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Formatted question description: https://leetcode.ca/all/1749.html

1749. Maximum Absolute Sum of Any Subarray

Level

Medium

Description

You are given an integer array nums. The absolute sum of a subarray [nums_l, nums_{l+1}, ..., nums_{r-1}, nums_r] is abs(nums_l + nums_{l+1} + ... + nums_{r-1} + nums_r).

Return the maximum absolute sum of any (possibly empty) subarray of nums.

Note that abs(x) is defined as follows:

  • If x is a negative integer, then abs(x) = -x.
  • If x is a non-negative integer, then abs(x) = x.

Example 1:

Input: nums = [1,-3,2,3,-4]

Output: 5

Explanation: The subarray [2,3] has absolute sum = abs(2+3) = abs(5) = 5.

Example 2:

Input: nums = [2,-5,1,-4,3,-2]

Output: 8

Explanation: The subarray [-5,1,-4] has absolute sum = abs(-5+1-4) = abs(-8) = 8.

Constraints:

  • 1 <= nums.length <= 10^5
  • -10^4 <= nums[i] <= 10^4

Solution

Calculate the maximum subarray sum and the minimum subarray sum of nums. Then calculate the absolute values of the two sums, and return the maximum value.

  • class Solution {
    public int maxAbsoluteSum(int[] nums) {
        int sumMax = nums[0], sumMin = nums[0];
        int max = Math.max(sumMax, 0), min = Math.min(sumMin, 0);
        int length = nums.length;
        for (int i = 1; i < length; i++) {
            int num = nums[i];
            sumMax = Math.max(sumMax + num, num);
            sumMin = Math.min(sumMin + num, num);
            max = Math.max(max, sumMax);
            min = Math.min(min, sumMin);
        }
        return Math.max(Math.abs(max), Math.abs(min));
    }
}

############

class Solution {
    public int maxAbsoluteSum(int[] nums) {
        int f = 0, g = 0;
        int ans = 0;
        for (int x : nums) {
            f = Math.max(f, 0) + x;
            g = Math.min(g, 0) + x;
            ans = Math.max(ans, Math.max(f, Math.abs(g)));
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/maximum-absolute-sum-of-any-subarray/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxAbsoluteSum(vector<int>& A) {
        int ans = 0, mn = 0, mx = 0, sum = 0;
        for (int n : A) {
            sum += n;
            ans = max({ ans, mx - sum, sum - mn });
            mx = max(mx, sum);
            mn = min(mn, sum);
        }
        return ans;
    }
};

  • # 1749. Maximum Absolute Sum of Any Subarray
# https://leetcode.com/problems/maximum-absolute-sum-of-any-subarray

class Solution:
    def maxAbsoluteSum(self, nums: List[int]) -> int:
        mmax = mmin = s = 0
        
        for num in nums:
            s += num
            mmax = max(mmax, s)
            mmin = min(mmin, s)

        return mmax - mmin


  • func maxAbsoluteSum(nums []int) (ans int) {
	var f, g int
	for _, x := range nums {
		f = max(f, 0) + x
		g = min(g, 0) + x
		ans = max(ans, max(f, abs(g)))
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

  • function maxAbsoluteSum(nums: number[]): number {
    let f = 0;
    let g = 0;
    let ans = 0;
    for (const x of nums) {
        f = Math.max(f, 0) + x;
        g = Math.min(g, 0) + x;
        ans = Math.max(ans, f, -g);
    }
    return ans;
}


  • impl Solution {
    pub fn max_absolute_sum(nums: Vec<i32>) -> i32 {
        let mut f = 0;
        let mut g = 0;
        let mut ans = 0;
        for x in nums {
            f = i32::max(f, 0) + x;
            g = i32::min(g, 0) + x;
            ans = i32::max(ans, f.max(-g));
        }
        ans
    }
}

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