Sync LeetCode submission Runtime - 103 ms (16.82%), Memory - 21.7 MB (88.86%)

Author: jimit105

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

@@ -0,0 +1,38 @@
+<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.py

@@ -0,0 +1,27 @@
+# Approach 1: Sliding Window with Two Pointers
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def minSwaps(self, data: List[int]) -> int:
+        ones = sum(data)
+        count_ones = max_ones = 0
+        left = right = 0
+
+        while right < len(data):
+            # updating the number of 1's by adding the new element
+            count_ones += data[right]
+            right += 1
+
+            # maintain the length of the window to ones
+            if right - left > ones:
+                # Remove oldest element
+                count_ones -= data[left]
+                left += 1
+
+            max_ones = max(max_ones, count_ones)
+
+        return ones - max_ones
+
+