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1107-minimum-swaps-to-group-all-1s-together/README.md

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+<p>Given a&nbsp;binary array <code>data</code>, return&nbsp;the minimum number of swaps required to group all <code>1</code>&rsquo;s present in the array together in <strong>any place</strong> in the array.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> data = [1,0,1,0,1]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> There are 3 ways to group all 1&#39;s together:
+[1,1,1,0,0] using 1 swap.
+[0,1,1,1,0] using 2 swaps.
+[0,0,1,1,1] using 1 swap.
+The minimum is 1.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> data = [0,0,0,1,0]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> Since there is only one 1 in the array, no swaps are needed.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> data = [1,0,1,0,1,0,0,1,1,0,1]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> One possible solution that uses 3 swaps is [0,0,0,0,0,1,1,1,1,1,1].
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= data.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>data[i]</code> is either <code>0</code> or <code>1</code>.</li>
+</ul>

1107-minimum-swaps-to-group-all-1s-together/solution.go

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+func minSwaps(data []int) int {
+	oneCounter, maximum, counter := 0, 0, 0
+
+	for _, i := range data {
+		oneCounter += i
+	}
+
+	for i := 0; i < oneCounter; i++ {
+		counter += data[i]
+	}
+	maximum = max(maximum, counter)
+
+	for i := oneCounter; i < len(data); i++ {
+		counter += data[i] - data[i - oneCounter]
+		maximum = max(maximum, counter)
+	}
+
+	return oneCounter - maximum
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}