Minimum Size Subarray Sum LeetCode Solution

Problem – Minimum Size Subarray Sum

Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [numsl, numsl+1, ..., numsr-1, numsr] of which the 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)).

Minimum Size Subarray Sum LeetCode Solution in Java

    public int minSubArrayLen(int s, int[] A) {
        int i = 0, n = A.length, res = n + 1;
        for (int j = 0; j < n; ++j) {
            s -= A[j];
            while (s <= 0) {
                res = Math.min(res, j - i + 1);
                s += A[i++];
            }
        }
        return res % (n + 1);
    }

Minimum Size Subarray Sum LeetCode Solution in C++

    int minSubArrayLen(int s, vector<int>& A) {
        int i = 0, n = A.size(), res = n + 1;
        for (int j = 0; j < n; ++j) {
            s -= A[j];
            while (s <= 0) {
                res = min(res, j - i + 1);
                s += A[i++];
            }
        }
        return res % (n + 1);
    }

Minimum Size Subarray Sum LeetCode Solution in Python

    def minSubArrayLen(self, s, A):
        i, res = 0, len(A) + 1
        for j in xrange(len(A)):
            s -= A[j]
            while s <= 0:
                res = min(res, j - i + 1)
                s += A[i]
                i += 1
        return res % (len(A) + 1)

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