README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1005.Maximize%20Sum%20Of%20Array%20After%20K%20Negations/README_EN.md	1274	Weekly Contest 127 Q1	Greedy Array Sorting

1005. Maximize Sum Of Array After K Negations

中文文档

Description

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

Solutions

Solution 1: Greedy + Counting

We observe that to maximize the sum of the array, we should try to turn the smallest negative numbers into positive numbers.

Given that the range of elements is $[-100, 100]$, we can use a hash table $\textit{cnt}$ to count the occurrences of each element in the array $\textit{nums}$. Then, starting from $-100$, we iterate through $x$. If $x$ exists in the hash table, we take $m = \min(\textit{cnt}[x], k)$ as the number of times to negate the element $x$. We then subtract $m$ from $\textit{cnt}[x]$, add $m$ to $\textit{cnt}[-x]$, and subtract $m$ from $k$. If $k$ becomes $0$, the operation is complete, and we exit the loop.

If $k$ is still odd and $\textit{cnt}[0] = 0$, we need to take the smallest positive number $x$ in $\textit{cnt}$, subtract $1$ from $\textit{cnt}[x]$, and add $1$ to $\textit{cnt}[-x]$.

Finally, we traverse the hash table $\textit{cnt}$ and sum the products of $x$ and $\textit{cnt}[x]$ to get the answer.

The time complexity is $O(n + M)$, and the space complexity is $O(M)$. Here, $n$ and $M$ are the length of the array $\textit{nums}$ and the size of the data range of $\textit{nums}$, respectively.

Python3

class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        cnt = Counter(nums)
        for x in range(-100, 0):
            if cnt[x]:
                m = min(cnt[x], k)
                cnt[x] -= m
                cnt[-x] += m
                k -= m
                if k == 0:
                    break
        if k & 1 and cnt[0] == 0:
            for x in range(1, 101):
                if cnt[x]:
                    cnt[x] -= 1
                    cnt[-x] += 1
                    break
        return sum(x * v for x, v in cnt.items())

Java

class Solution {
    public int largestSumAfterKNegations(int[] nums, int k) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : nums) {
            cnt.merge(x, 1, Integer::sum);
        }
        for (int x = -100; x < 0 && k > 0; ++x) {
            if (cnt.getOrDefault(x, 0) > 0) {
                int m = Math.min(cnt.get(x), k);
                cnt.merge(x, -m, Integer::sum);
                cnt.merge(-x, m, Integer::sum);
                k -= m;
            }
        }
        if ((k & 1) == 1 && cnt.getOrDefault(0, 0) == 0) {
            for (int x = 1; x <= 100; ++x) {
                if (cnt.getOrDefault(x, 0) > 0) {
                    cnt.merge(x, -1, Integer::sum);
                    cnt.merge(-x, 1, Integer::sum);
                    break;
                }
            }
        }
        int ans = 0;
        for (var e : cnt.entrySet()) {
            ans += e.getKey() * e.getValue();
        }
        return ans;
    }
}

C++

class Solution {
public:
    int largestSumAfterKNegations(vector<int>& nums, int k) {
        unordered_map<int, int> cnt;
        for (int& x : nums) {
            ++cnt[x];
        }
        for (int x = -100; x < 0 && k > 0; ++x) {
            if (cnt[x]) {
                int m = min(cnt[x], k);
                cnt[x] -= m;
                cnt[-x] += m;
                k -= m;
            }
        }
        if ((k & 1) && !cnt[0]) {
            for (int x = 1; x <= 100; ++x) {
                if (cnt[x]) {
                    --cnt[x];
                    ++cnt[-x];
                    break;
                }
            }
        }
        int ans = 0;
        for (auto& [x, v] : cnt) {
            ans += x * v;
        }
        return ans;
    }
};

Go

func largestSumAfterKNegations(nums []int, k int) (ans int) {
	cnt := map[int]int{}
	for _, x := range nums {
		cnt[x]++
	}
	for x := -100; x < 0 && k > 0; x++ {
		if cnt[x] > 0 {
			m := min(k, cnt[x])
			cnt[x] -= m
			cnt[-x] += m
			k -= m
		}
	}
	if k&1 == 1 && cnt[0] == 0 {
		for x := 1; x <= 100; x++ {
			if cnt[x] > 0 {
				cnt[x]--
				cnt[-x]++
				break
			}
		}
	}
	for x, v := range cnt {
		ans += x * v
	}
	return
}

TypeScript

function largestSumAfterKNegations(nums: number[], k: number): number {
    const cnt: Map<number, number> = new Map();
    for (const x of nums) {
        cnt.set(x, (cnt.get(x) || 0) + 1);
    }
    for (let x = -100; x < 0 && k > 0; ++x) {
        if (cnt.get(x)! > 0) {
            const m = Math.min(cnt.get(x) || 0, k);
            cnt.set(x, (cnt.get(x) || 0) - m);
            cnt.set(-x, (cnt.get(-x) || 0) + m);
            k -= m;
        }
    }
    if ((k & 1) === 1 && (cnt.get(0) || 0) === 0) {
        for (let x = 1; x <= 100; ++x) {
            if (cnt.get(x)! > 0) {
                cnt.set(x, (cnt.get(x) || 0) - 1);
                cnt.set(-x, (cnt.get(-x) || 0) + 1);
                break;
            }
        }
    }
    return Array.from(cnt.entries()).reduce((acc, [k, v]) => acc + k * v, 0);
}