README_EN.md

209. Minimum Size Subarray Sum

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Description

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)).

Solutions

Python3

Presum and binary search.

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        n = len(nums)
        pre_sum = [0] * (n + 1)
        for i in range(1, n + 1):
            pre_sum[i] = pre_sum[i - 1] + nums[i - 1]
        res = n + 1
        for i in range(1, n + 1):
            t = pre_sum[i - 1] + target
            left, right = 0, n
            while left < right:
                mid = (left + right) >> 1
                if pre_sum[mid] >= t:
                    right = mid
                else:
                    left = mid + 1
            if pre_sum[left] - pre_sum[i - 1] >= target:
                res = min(res, left - i + 1)
        return 0 if res == n + 1 else res

Slide window.

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        n = len(nums)
        left = right = 0
        sum, res = 0, n + 1
        while right < n:
            sum += nums[right]
            while sum >= target:
                res = min(res, right - left + 1)
                sum -= nums[left]
                left += 1
            right += 1
        return 0 if res == n + 1 else res

Java

Presum and binary search.

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        int[] preSum = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            preSum[i] = preSum[i - 1] +nums[i - 1];
        }
        int res = n + 1;
        for (int i = 1; i <= n; ++i) {
            int t = preSum[i - 1] + target;
            int left = 0, right = n;
            while (left < right) {
                int mid = (left + right) >> 1;
                if (preSum[mid] >= t) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            if (preSum[left] - preSum[i - 1] >= target) {
                res = Math.min(res, left - i + 1);
            }
        }
        return res == n + 1 ? 0 : res;
    }
}

Slide window.

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        int left = 0, right = 0;
        int sum = 0, res = n + 1;
        while (right < n) {
            sum += nums[right];
            while (sum >= target) {
                res = Math.min(res, right - left + 1);
                sum -= nums[left++];
            }
            ++right;
        }
        return res == n + 1 ? 0 : res;
    }
}

C++

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int left = 0, right;
        int sum = 0;
        int minlen = INT_MAX;

        for (right = 0; right < nums.size(); right++) {
            sum += nums[right];
            while (left <= right && sum >= target) {
                minlen = min(minlen, right - left + 1);
                sum -= nums[left++];
            }
        }

        return minlen == INT_MAX ? 0 : minlen;
    }
};

C#

public class Solution {
    public int MinSubArrayLen(int target, int[] nums) {
        int n = nums.Length;
        int left = 0, right = 0;
        int sum = 0, res = n + 1;
        while (right < n)
        {
            sum += nums[right];
            while (sum >= target)
            {
                res = Math.Min(res, right - left + 1);
                sum -= nums[left++];
            }
            ++right;
        }
        return res == n + 1 ? 0 : res;
    }
}

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