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Separate Black and White Balls Problem and sol.
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Ghanshyam-07 committed Oct 18, 2024
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#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
long long minimumSteps(string s) {
int n = s.length();
int ones = 0, zeros = 0;
long long count = 0;
for (int i = n - 1; i >= 0; --i) {
if (s[i] == '0') {
// Count how many ones are after this zero
zeros++;
} else {
// When encountering '1', we count it to match future '0's
count += zeros;
ones++;
}
}
return count;
}
};

int main() {
Solution solution;
string s;

// Taking input string from user
cout << "Enter a binary string (consisting of 0's and 1's): ";
cin >> s;

// Calculating the minimum steps
long long result = solution.minimumSteps(s);

// Output the result
cout << "Minimum steps to rearrange the string: " << result << endl;

return 0;
}
37 changes: 37 additions & 0 deletions Beginner Level πŸ“/Separate Black and White Balls/question.md
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### Separate Black and White Balls

There are n balls on a table, each ball has a color black or white.
You are given a 0-indexed binary string s of length n, where 1 and 0 represent black and white balls, respectively.
In each step, you can choose two adjacent balls and swap them.
Return the minimum number of steps to group all the black balls to the right and all the white balls to the left.

## Example 1:

Input: s = "101"
Output: 1
Explanation: We can group all the black balls to the right in the following way:
- Swap s[0] and s[1], s = "011".
Initially, 1s are not grouped together, requiring at least 1 step to group them to the right.

## Example 2:

Input: s = "100"
Output: 2
Explanation: We can group all the black balls to the right in the following way:
- Swap s[0] and s[1], s = "010".
- Swap s[1] and s[2], s = "001".
It can be proven that the minimum number of steps needed is 2.

## Example 3:

Input: s = "0111"
Output: 0
Explanation: All the black balls are already grouped to the right.



## Constraints:

- 1 <= n == s.length <= 105
- s[i] is either '0' or '1'.

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