1005 Maximize Sum Of Array After K Negations.py

Author: rajeevranjancom

1005 Maximize Sum Of Array After K Negations.py

@@ -0,0 +1,61 @@
+#!/usr/bin/python3
+"""
+Given an array A of integers, we must modify the array in the following way: we
+choose an i and replace A[i] with -A[i], and we repeat this process K times in
+total.  (We may choose the same index i multiple times.)
+
+Return the largest possible sum of the array after modifying it in this way.
+
+
+
+Example 1:
+
+Input: A = [4,2,3], K = 1
+Output: 5
+Explanation: Choose indices (1,) and A becomes [4,-2,3].
+Example 2:
+
+Input: A = [3,-1,0,2], K = 3
+Output: 6
+Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].
+Example 3:
+
+Input: A = [2,-3,-1,5,-4], K = 2
+Output: 13
+Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].
+
+
+Note:
+
+1 <= A.length <= 10000
+1 <= K <= 10000
+-100 <= A[i] <= 100
+"""
+from typing import List
+
+
+class Solution:
+    def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
+        """
+        revert all the negative, if positive, revert multiple times the smallest
+
+        since -100 <= A[i] <= 100, then sort can be done in linear time
+        """
+        A.sort()
+        for i in range(len(A)):
+            if K == 0:
+                break
+
+            if A[i] < 0:
+                A[i] *= -1
+                prev = A[i]
+                K -= 1
+            else:
+                if K % 2 != 0:
+                    if i - 1 >= 0 and A[i-1] < A[i]:
+                        A[i-1] *= -1
+                    else:
+                        A[i] *= -1
+                break
+
+        return sum(A)