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Shortest String ✅

Unique subarray sum solution

Sending Largest Array with Equal Number of zeroes code✅ All test Cases Passed ✅ share @coding_000❤️

Unique subarray code ✅

eqalibrium python 3

largest subarray equal number Infosys ✅

Split string code

#include <bits/stdc++.h> using namespace std; int equalzeroandone(vector<int>v){     int n=v.size();     for(int i=0;i<n;i++){         if(v[i]==0){             v[i]=-1;         }     }     int sum=0;     int ans=-1;    map<int,int>mp;     for(int i=0;i<n;i++){         sum+=v[i];         if(sum==0){             ans=i+1;         }         if(mp.find(sum)!=mp.end()){             ans=max(ans,i-mp[sum]);         }         else{             mp[sum]=i;         }     }     return ans; } int main() {     int n;     cin>>n;     vector<int>v(n);     for(int i=0;i<n;i++){         cin>>v[i];     }     cout<<equalzeroandone(v); } Equal number of zero Infosys

def split_string_cost(S):     # Length of the string S     len_S = len(S)         # To store the cost of the split parts     max_cost = 0         # Set to keep track of distinct characters in the first part     distinct_chars_A = set()         # List to keep track of the cost for the second part from each split position     cost_B = [0] * len_S         # Set to keep track of distinct characters in the second part     distinct_chars_B = set()         # Calculate cost for second part from the end     for i in range(len_S - 1, -1, -1):         distinct_chars_B.add(S[i])         cost_B[i] = len(distinct_chars_B)         # Calculate maximum sum of cost for parts A and B     for i in range(len_S - 1):         distinct_chars_A.add(S[i])         cost_A = len(distinct_chars_A)         cost = cost_A + cost_B[i + 1]         max_cost = max(max_cost, cost)         # Calculate the result as |S| - X     result = len_S - max_cost     return result # Example usage S = "aaabbb" print(split_string_cost(S))  # Output: 3

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string operations python 3✅

def is_palindrome(s):     return s == s[::-1] def longest_palindrome_from_substrings(A):     palindromes = []     pairs = []     max_single_palindrome = ""     for s in A:         if is_palindrome(s):             palindromes.append(s)             if len(s) > len(max_single_palindrome):                 max_single_palindrome = s     for i in range(len(A)):         for j in range(i + 1, len(A)):             combined1 = A[i] + A[j]             combined2 = A[j] + A[i]             if is_palindrome(combined1):                 pairs.append(combined1)             if is_palindrome(combined2):                 pairs.append(combined2)     longest_palindrome = max_single_palindrome     for p in pairs:         if len(p) > len(longest_palindrome):             longest_palindrome = p     return longest_palindrome Palindromic String✅👨‍💻 Share Our channel Fast @coding_000✅

#include <bits/stdc++.h> #define int long long using namespace std; #define ll long long vector<vector<ll>> solve(ll n,ll k,vector<ll>&a,vector<ll>&b) {     priority_queue<pair<double,pair<ll,ll>>> pq;     for(int i=0;i<n;i++)     {         double x=a[i];         double y=b[i];         double dis=sqrt(x+y);         pq.push({dis,{x,y}});         if(pq.size()>k)  pq.pop();     }     vector<vector<ll>>ans;     while(!pq.empty())     {         ans.push_back({pq.top().second.first,pq.top().second.second});         pq.pop();     }     sort(begin(ans),end(ans));     return ans; } signed main() {                ll n,k; cin>>n>>k;         vector<ll>a(n),b(n);         for(ll i=0;i<n;i++) cin>>a[i];         for(ll i=0;i<n;i++) cin>>b[i];         vector<vector<ll>>ans=solve(n,k,a,b);         for(auto it:ans) cout<<it[0]<<" "<<it[1]<<endl;                 return 0; } Closet K Points Share our channel fast @coding_000✅

Code Name - cost of string s int solve(const string& S) {     vector<int> freq(26, 0);         for (char c : S) {         freq[c - 'a']++;     }     int total = 0;     for (int i = 0; i < 26; ++i) {         for (int j = i + 1; j < 26; ++j) {             total += freq[i] * freq[j] * abs(i - j);         }     }     return total; } 100% Running ✅✅✅ Share our channel @coding_000👨‍💻😊

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