209

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)).

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

Similar Questions:

• Minimum Window Substring (Hard)
• Maximum Size Subarray Sum Equals k (Hard)
• Maximum Length of Repeated Subarray (Hard)
• Minimum Operations to Reduce X to Zero (Hard)
• K Radius Subarray Averages (Hard)
• Maximum Product After K Increments (Hard)

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;
    }
};

Solution 3. Binary Search

valid(len) returns true if whether there exists a subarray of length len that has sum greater than or equal to target.

We binary search the length range [1, N] because this predicate valid(len) is monotonic. That is, there must exist an index K that for any len >= K, valid(len) is true; and for any len < K, valid(len) is false.

// OJ: https://leetcode.com/problems/minimum-size-subarray-sum
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int minSubArrayLen(int target, vector<int>& A) {
        int N = A.size(), L = 1, R = N;
        auto valid = [&](int len) {
            int sum = 0;
            for (int i = 0; i < N; ++i) {
                sum += A[i];
                if (i - len >= 0) sum -= A[i - len];
                if (sum >= target) return true;
            }
            return false;
        };
        while (L <= R) {
            int M = (L + R) / 2;
            if (valid(M)) R = M - 1;
            else L = M + 1;
        }
        return L > N ? 0 : L;
    }
};