# 1918. Kth Smallest Subarray Sum

## Description

Given an integer array nums of length n and an integer k, return the kth smallest subarray sum.

A subarray is defined as a non-empty contiguous sequence of elements in an array. A subarray sum is the sum of all elements in the subarray.

Example 1:

Input: nums = [2,1,3], k = 4
Output: 3
Explanation: The subarrays of [2,1,3] are:
- [2] with sum 2
- [1] with sum 1
- [3] with sum 3
- [2,1] with sum 3
- [1,3] with sum 4
- [2,1,3] with sum 6
Ordering the sums from smallest to largest gives 1, 2, 3, 3, 4, 6. The 4th smallest is 3.


Example 2:

Input: nums = [3,3,5,5], k = 7
Output: 10
Explanation: The subarrays of [3,3,5,5] are:
- [3] with sum 3
- [3] with sum 3
- [5] with sum 5
- [5] with sum 5
- [3,3] with sum 6
- [3,5] with sum 8
- [5,5] with sum 10
- [3,3,5], with sum 11
- [3,5,5] with sum 13
- [3,3,5,5] with sum 16
Ordering the sums from smallest to largest gives 3, 3, 5, 5, 6, 8, 10, 11, 13, 16. The 7th smallest is 10.


Constraints:

• n == nums.length
• 1 <= n <= 2 * 104
• 1 <= nums[i] <= 5 * 104
• 1 <= k <= n * (n + 1) / 2

## Solutions

• class Solution {
public int kthSmallestSubarraySum(int[] nums, int k) {
int l = 1 << 30, r = 0;
for (int x : nums) {
l = Math.min(l, x);
r += x;
}
while (l < r) {
int mid = (l + r) >> 1;
if (f(nums, mid) >= k) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}

private int f(int[] nums, int s) {
int t = 0, j = 0;
int cnt = 0;
for (int i = 0; i < nums.length; ++i) {
t += nums[i];
while (t > s) {
t -= nums[j++];
}
cnt += i - j + 1;
}
return cnt;
}
}

• class Solution {
public:
int kthSmallestSubarraySum(vector<int>& nums, int k) {
int l = 1 << 30, r = 0;
for (int& x : nums) {
l = min(l, x);
r += x;
}
auto f = [&](int s) {
int cnt = 0, t = 0;
for (int i = 0, j = 0; i < nums.size(); ++i) {
t += nums[i];
while (t > s) {
t -= nums[j++];
}
cnt += i - j + 1;
}
return cnt;
};
while (l < r) {
int mid = (l + r) >> 1;
if (f(mid) >= k) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
};

• class Solution:
def kthSmallestSubarraySum(self, nums: List[int], k: int) -> int:
def f(s):
t = j = 0
cnt = 0
for i, x in enumerate(nums):
t += x
while t > s:
t -= nums[j]
j += 1
cnt += i - j + 1
return cnt >= k

l, r = min(nums), sum(nums)
return l + bisect_left(range(l, r + 1), True, key=f)


• func kthSmallestSubarraySum(nums []int, k int) int {
l, r := 1<<30, 0
for _, x := range nums {
l = min(l, x)
r += x
}
f := func(s int) (cnt int) {
t := 0
for i, j := 0, 0; i < len(nums); i++ {
t += nums[i]
for t > s {
t -= nums[j]
j++
}
cnt += i - j + 1
}
return
}
for l < r {
mid := (l + r) >> 1
if f(mid) >= k {
r = mid
} else {
l = mid + 1
}
}
return l
}