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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

• Java
• C++
• Python
• Go

• 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
  }
  

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