Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 112.2 MB (47.28%)

Author: koralkulacoglu

LeetCode/1037-minimum-number-of-k-consecutive-bit-flips/README.md

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+<p>You are given a binary array <code>nums</code> and an integer <code>k</code>.</p>
+
+<p>A <strong>k-bit flip</strong> is choosing a <strong>subarray</strong> of length <code>k</code> from <code>nums</code> and simultaneously changing every <code>0</code> in the subarray to <code>1</code>, and every <code>1</code> in the subarray to <code>0</code>.</p>
+
+<p>Return <em>the minimum number of <strong>k-bit flips</strong> required so that there is no </em><code>0</code><em> in the array</em>. If it is not possible, return <code>-1</code>.</p>
+
+<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,1,0], k = 1
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> Flip nums[0], then flip nums[2].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,1,0], k = 2
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,0,0,1,0,1,1,0], k = 3
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> 
+Flip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0]
+Flip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0]
+Flip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= k &lt;= nums.length</code></li>
+</ul>

LeetCode/1037-minimum-number-of-k-consecutive-bit-flips/solution.cpp

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+class Solution {
+public:
+    int minKBitFlips(vector<int>& nums, int k) {
+        int n = nums.size();
+        int ans = 0;
+        int flips = 0;
+        queue<int> indices;
+        for (int i=0; i<n; i++) {
+            while (!indices.empty() && i > indices.front()) {
+                indices.pop();
+                flips--;
+            }
+
+            if ((nums[i] & 1) == (flips & 1)) {
+                if (i + k > n) return -1;
+                
+                indices.push(i + k - 1);
+                flips++;
+                ans++;
+            }
+        }
+
+        return ans;
+    }
+};