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List duplicates = new ArrayList<>();         for (int i = 0; i < inputchar.size(); i++) {             for (int j = i + 1; j < inputchar.size(); j++) {                 if (inputchar.get(i).equals(inputchar.get(j)) && !duplicates.contains(inputchar.get(i))) {                     duplicates.add(inputchar.get(i));                     break;                 }             }         }         return duplicates;     } Program to find duplicate in element Array IBM

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#include<bits/stdc++.h> using namespace std; int main() {     int n,c,d ;     cin>>n>>c>>d ;     int b[n],p[n],t[n] ;     for(int i=0;i<n;i++)         cin>>b[i]>>p[i]>>t[i] ;     vector<pair<int,int>>type_0,type_1 ;     for(int i=0;i<n;i++)     {         if(t[i]==0)         {             type_0.push_back({p[i],b[i]}) ;         }         else         {             type_1.push_back({p[i],b[i]}) ;         }     }     sort(type_0.begin(),type_0.end()) ;     sort(type_1.begin(),type_1.end()) ;     // One using coins and one using diamonds     int max_0=0,max_1=0 ;     int x=type_0.size() ;     int y=type_1.size() ;     for(int i=0;i<x;i++)     {         if(type_0[i].first<=c)             max_0=max(max_0,type_0[i].second) ;     }     for(int i=0;i<y;i++)     {         if(type_1[i].first<=d)             max_1=max(max_1,type_1[i].second) ;     }     int ans=0 ;     if(max_0&&max_1)         ans=max(ans,max_0+max_1) ;       // Both using coins     multiset<int>m;                             for(int i=0;i<x;i++)     {        m.insert(type_0[i].second) ;     }     int j=type_0.size()-1 ;     for(int i=0;i<x;i++)     {         if(j<=i)             break ;         auto it=m.find(type_0[i].second) ;         m.erase(it) ;         int flag=0 ;            while(j>i)         {             if(type_0[j].first+type_0[i].first<=c)             {                 flag=1 ;                 break ;             }             auto it=m.find(type_0[j].second) ;             m.erase(it) ;             j-- ;         }         if(flag==0)             break ;         if(m.size())         ans=max(ans,type_0[i].second+*m.rbegin()) ;       }     // Both using diamonds     m.clear() ;     for(int i=0;i<y;i++)     {        m.insert(type_1[i].second) ;     }      j=type_1.size()-1 ;     for(int i=0;i<y;i++)     {         if(j<=i)             break ;         auto it=m.find(type_1[i].second) ;         m.erase(it) ;         int flag=0 ;         while(j>i)         {             if(type_1[j].first+type_1[i].first<=d)             {                 flag=1 ;                 break ;             }             auto it=m.find(type_1[j].second) ;             m.erase(it) ;             j-- ;         }         if(flag==0)             break ;         if(m.size())         ans=max(ans,type_1[i].second+*m.rbegin()) ;     }     cout<<ans<<"\n" ;     return 0; } Uber

#include<bits/stdc++.h> using namespace std; string str, bad_string; struct node{     bool end_mark;     node *next[10];     node()     {         end_mark = false;         for(int i = 0; i<10; i++)             next[i] = NULL;     } }*root; bool add(string s) {     node *current = root;     for(int i = 0; i<s.size(); i++)     {         int nw = s[i] - 'a';         if(i == (s.size()-1) && current->next[nw] != NULL)             return false;         if(current->next[nw] == NULL)             current->next[nw] = new node();         current = current->next[nw];         if(current->end_mark)             return false;     }     current->end_mark = true;     return true; } int main() {     int i, N;     bool ok = true;     cin >> N;     root = new node();     for(i = 1; i<=N; i++)     {         cin >> str;         if(!ok)             continue;         ok = add(str);         if(!ok)             bad_string = str;     }     if(ok)         printf("GOOD SET\n");     else     {         printf("BAD SET\n");         cout << bad_string << endl;     } } Good Bad String Uber

#include <bits/stdc++.h> using namespace std; const int MOD = 1e9 + 7; long long solution(int n, vector<long long>& a, vector<long long>& b) {     long long ans = 0;     int minP = min_element(a.begin(), a.end()) - a.begin();     int minQ = min_element(b.begin(), b.end()) - b.begin();     for (int i = 0; i < n; i++) {         if (i == minP || i == minQ)             continue;         ans += min(a[minP] * b[i], b[minQ] * a[i]);         ans %= MOD;     }     if (minP != minQ) {         ans += a[minP] * b[minQ];         ans %= MOD;     }     return ans; }. professor code Uber

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#include <vector> #include <unordered_map> #include <iostream> using namespace std; int getMinTransactions(int n, vector<vector<int>>& debt) {     unordered_map<int, int> balance     for (const auto& d : debt) {         balance[d[0]] -= d[2];         balance[d[1]] += d[2];     }         vector<int> transactions;     for (const auto& entry : balance) {         if (entry.second != 0) {             transactions.push_back(entry.second);         }     }     int count = 0;     int i = 0, j = transactions.size() - 1;     while (i < j) {         if (transactions[i] + transactions[j] == 0) {             count++;             i++;             j--;         } else if (transactions[i] + transactions[j] > 0) {             transactions[i] += transactions[j];             j--;             count++;         } else {             transactions[j] += transactions[i];             i++;             count++;         }     }     return count; } Transaction Simplification

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int solve(vector tc) {     int n = tc.size();     vector p(n), s(n);     p[0] = tc[0];     for (int i = 1; i < n; ++i) {         p[i] = p[i - 1] + tc[i];     }     s[n - 1] = tc[n - 1];     for (int i = n - 2; i >= 0; --i) {         s[i] = s[i + 1] + tc[i];     }     int m = max(p[0], s[0]);     for (int i = 1; i < n; ++i) {         m = max(m, max(p[i], s[i]));     }     return m; } Amazon

#include <iostream> #include <string> #include <unordered_map> using namespace std; long getkRepValue(string user_history, long k) {     long n = user_history.size();     unordered_map<char, long> count;     long left = 0, right = 0, ans = 0;     while (right < n) {         count[user_history[right]]++;         while (count[user_history[right]] >= k && left <= right) {             ans += n - right;             count[user_history[left]]--;             left++;         }         right++;     }     return ans; }  Machine learning Amazon