# 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.

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}


• 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;
}
}