LeetCode 1749. Maximum Absolute Sum of Any Subarray Solution in Java, C++, Python & More | Explanation + Code

Description

You are given an integer array nums. The absolute sum of a subarray [numsl, numsl+1, ..., numsr-1, numsr] is abs(numsl + numsl+1 + ... + numsr-1 + numsr).

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 <= 105
• -104 <= nums[i] <= 104

Solutions

Solution 1: Dynamic Programming

We define f[i] to represent the maximum value of the subarray ending with nums[i], and define g[i] to represent the minimum value of the subarray ending with nums[i]. Then the state transition equation of f[i] and g[i] is as follows:

\begin{aligned} f[i] &= max(f[i - 1], 0) + nums[i] \ g[i] &= min(g[i - 1], 0) + nums[i] \end{aligned}

The final answer is the maximum value of max(f[i], |g[i]|).

Since f[i] and g[i] are only related to f[i - 1] and g[i - 1], we can use two variables to replace the array, reducing the space complexity to O(1).

Time complexity O(n), space complexity O(1), where n is the length of the array nums.

PythonJavaC++GoTypeScriptRust

class Solution: def maxAbsoluteSum(self, nums: List[int]) -> int: f = g = 0 ans = 0 for x in nums: f = max(f, 0) + x g = min(g, 0) + x ans = max(ans, f, abs(g)) return ans(code-box)

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; } }(code-box)

class Solution { public: int maxAbsoluteSum(vector<int>& nums) { int f = 0, g = 0; int ans = 0; for (int& x : nums) { f = max(f, 0) + x; g = min(g, 0) + x; ans = max({ans, f, abs(g)}); } return ans; } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)