Maximize sum of array after k negation [Quick Select]

Time: O(N) on average; Space: O(1); easy

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].

Notes:

• 1 <= len(A) <= 10000

• 1 <= K <= 10000

• -100 <= A[i] <= 100

1. Quick select solution

import random

class Solution1(object):
    """
    Time: O(N)~O(N^2), O(N) on average.
    Space: O(1)
    """
    def largestSumAfterKNegations(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        def kthElement(nums, k, compare):
            def PartitionAroundPivot(left, right, pivot_idx, nums, compare):
                new_pivot_idx = left
                nums[pivot_idx], nums[right] = nums[right], nums[pivot_idx]
                for i in range(left, right):
                    if compare(nums[i], nums[right]):
                        nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i]
                        new_pivot_idx += 1

                nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right]
                return new_pivot_idx

            left, right = 0, len(nums) - 1
            while left <= right:
                pivot_idx = random.randint(left, right)
                new_pivot_idx = PartitionAroundPivot(left, right, pivot_idx, nums, compare)
                if new_pivot_idx == k:
                    return
                elif new_pivot_idx > k:
                    right = new_pivot_idx - 1
                else:                     # new_pivot_idx < k.
                    left = new_pivot_idx + 1

        kthElement(A, K, lambda a, b: a < b)

        remain = K

        for i in range(K):
            if A[i] < 0:
                A[i] = -A[i]
                remain -= 1

        return sum(A) - ((remain)%2) * min(A) * 2

s = Solution1()
A = [4,2,3]
K = 1
assert s.largestSumAfterKNegations(A, K) == 5
A = [3,-1,0,2]
K = 3
assert s.largestSumAfterKNegations(A, K) == 6
A = [2,-3,-1,5,-4]
K = 2
assert s.largestSumAfterKNegations(A, K) == 13

class Solution2(object):
    def largestSumAfterKNegations(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        A.sort()
        remain = K

        for i in range(K):
            if A[i] >= 0:
                break
            A[i] = -A[i]
            remain -= 1

        return sum(A) - (remain % 2) * min(A) * 2

s = Solution2()
A = [4,2,3]
K = 1
assert s.largestSumAfterKNegations(A, K) == 5
A = [3,-1,0,2]
K = 3
assert s.largestSumAfterKNegations(A, K) == 6
A = [2,-3,-1,5,-4]
K = 2
assert s.largestSumAfterKNegations(A, K) == 13

See also:

https://leetcode.com/problems/maximize-sum-of-array-after-k-negations