209. Minimum Size Subarray Sum

209. Minimum Size Subarray Sum (Medium)

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] <= 105

 

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)).

Companies:
Goldman Sachs, Facebook, Bloomberg, ByteDance

Related Topics:
Array, Binary Search, Sliding Window, Prefix Sum

Similar Questions:

• Minimum Window Substring (Hard)
• Maximum Size Subarray Sum Equals k (Medium)
• Maximum Length of Repeated Subarray (Medium)
• Minimum Operations to Reduce X to Zero (Medium)

Solution 1. Sliding Window

// OJ: https://leetcode.com/problems/minimum-size-subarray-sum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int sum = 0, i = 0, j = 0, N = nums.size(), ans = INT_MAX;
        while (j < N) {
            while (j < N && sum < s) sum += nums[j++];
            if (sum < s) break;
            while (i < j && sum >= s) sum -= nums[i++];
            ans = min(ans, j - i + 1);
        }
        return ans == INT_MAX ? 0 : ans;
    }
};

Solution 2. Sliding Window

// OJ: https://leetcode.com/problems/minimum-size-subarray-sum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minSubArrayLen(int target, vector<int>& A) {
        int sum = 0, N = A.size(), i = 0, j = 0, ans = INT_MAX;
        while (j < N) {
            sum += A[j++];
            while (sum >= target) {
                ans = min(ans, j - i);
                sum -= A[i++];
            }
        }
        return ans == INT_MAX ? 0 : ans;
    }
};