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1918. Kth Smallest Subarray Sum πŸ”’

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

Solution 1

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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)

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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;
    }
}

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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;
    }
};

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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
}

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