# CP01 Editorial

**URL:** <https://discuss.codechef.com/t/cp01-editorial/100459>\
**Category:** editorial\
**Tags:** ad-hoc, greedy\
**Created:** [April 6, 2022, 7:29pm UTC](https://discuss.codechef.com/t/cp01-editorial/100459 "2022-04-06T19:29:10Z")\
**Posts on this page:** 1\
**Page:** 1

<div class="post-metadata">

**Author:** ![saurav1717](https://www.codechef.com/sites/all/themes/abessive/images/user_default_thumb.jpg) [@saurav1717](https://discuss.codechef.com/u/saurav1717)\
**Post date:** [April 6, 2022, 7:29pm UTC](https://discuss.codechef.com/t/cp01-editorial/100459/1 "2022-04-06T19:29:10Z")

</div>

**Problem Link**

[Click here](https://www.codechef.com/SPY22022/problems/CP01)

**Difficulty level**

Easy

**Solution**

It is always optimal to create a partition as small as possible in which ‘1’ appears exactly once more than ‘0’. So if cnt1 is the number of times ‘1’ appears in s and cnt0 is the number of times ‘0’ appears in s, then our answer will be cnt1-cnt0.

**Code**

```auto
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int T;
	cin>>T;
	while(T--)
	{
		string s;
		cin>>s;
		int n=s.length();
		int cnt1=0,cnt0=0;
		for(int i=0; i<n; i++){
			if(s[i]=='1')
			 cnt1++;
			else
			 cnt0++;
		}
		int var=cnt1-cnt0;
		cout<<cnt1-cnt0<<"\n";
	}
	return 0;
}

```