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int MinimumCost(int n, int m, vector<vector<int>>& grid) {     vector<int> ele;     vector<vector<int>> grid_inv(m, vector<int>(n));     for (int j = 0; j < m; j += 1) {         for (int i = 0; i < n; i += 1) {             grid_inv[j][i] = grid[i][j];         }     }     for (int j = 0; j < m; j += 1) {         sort(grid_inv[j].begin(), grid_inv[j].end());     }     for (int i = 0; i < n; i += 1) {         for (int j = 0; j < m; j += 1) {             ele.push_back(grid[i][j]);         }     }     sort(ele.begin(), ele.end());     int sz = ele.size();     int ans = 1e9; // Assuming int limit is sufficient for ans     for (int i = 0; i < ele.size(); i += 1) {         int mx = ele[i];         bool f = 1;         int mn = 1e9; // Assuming int limit is sufficient for mn         for (int j = 0; j < m; j += 1) {            int idx = lower_bound(grid_inv[j].begin(), grid_inv[j].end(), mx) - grid_inv[j].begin();            if (idx >= n || grid_inv[j][idx] > mx) idx--;            if (idx < 0) {             f = 0;             break;            }            mn = min(mn, grid_inv[j][idx]);         }         if (f) ans = min(ans, mx - mn);     }     return ans; } Minimum Cost path✅

int countKSubsequencesWithMaxBeauty(string s, int k) {     vector<int> count(26);     for (char& c : s)         count[c - 'a']++;          nth_element(count.begin(), count.begin() + 26 - k, count.end());          if (k > 26 || count[26 - k] == 0)         return 0;          long long res = 1, comb = 1, bar = count[26 - k], mod = 1e9 + 7, pend = 0;          for (int& freq : count) {         if (freq > bar) {             k--;             res = res * freq % mod;         }         if (freq == bar)             pend++;     }          for (int i = 0; i < k; ++i) {         comb = comb * (pend - i) / (i + 1);         res = res * bar % mod;     }          return res * comb % mod; } SPECIAL-SUBSEQUENCE CODE

int solve(string S) { set decimalValues; int n = S.length(); for (int i = 1; i < (1 << n); i++) { int subsequenceDecimal = 0; for (int j = 0; j < n; j++) { if (i & (1 << j)) { subsequenceDecimal = subsequenceDecimal * 2 + (S[j] - '0'); } } decimalValues.insert(subsequenceDecimal); } int distinctDecimalValues = decimalValues.size(); int moduloResult = distinctDecimalValues % (int)(1e9 + 7); return moduloResult; } Distinct Subsequence

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int MinimumCost(int n, int m, vector<vector<int>>& grid) {     vector<int> ele;     vector<vector<int>> grid_inv(m, vector<int>(n));     for (int j = 0; j < m; j += 1) {         for (int i = 0; i < n; i += 1) {             grid_inv[j][i] = grid[i][j];         }     }     for (int j = 0; j < m; j += 1) {         sort(grid_inv[j].begin(), grid_inv[j].end());     }     for (int i = 0; i < n; i += 1) {         for (int j = 0; j < m; j += 1) {             ele.push_back(grid[i][j]);         }     }     sort(ele.begin(), ele.end());     int sz = ele.size();     int ans = 1e9; // Assuming int limit is sufficient for ans     for (int i = 0; i < ele.size(); i += 1) {         int mx = ele[i];         bool f = 1;         int mn = 1e9; // Assuming int limit is sufficient for mn         for (int j = 0; j < m; j += 1) {            int idx = lower_bound(grid_inv[j].begin(), grid_inv[j].end(), mx) - grid_inv[j].begin();            if (idx >= n || grid_inv[j][idx] > mx) idx--;            if (idx < 0) {             f = 0;             break;            }            mn = min(mn, grid_inv[j][idx]);         }         if (f) ans = min(ans, mx - mn);     }     return ans; } Minimum Cost path

int solve(string S) { set<int> decimalValues; int n = S.length(); for (int i = 1; i < (1 << n); i++) { int subsequenceDecimal = 0; for (int j = 0; j < n; j++) { if (i & (1 << j)) { subsequenceDecimal = subsequenceDecimal * 2 + (S[j] - '0'); } } decimalValues.insert(subsequenceDecimal); } int distinctDecimalValues = decimalValues.size(); int moduloResult = distinctDecimalValues % (int)(1e9 + 7); return moduloResult; } Distinct Subsequence

int countKSubsequencesWithMaxBeauty(string s, int k) {     vector<int> count(26);     for (char& c : s)         count[c - 'a']++;          nth_element(count.begin(), count.begin() + 26 - k, count.end());          if (k > 26 || count[26 - k] == 0)         return 0;          long long res = 1, comb = 1, bar = count[26 - k], mod = 1e9 + 7, pend = 0;          for (int& freq : count) {         if (freq > bar) {             k--;             res = res * freq % mod;         }         if (freq == bar)             pend++;     }          for (int i = 0; i < k; ++i) {         comb = comb * (pend - i) / (i + 1);         res = res * bar % mod;     }          return res * comb % mod; } GOOGLE SPECIAL-SUBSEQUENCE CODE

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map mp; long long  dp[100005]; long long solve(vector&a,int i,int n) {     if(i>=n)     return 0;         if(dp[i]!=-1)     return dp[i];         int idx=upper_bound(a.begin(),a.end(),a[i]+1)-a.begin();         return dp[i]=max(solve(a,i+1,n),mp[a[i]]*a[i]+solve(a,idx,n)); } long getMaximumPower(vector a) {    int n=a.size();    memset(dp,-1,sizeof(dp));    sort(a.begin(),a.end());    vector res;    for(int i=0;i

int getMaxSubarrayLen(vector team_a, vector team_b) { int n=team_a.size(); vector > dp(n+1,vector (2,0)); int result=0; for(int i=1;i= team_a[i - 1] ? max(dp[i][0], 1 + dp[i - 1][0]):dp[i][0]; dp[i][0] = team_a[i] >= team_b[i - 1] ? max(dp[i][0], 1 + dp[i - 1][1]):dp[i][0]; dp[i][1] = team_b[i] >= team_a[i - 1] ? max(dp[i][1], 1 + dp[i - 1][0]):dp[i][1]; dp[i][1] = team_b[i] >= team_b[i - 1] ? max(dp[i][1], 1 + dp[i - 1][1]):dp[i][1]; result = max({dp[i][0],dp[i][1], result}); } return result+1; }

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