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no of balanced substrings

Dream FTBL answered on July 22, 2022 Popularity 3/10 Helpfulness 1/10

answer no of balanced substrings

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no of balanced substrings

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#include <bits/stdc++.h> 
 using namespace std; 
 int dp[100005]; 
 int main() 
 { 
       int t,n; 
      long long ans; 
       string s; 
       map <int,long long> m; 
      cin >> t; 
      while ( t-- ) 
     { 
              cin >> s; 
             n = (int)s.size();//calculates length of string 
              assert(n<=100000);//checks if n<=100000 
              m.clear(); 
              dp[0] = 0; 
             ans = 0; 
              for ( int i = 1; i <= n; i++ )  
                     dp[i] = dp[i-1]^(1<<(s[i-1]-97)); 
              m[dp[0]]++; 
              for ( int i = 1; i <= n; i++ ) 
             { 
                     ans += m[dp[i]]; 
                    m[dp[i]]++; 
             } 
            cout << ans << endl; 
      } 
      return 0; 
}

Popularity 3/10 Helpfulness 1/10 Language cpp

Source: www.quora.com

Tags: c++

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Contributed on Jul 24 2022

Dream FTBL

0 Answers Avg Quality 2/10

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import java.util.HashMap;
 import java.util.LinkedHashMap;
 import java.util.Scanner;
 import java.util.Set;

    public class BalancedStrings {
    static int  returnbalance(String input){
    HashMap<Character, Integer> characters=new LinkedHashMap<>();
    int count=0;
    Character c=null;
    for(int i=0;i<input.length();i++){
        c=input.charAt(i);
        if(characters.containsKey(c)==true){
            count=characters.get(c);
            characters.put(c, count+1);
        }
        else
            characters.put(c,1);        
    }
    int countunique=0;
    Set<Character> s=characters.keySet();       
    for(Character cc:s){            
        count=characters.get(cc);
        if(count%2==0)
            countunique++;
    }
    if(countunique!=s.size())
        return 0;
    Object[] sar= s.toArray();
    int len=0;
    for(int i=sar.length-1;i>=0;i--){
        len=len+i;
    }

    return len+countunique;

}
public static void main(String[] args) {
    Scanner in=new Scanner(System.in);
    int numberOfInputs=in.nextInt();
    in.nextLine();
    for(int i=0;i<numberOfInputs;i++){
        System.out.println(BalancedStrings.returnbalance(in.nextLine()));
    }

}

}

Popularity 3/10 Helpfulness 1/10 Language whatever

Source: codereview.stackexchange.com

Tags: whatever substring

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Contributed on Jul 22 2022

Dream FTBL

0 Answers Avg Quality 2/10

no of balanced substrings

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