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# 1005. Maximize Sum Of Array After K Negations

LeetCode problem link (opens new window)

Given an integer array A, you can perform the following operation K times at most: Choose any index i, and replace A[i] with -A[i]. Return the maximum possible sum of the array after modifying it in this way.

Example 1:

  • Input: A = [4,2,3], K = 1
  • Output: 5
  • Explanation: Choose index (1), then A becomes [4, -2, 3].

Example 2:

  • Input: A = [3,-1,0,2], K = 3
  • Output: 6
  • Explanation: Choose index (1, 2, 2), then A becomes [3, 1, 0, 2].

Example 3:

  • Input: A = [2,-3,-1,5,-4], K = 2
  • Output: 13
  • Explanation: Choose index (1, 4), then A becomes [2, 3, -1, 5, 4].

Constraints:

  • 1 <= A.length <= 10000
  • 1 <= K <= 10000
  • -100 <= A[i] <= 100

# Approach

The approach to this problem is straightforward. How can we maximize the sum of the array?

Using a greedy strategy, find the local optimal: converting negative numbers with larger absolute values into positives maximizes the current number, leading to a global optimal solution maximizing the entire array.

The local optimal converts to a global optimal.

If all negative numbers are converted and K is still greater than zero, we now face a sequence of positive integers. How can we perform K flips on positive integers to maximize the sum?

Once again, a greedy strategy: local optimal is achieved by flipping the smallest positive number to maximize the current sum (for instance, among {5, 3, 1}, flipping 1 to -1 results in a higher number than flipping 5 to -5), resulting in the global optimal of a maximum array sum.

Even though you might easily solve this problem without considering what a greedy algorithm is, it's essential to realize that this problem involves two separate greedy strategies!

By adopting a mindset for greedy strategies (local optimal, global optimal), it becomes easier to handle both simple and difficult greedy algorithm problems.

Important to note: A thoughtful approach is crucial for solving this problem effectively.

Thus, the steps to solve this problem are:

  • Step 1: Sort the array in descending order by absolute value. Ensure it's sorted by absolute value.
  • Step 2: Traverse from the front, flipping negative numbers to positive while decreasing K.
  • Step 3: If K remains greater than zero, repeatedly flip the smallest element to use up K.
  • Step 4: Calculate the sum.

Here is the corresponding code in C++:

class Solution {
static bool cmp(int a, int b) {
    return abs(a) > abs(b);
}
public:
    int largestSumAfterKNegations(vector<int>& A, int K) {
        sort(A.begin(), A.end(), cmp);       // Step 1
        for (int i = 0; i < A.size(); i++) { // Step 2
            if (A[i] < 0 && K > 0) {
                A[i] *= -1;
                K--;
            }
        }
        if (K % 2 == 1) A[A.size() - 1] *= -1; // Step 3
        int result = 0;
        for (int a : A) result += a;        // Step 4
        return result;
    }
};
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  • Time complexity: O(nlogn)
  • Space complexity: O(1)

# Conclusion

When confronted with simple greedy problems, you may think it's just common sense and question if it's indeed an algorithm. Remember, recognizing the greedy approach is critical!

Developing a mindset for solving problems with a greedy approach is always beneficial.

# Other Language Versions

# Java

class Solution {
    public int largestSumAfterKNegations(int[] nums, int K) {
    	// Sorting the array by absolute value in descending order.
	nums = IntStream.of(nums)
		     .boxed()
		     .sorted((o1, o2) -> Math.abs(o2) - Math.abs(o1))
		     .mapToInt(Integer::intValue).toArray();
	int len = nums.length;	    
	for (int i = 0; i < len; i++) {
	    if (nums[i] < 0 && K > 0) {
	    	nums[i] = -nums[i];
	    	K--;
	    }
	}
	// If K is greater than 0, keep flipping the smallest positive element till K is zero.

	if (K % 2 == 1) nums[len - 1] = -nums[len - 1];
	return Arrays.stream(nums).sum();

    }
}

// Second Version: Sort the array and greedily negate negative numbers into positives as much as possible, then adjust the smallest negative or positive integer according to remaining K to attain the maximal total sum.
class Solution {
    public int largestSumAfterKNegations(int[] nums, int k) {
        if (nums.length == 1) return nums[0];

        // Sort: Initially deal with negative numbers.
        Arrays.sort(nums); 

        for (int i = 0; i < nums.length && k > 0; i++) { // Greedily flips to use as much k by converting negative values to positive ones.
            if (nums[i] < 0) {
                nums[i] = -nums[i];
                k--;
            }
        }

        // Exit the loop with k > 0 || k < 0 (discussion not required if k is used up)
        if (k % 2 == 1) { // k > 0 && k is odd: Flip the smallest remaining negative number or smallest positive integer.
            Arrays.sort(nums); // Again sort the array to get the smallest remaining negative or smallest positive integer.
            nums[0] = -nums[0];
        }
        // k > 0 && k is even: Flipping does not affect as flipping twice will result in the same number: For negative: - => + => -; For positive: + => - => + 

        int sum = 0;
        for (int num : nums) { // Calculate the maximal sum
            sum += num;
        }
        return sum;
    }
}
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# Python

Greedy Approach

class Solution:
    def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
        A.sort(key=lambda x: abs(x), reverse=True)  # Step 1: Sort the array in descending order by absolute value.

