1004 Max Consecutive Ones III.py

Author: rajeevranjancom

1004 Max Consecutive Ones III.py

@@ -0,0 +1,65 @@
+#!/usr/bin/python3
+"""
+Given an array A of 0s and 1s, we may change up to K values from 0 to 1.
+
+Return the length of the longest (contiguous) subarray that contains only 1s.
+
+
+
+Example 1:
+
+Input: A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
+Output: 6
+Explanation:
+[1,1,1,0,0,1,1,1,1,1,1]
+Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.
+Example 2:
+
+Input: A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
+Output: 10
+Explanation:
+[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
+Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.
+
+
+Note:
+1 <= A.length <= 20000
+0 <= K <= A.length
+A[i] is 0 or 1
+"""
+from typing import List
+
+
+class Solution:
+    def longestOnes(self, A: List[int], K: int) -> int:
+        """
+        len(gap)
+        But there is multiple gap need to fill, and which gaps to fill is
+        undecided. Greedy? No. DP?
+
+        Sliding Window: Find the longest subarray with at most K zeros.
+        """
+        i, j = 0, 0
+        cnt_0 = 0
+        n = len(A)
+        ret = 0
+        while i < n and j < n:
+            while j < n:
+                if A[j] == 0 and cnt_0 < K:
+                    j += 1
+                    cnt_0 += 1
+                elif A[j] == 1:
+                    j += 1
+                else:
+                    break
+
+            ret = max(ret, j - i)
+            if A[i] == 0:
+                cnt_0 -= 1
+            i += 1
+
+        return ret
+
+
+if __name__ == "__main__":
+    assert Solution().longestOnes([1,1,1,0,0,0,1,1,1,1,0], 2) == 6