LeetCode 1918: Kth Smallest Subarray Sum Solution

Master LeetCode problem 1918 (Kth Smallest Subarray Sum), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1918. Kth Smallest Subarray Sum

Medium

Array Binary Search Sliding Window

Problem Explanation

Explanation

The problem asks us to find the kth smallest subarray sum of a given array of integers. To solve this problem, we can use binary search to find the answer.

Algorithm:

We initialize the lower bound as the minimum element in the array and the upper bound as the sum of all elements in the array.
We perform a binary search on the range [lower, upper]. For each middle value, we check how many subarrays have a sum less than or equal to this middle value.
If the count of subarrays with sum less than or equal to the middle value is less than k, we update the lower bound to middle + 1.
If the count is greater than or equal to k, we update the upper bound to middle.
We continue this process until the lower bound is less than the upper bound. At this point, the lower bound will be the kth smallest subarray sum.

Time Complexity:

The time complexity of this solution is O(n * log(sum of array)), where n is the number of elements in the array.

Space Complexity:

The space complexity of this solution is O(1).

Solution Code

class Solution {
    public int kthSmallestSubarraySum(int[] nums, int k) {
        int left = Integer.MAX_VALUE, right = 0;
        for (int num : nums) {
            left = Math.min(left, num);
            right += num;
        }
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (countSubarrays(nums, mid) < k) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return left;
    }
    
    private int countSubarrays(int[] nums, int target) {
        int count = 0;
        int sum = 0;
        int left = 0;
        
        for (int right = 0; right < nums.length; right++) {
            sum += nums[right];
            while (sum > target) {
                sum -= nums[left];
                left++;
            }
            count += right - left + 1;
        }
        
        return count;
    }
}

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