Largest Sum After K Negations - Amazon Top Interview Questions


Problem Statement :


You are given a list of integers nums and an integer k. Consider an operation where you pick an element in nums and negate it. Given that you must make exactly k operations, return the maximum resulting sum that can be obtained.

Constraints

n ≤ 100,000 where n is the length of nums
k < 2 ** 31

Example 1

Input

nums = [1, 0, -5, -3]
k = 4


Output
9

Explanation

We can negate -5 and -3 once each and 0 twice to get [1, 0, 5, 3] and its sum is 9.



Solution :



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                        Solution in C++ :

int solve(vector<int>& nums, int k) {
    int l = nums.size();
    if (l < 1) return 0;
    sort(nums.begin(), nums.end());
    int i, sum = 0;
    for (i = 0; i < l; i++) {
        if (nums[i] >= 0 || k <= 0) break;
        nums[i] *= -1;
        k--;
        sum += nums[i];
    }
    if (k % 2) {
        if (i == l)
            sum -= 2 * nums[i - 1];
        else if (i == 0)
            sum -= 2 * nums[i];
        else
            sum -= 2 * min(nums[i - 1], nums[i]);
    }
    for (; i < l; i++) sum += nums[i];
    return sum;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums, int k) {
        Arrays.sort(nums);
        int min = Integer.MAX_VALUE;
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] < 0 && k > 0) {
                nums[i] = Math.abs(nums[i]);
                k--;
            } else if (nums[i] == 0)
                k = 0;
            sum += nums[i];
            min = Math.min(nums[i], min);
        }
        if (k != 0) {
            if (k % 2 != 0) {
                sum -= min * 2;
            }
        }
        return sum;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, k):
        if not nums:
            return 0
        heap = []
        for n in nums:
            heapq.heappush(heap, n)
        while k > 0:
            minElement = heapq.heappop(heap)
            if minElement == 0:
                break
            minElement = minElement * (-1)
            heapq.heappush(heap, minElement)
            k -= 1
        return sum(heap)
                    


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