){       bb.push_back(val);   }         int n = a.size();   int count=0;     for(int i=0;i<n;i++){       if(a[i]!=b[i]) count++;   }   //   cout<<count<<endl;     map<int,int> m;     int timer=1;     while(q--){             int qqq;       cin>>qqq;             if(m[timer]!=0){           count++;       }                               if(qqq==1){                     int pos;           cin>>pos;           pos--;                     if(aa[pos]!=bb[pos]){               m[timer+t]++;               count--;           }                           }       else if(qqq==2){                     int qq,ww,pos1,pos2;           cin>>qq>>pos1>>ww>>pos2;                     pos1--;           pos2--;                               if(aa[pos1]==bb[pos1]) count++;                     if(aa[pos2]==bb[pos2]) count++;                               char s1;           if(qq==1){             //   cout<<"here"<<endl;               s1 = aa[pos1];           }           else{               s1 = bb[pos1];           }                     char s2;                     if(ww==1){               s2 = aa[pos2];           }           else{               s2 = bb[pos2];           }                                         if(qq==1){               aa[pos1] = s2;           }           else{               bb[pos1] = s2;           }                     if(ww==1){               aa[pos2]=s1;           }           else{               bb[pos2]=s1;           }                                if(aa[pos1]==bb[pos1]) count--;                     if(aa[pos2]==bb[pos2]) count--;                   //   cout<<s1<<" "<<s2<<" "<<count<<endl;                                               }       else{           if(count==0){               cout<<"yes"<<endl;           }           else{               cout<<"no"<<endl;           }       }                     //   cout<<count<<endl;                                                       timer++;   }                                       return; } signed main() {   int test=1;   cin>>test;     while(test--){       sol();   }         return 0;     } #include <string> #include <unordered_set> class Solution { public:     bool isFascinating(int num) {         std::string concatenated = std::to_string(num) + std::to_string(2 * num) + std::to_string(3 * num);         if (concatenated.length() != 9) {             return false;         }                 std::unordered_set<char> uniqueDigits(concatenated.begin(), concatenated.end());         return uniqueDigits.size() == 9 && uniqueDigits.find('0') == uniqueDigits.end();     } };

We suggest you to apply for this event as it held only once in the year. All the batches must apply and specially 2024 / 2025 / 2026 batch must apply in this as it will help you practice for your upcoming hiring challenges. Must Apply - CodeKaze ✅ Register Link: https://bit.ly/Mega_Hiring_June Registration Deadline: 10 June (Tomorrow)