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3026. Maximum Good Subarray Sum

Description

You are given an array nums of length n and a positive integer k.

A subarray of nums is called good if the absolute difference between its first and last element is exactly k, in other words, the subarray nums[i..j] is good if |nums[i] - nums[j]| == k.

Return the maximum sum of a good subarray of nums. If there are no good subarrays, return 0.

 

Example 1:

Input: nums = [1,2,3,4,5,6], k = 1
Output: 11
Explanation: The absolute difference between the first and last element must be 1 for a good subarray. All the good subarrays are: [1,2], [2,3], [3,4], [4,5], and [5,6]. The maximum subarray sum is 11 for the subarray [5,6].

Example 2:

Input: nums = [-1,3,2,4,5], k = 3
Output: 11
Explanation: The absolute difference between the first and last element must be 3 for a good subarray. All the good subarrays are: [-1,3,2], and [2,4,5]. The maximum subarray sum is 11 for the subarray [2,4,5].

Example 3:

Input: nums = [-1,-2,-3,-4], k = 2
Output: -6
Explanation: The absolute difference between the first and last element must be 2 for a good subarray. All the good subarrays are: [-1,-2,-3], and [-2,-3,-4]. The maximum subarray sum is -6 for the subarray [-1,-2,-3].

 

Constraints:

  • 2 <= nums.length <= 105
  • -109 <= nums[i] <= 109
  • 1 <= k <= 109

Solutions

Solution 1: Prefix Sum + Hash Table

We use a hash table \(p\) to record the sum \(s\) of the prefix array \(nums[0..i-1]\) for \(nums[i]\). If there are multiple identical \(nums[i]\), we only keep the smallest \(s\). Initially, we set \(p[nums[0]]\) to \(0\). In addition, we use a variable \(s\) to record the current prefix sum, initially \(s = 0\). Initialize the answer \(ans\) to \(-\infty\).

Next, we enumerate \(nums[i]\), and maintain a variable \(s\) to represent the sum of \(nums[0..i]\). If \(nums[i] - k\) is in \(p\), then we have found a good subarray, and update the answer to \(ans = \max(ans, s - p[nums[i] - k])\). Similarly, if \(nums[i] + k\) is in \(p\), then we have also found a good subarray, and update the answer to \(ans = \max(ans, s - p[nums[i] + k])\). Then, if \(i + 1 \lt n\) and \(nums[i + 1]\) is not in \(p\), or \(p[nums[i + 1]] \gt s\), we set \(p[nums[i + 1]]\) to \(s\).

Finally, if \(ans = -\infty\), then we return \(0\), otherwise return \(ans\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the array.

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class Solution:
    def maximumSubarraySum(self, nums: List[int], k: int) -> int:
        ans = -inf
        p = {nums[0]: 0}
        s, n = 0, len(nums)
        for i, x in enumerate(nums):
            s += x
            if x - k in p:
                ans = max(ans, s - p[x - k])
            if x + k in p:
                ans = max(ans, s - p[x + k])
            if i + 1 < n and (nums[i + 1] not in p or p[nums[i + 1]] > s):
                p[nums[i + 1]] = s
        return 0 if ans == -inf else ans

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class Solution {
    public long maximumSubarraySum(int[] nums, int k) {
        Map<Integer, Long> p = new HashMap<>();
        p.put(nums[0], 0L);
        long s = 0;
        int n = nums.length;
        long ans = Long.MIN_VALUE;
        for (int i = 0; i < n; ++i) {
            s += nums[i];
            if (p.containsKey(nums[i] - k)) {
                ans = Math.max(ans, s - p.get(nums[i] - k));
            }
            if (p.containsKey(nums[i] + k)) {
                ans = Math.max(ans, s - p.get(nums[i] + k));
            }
            if (i + 1 < n && (!p.containsKey(nums[i + 1]) || p.get(nums[i + 1]) > s)) {
                p.put(nums[i + 1], s);
            }
        }
        return ans == Long.MIN_VALUE ? 0 : ans;
    }
}

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class Solution {
public:
    long long maximumSubarraySum(vector<int>& nums, int k) {
        unordered_map<int, long long> p;
        p[nums[0]] = 0;
        long long s = 0;
        const int n = nums.size();
        long long ans = LONG_LONG_MIN;
        for (int i = 0;; ++i) {
            s += nums[i];
            auto it = p.find(nums[i] - k);
            if (it != p.end()) {
                ans = max(ans, s - it->second);
            }
            it = p.find(nums[i] + k);
            if (it != p.end()) {
                ans = max(ans, s - it->second);
            }
            if (i + 1 == n) {
                break;
            }
            it = p.find(nums[i + 1]);
            if (it == p.end() || it->second > s) {
                p[nums[i + 1]] = s;
            }
        }
        return ans == LONG_LONG_MIN ? 0 : ans;
    }
};

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func maximumSubarraySum(nums []int, k int) int64 {
    p := map[int]int64{nums[0]: 0}
    var s int64 = 0
    n := len(nums)
    var ans int64 = math.MinInt64
    for i, x := range nums {
        s += int64(x)
        if t, ok := p[nums[i]-k]; ok {
            ans = max(ans, s-t)
        }
        if t, ok := p[nums[i]+k]; ok {
            ans = max(ans, s-t)
        }
        if i+1 == n {
            break
        }
        if t, ok := p[nums[i+1]]; !ok || s < t {
            p[nums[i+1]] = s
        }
    }
    if ans == math.MinInt64 {
        return 0
    }
    return ans
}

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function maximumSubarraySum(nums: number[], k: number): number {
    const p: Map<number, number> = new Map();
    p.set(nums[0], 0);
    let ans: number = -Infinity;
    let s: number = 0;
    const n: number = nums.length;
    for (let i = 0; i < n; ++i) {
        s += nums[i];
        if (p.has(nums[i] - k)) {
            ans = Math.max(ans, s - p.get(nums[i] - k)!);
        }
        if (p.has(nums[i] + k)) {
            ans = Math.max(ans, s - p.get(nums[i] + k)!);
        }
        if (i + 1 < n && (!p.has(nums[i + 1]) || p.get(nums[i + 1])! > s)) {
            p.set(nums[i + 1], s);
        }
    }
    return ans === -Infinity ? 0 : ans;
}

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public class Solution {
    public long MaximumSubarraySum(int[] nums, int k) {
        Dictionary<int, long> p = new Dictionary<int, long>();
        p[nums[0]] = 0L;
        long s = 0;
        int n = nums.Length;
        long ans = long.MinValue;
        for (int i = 0; i < n; ++i) {
            s += nums[i];
            if (p.ContainsKey(nums[i] - k)) {
                ans = Math.Max(ans, s - p[nums[i] - k]);
            }
            if (p.ContainsKey(nums[i] + k)) {
                ans = Math.Max(ans, s - p[nums[i] + k]);
            }
            if (i + 1 < n && (!p.ContainsKey(nums[i + 1]) || p[nums[i + 1]] > s)) {
                p[nums[i + 1]] = s;
            }
        }
        return ans == long.MinValue ? 0 : ans;
    }
}

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