209. Minimum Size Subarray Sum¶

Difficulty: Medium

209. Minimum Size Subarray Sum

Medium

Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

Constraints:

1 <= target <= 10⁹
1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁴

Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).

Solution¶

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        int mini = Integer.MAX_VALUE;
        int i = 0, j = 0, currSum = 0;
        while (i < n) {
            while (j < n && currSum < target) {
                currSum += nums[j++]; 
            }
            if (currSum >= target) {
                mini = Math.min(mini, j - i);
            }
            currSum -= nums[i];
            i++;
        }
        if (mini == Integer.MAX_VALUE) 
            return 0;
        return mini;
    }
    private boolean ok(int mid, int arr[], int target) {
        int n = arr.length;
        int currSum = 0;
        for (int i = 0; i < mid; i++) {
            currSum += arr[i]; 
        }
        if (currSum >= target) 
            return true;
        int start = 0;
        for (int i = mid; i < n; i++) {
            currSum -= arr[start++];
            currSum += arr[i];
            if (currSum >= target) 
                return true;
        }
        return false;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here