Separate Black and White Balls
<p>There are <code>n</code> balls on a table, each ball has a color black or white.</p>

<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> of length <code>n</code>, where <code>1</code> and <code>0</code> represent black and white balls, respectively.</p>

<p>In each step, you can choose two adjacent balls and swap them.</p>

<p>Return <em>the <strong>minimum</strong> number of steps to group all the black balls to the right and all the white balls to the left</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0111&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> All the black balls are already grouped to the right.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n == s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>
Greedy; Two Pointers; String
class Solution {
public:
    long long minimumSteps(string s) {
        long long ans = 0;
        int cnt = 0;
        int n = s.size();
        for (int i = n - 1; i >= 0; --i) {
            if (s[i] == '1') {
                ++cnt;
                ans += n - i - cnt;
            }
        }
        return ans;
    }
};