Help me in solving PRIMEREVERSE problem

help
,
1

My issue

check my code

My code

#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
int t,l,k,h;
 cin>>t;
 while(t--)
 {long int n ;
     cin>>n ;
     char a[n+1];
     cin>>a;
     char b[n+1];
     cin>> b;
     int i ,j,l=0;
     if(j=='0'||j=='1')
     {k=0,h=0;
     for(i=0;a[i]!='\0';i++)
     {
     if(a[i]==j)
        { k++;}
     }
     for(i=0;b[i]!='\0';i++)
     {
     if( b[i]==j)
     {h++;}
     }
     if(k!=h)
     {l=1;}
     }
     if(l==0)
     {cout<<"yes\n";
     }else
     {cout<<"no\n";}
}
}

Problem Link: Prime Reversal Practice Coding Problem

2

@ayushbansal07
plz refer the following code for better understanding of the logic

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--)
	{
	    int n;
        cin>>n;
        string s,s1;
        cin>>s>>s1;
        int o=0,z=0;
        for(int i=0;i<n;i++)
        {
            if(s[i]=='1')
                o++;
            if(s1[i]=='1')
                o--;
            if(s[i]=='0')
                z++;
            if(s1[i]=='0')
                z--;
        }
        if(o==0&&z==0)
            cout<<"YES";
        else
            cout<<"NO";
        cout<<endl;
	}
	return 0;
}
1 Like
3

The code checks whether the two binary strings contain the same number of 0s and 1s.

Logic

For each test case:

  1. Read n, s, and s1.

  2. Count the difference in the number of 1s between the two strings using o.

    • If s[i] == '1' → o++

    • If s1[i] == '1' → o--

  3. Count the difference in the number of 0s between the two strings using z.

    • If s[i] == '0' → z++

    • If s1[i] == '0' → z--

  4. If both o == 0 and z == 0, then both strings have identical counts of 0s and 1s, so print "YES".

  5. Otherwise print "NO".

Simpler Observation

Since the strings are binary, checking only the count of 1s is sufficient:



int cnt1 = 0, cnt2 = 0;

for(int i = 0; i < n; i++) {
    if(s[i] == '1') cnt1++;
    if(s1[i] == '1') cnt2++;
}

if(cnt1 == cnt2)
    cout << "YES\n";
else
    cout << "NO\n";

Example

Input:



1
4
1100
1010

Counts:

  • First string: 2 ones, 2 zeros

  • Second string: 2 ones, 2 zeros

Output:



YES