Question

1749. Maximum Absolute Sum of Any Subarray

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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 $nums$.

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

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;
  }
}

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;
  }
};

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
}

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
  }
}