        for i in range(len(A)):  # Step 2: Execute K negation operations
            if A[i] < 0 and K > 0:
                A[i] *= -1
                K -= 1

        if K % 2 == 1:  # Step 3: If K is still available, flip the smallest absolute value element.
            A[-1] *= -1

        result = sum(A)  # Step 4: Calculate the sum of array A
        return result
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# Go

func largestSumAfterKNegations(nums []int, K int) int {
	sort.Slice(nums, func(i, j int) bool {
		return math.Abs(float64(nums[i])) > math.Abs(float64(nums[j]))
	})
  
	for i := 0; i < len(nums); i++ {
		if K > 0 && nums[i] < 0 {
			nums[i] = -nums[i]
			K--
		}
	}

	if K%2 == 1 {
		nums[len(nums)-1] = -nums[len(nums)-1]
	}

	result := 0
	for i := 0; i < len(nums); i++ {
		result += nums[i]
	}
	return result
}
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# JavaScript

var largestSumAfterKNegations = function(nums, k) {
    nums.sort((a,b) => Math.abs(b) - Math.abs(a))
    
    for(let i = 0 ;i < nums.length; i++){
        if(nums[i] < 0 && k > 0){
            nums[i] = - nums[i];
            k--;
        }
    }

    // If k is still greater than 0, find the smallest number and keep flipping it.
    while( k > 0 ){
        nums[nums.length-1] = - nums[nums.length-1]
        k--;
    }

    // Use of arrow function's implicit return - if using written shorthand, remove brackets, else include return before a + b.
    return nums.reduce((a, b) => a + b)
};

// Second version (optimized: one pass)
var largestSumAfterKNegations = function(nums, k) {
    nums.sort((a, b) => Math.abs(b) - Math.abs(a)); // Sorting
    let sum = 0;
    for(let i = 0; i < nums.length; i++) {
        if(nums[i] < 0 && k-- > 0) { // Greedy approach, flip negative to positive as much as possible
            nums[i] = -nums[i];
        }
        sum += nums[i]; // Sum calculation
    }
    if(k % 2 > 0) { // Remaining k (>0 & odd): k flips result to a negative
        sum -= 2 * nums[nums.length - 1]; // Subtract twice the smallest value (added once already)
    }
    return sum;
};
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# Rust

impl Solution {
    pub fn largest_sum_after_k_negations(mut nums: Vec<i32>, mut k: i32) -> i32 {
        nums.sort_by_key(|b| std::cmp::Reverse(b.abs()));
        for v in nums.iter_mut() {
            if *v < 0 && k > 0 {
                *v *= -1;
                k -= 1;
            }
        }
        if k % 2 == 1 {
            *nums.last_mut().unwrap() *= -1;
        }
        nums.iter().sum()
    }
}
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# C

#define abs(a) (((a) > 0) ? (a) : (-(a)))

// Sum of the array
int sum(int *nums, int numsSize) {
    int sum = 0;

    int i;
    for(i = 0; i < numsSize; ++i) {
        sum += nums[i];
    }
    return sum;
}

int cmp(const void* v1, const void* v2) {
    return abs(*(int*)v2) - abs(*(int*)v1);
}

int largestSumAfterKNegations(int* nums, int numsSize, int k){
    qsort(nums, numsSize, sizeof(int), cmp);

    int i;
    for(i = 0; i < numsSize; ++i) {
        // Traverse the array, if the current element < 0, flip it, reduce k by 1
        if(nums[i] < 0 && k > 0) {
            nums[i] *= -1;
            --k;
        }
    }

    // If there are remaining k after traversing, the smallest absolute element nums[numsSize - 1] is flipped.
    if(k % 2 == 1)
        nums[numsSize - 1] *= -1;

    return sum(nums, numsSize);
}
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# TypeScript

function largestSumAfterKNegations(nums: number[], k: number): number {
    nums.sort((a, b) => Math.abs(b) - Math.abs(a));
    let curIndex: number = 0;
    const length = nums.length;
    while (curIndex < length && k > 0) {
        if (nums[curIndex] < 0) {
            nums[curIndex] *= -1;
            k--;
        }
        curIndex++;
    }
    while (k > 0) {
        nums[length - 1] *= -1;
        k--;
    }
    return nums.reduce((pre, cur) => pre + cur, 0);
};
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# Scala

object Solution {
  def largestSumAfterKNegations(nums: Array[Int], k: Int): Int = {
    var num = nums.sortWith(math.abs(_) > math.abs(_))

    var kk = k // Because k is immutable, assign it to a mutable variable
    for (i <- num.indices) {
      if (num(i) < 0 && kk > 0) {
        num(i) *= -1 // Flip
        kk -= 1
      }
    }

    // If kk is odd, flip the smallest value (since it's previously added once)
    if (kk % 2 == 1) num(num.size - 1) *= -1

    num.sum // Return the sum of numbers
  }
}
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# C#

public class Solution
{
    public int LargestSumAfterKNegations(int[] nums, int k)
    {
        int res = 0;
        Array.Sort(nums, (a, b) => Math.Abs(b) - Math.Abs(a));
        for (int i = 0; i < nums.Length; i++)
        {
            if (nums[i] < 0 && k > 0)
            {
                nums[i] *= -1;
                k--;
            }
        }
        if (k % 2 == 1) nums[nums.Length - 1] *= -1;
        foreach (var item in nums) res += item;
        return res;
    }
}
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Last updated:: 9/4/2025, 3:19:38 PM

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