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209. Minimum Size Subarray Sum

Description

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)).

Solutions

First, we preprocess the prefix sum array $s$ of the array $nums$, where $s[i]$ represents the sum of the first $i$ elements of the array $nums$. Since all elements in the array $nums$ are positive integers, the array $s$ is also monotonically increasing. Also, we initialize the answer $ans = n + 1$, where $n$ is the length of the array $nums$.

Next, we traverse the prefix sum array $s$. For each element $s[i]$, we can find the smallest index $j$ that satisfies $s[j] \geq s[i] + target$ by binary search. If $j \leq n$, it means that there exists a subarray that satisfies the condition, and we can update the answer, i.e., $ans = min(ans, j - i)$.

Finally, if $ans \leq n$, it means that there exists a subarray that satisfies the condition, return $ans$, otherwise return $0$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

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class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        n = len(nums)
        s = list(accumulate(nums, initial=0))
        ans = n + 1
        for i, x in enumerate(s):
            j = bisect_left(s, x + target)
            if j <= n:
                ans = min(ans, j - i)
        return ans if ans <= n else 0

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class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        int ans = n + 1;
        for (int i = 0; i <= n; ++i) {
            int j = search(s, s[i] + target);
            if (j <= n) {
                ans = Math.min(ans, j - i);
            }
        }
        return ans <= n ? ans : 0;
    }

    private int search(long[] nums, long x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

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class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int n = nums.size();
        vector<long long> s(n + 1);
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        int ans = n + 1;
        for (int i = 0; i <= n; ++i) {
            int j = lower_bound(s.begin(), s.end(), s[i] + target) - s.begin();
            if (j <= n) {
                ans = min(ans, j - i);
            }
        }
        return ans <= n ? ans : 0;
    }
};

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