Algorithm

Problem Name: 1005. Maximize Sum Of Array After K Negations

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

Given an integer array nums and an integer k, modify the array in the following way:

  • choose an index i and replace nums[i] with -nums[i].

You should apply this process exactly k times. You 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: nums = [4,2,3], k = 1
Output: 5
Explanation: Choose index 1 and nums becomes [4,-2,3].

Example 2:

Input: nums = [3,-1,0,2], k = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and nums becomes [3,1,0,2].

Example 3:

Input: nums = [2,-3,-1,5,-4], k = 2
Output: 13
Explanation: Choose indices (1, 4) and nums becomes [2,3,-1,5,4].

 

Constraints:

  • 1 <= nums.length <= 104
  • -100 <= nums[i] <= 100
  • 1 <= k <= 104

Code Examples

#1 Code Example with Java Programming

Code - Java Programming


class Solution {
  public int largestSumAfterKNegations(int[] nums, int k) {
    PriorityQueue pq = new PriorityQueue<>();
    int sum = 0;
    for (int num : nums) {
      pq.add(num);
      sum += num;
    }
    while (k-- > 0) {
      sum -= pq.peek();
      sum += -1 * pq.peek();
      pq.add(-1 * pq.poll());
    }
    return sum;
  }
}
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Input

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nums = [4,2,3], k = 1

Output

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5

#2 Code Example with Javascript Programming

Code - Javascript Programming


const largestSumAfterKNegations = function(A, K) {
  if (A.length === 0) return 0;
  A.sort((a, b) => a - b);
  let res = 0;
  let posIdx;
  for (let i = 0, num = 0; i  <  A.length; i++) {
    if (num < K) {
      if (A[i] < 0) {
        A[i] = -A[i];
      } else {
        if (posIdx == null) {
          posIdx = Math.abs(A[i]) - Math.abs(A[i - 1]) > 0 ? i - 1 : i;
        }
        A[posIdx] = -A[posIdx];
      }
      num++;
    }
  }
  res = A.reduce((ac, el) => ac + el, 0);
  return res;
};
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Input

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nums = [4,2,3], k = 1

Output

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5

#3 Code Example with Python Programming

Code - Python Programming


class Solution:
    def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
        heapq.heapify(A)
        for _ in range(K):
            val = heapq.heappop(A)
            heapq.heappush(A, -val)
        return sum(A)
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Input

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nums = [3,-1,0,2], k = 3

Output

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6

#4 Code Example with C# Programming

Code - C# Programming


using System.Collections.Generic;
using System.Linq;

namespace LeetCode
{
    public class _1005_MaximizeSumOfArrayAfterKNegations
    {
        public int LargestSumAfterKNegations(int[] A, int K)
        {
            var list = new SortedList < int, int>(A.Length);
            var sum = 0;
            foreach (var num in A)
            {
                if (!list.ContainsKey(num))
                    list.Add(num, 1);
                else
                    list[num]++;
                sum += num;
            }

            for (int i = 0; i  <  K; i++)
            {
                var num = list.First().Key;
                if (num == 0) break;

                sum -= num * 2;
                list[num]--;
                if (list[num] == 0)
                    list.Remove(num);

                if (!list.ContainsKey(-num))
                    list.Add(-num, 1);
                else
                    list[-num]++;
            }

            return sum;
        }
    }
}
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Input

x
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nums = [3,-1,0,2], k = 3

Output

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6
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