Minimum Size Subarray Sum LeetCode Solution | Easy Approach

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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] <= 105

Minimum Size Subarray Sum Solutions

Time: O(n)
Space: O(1)

C++

class Solution {
 public:
  int minSubArrayLen(int s, vector<int>& nums) {
    int ans = INT_MAX;
    int sum = 0;

    for (int l = 0, r = 0; r < nums.size(); ++r) {
      sum += nums[r];
      while (sum >= s) {
        ans = min(ans, r - l + 1);
        sum -= nums[l++];
      }
    }

    return ans < INT_MAX ? ans : 0;
  }
};

Java

 
class Solution {
  public int minSubArrayLen(int s, int[] nums) {
    int ans = Integer.MAX_VALUE;
    int sum = 0;

    for (int l = 0, r = 0; r < nums.length; ++r) {
      sum += nums[r];
      while (sum >= s) {
        ans = Math.min(ans, r - l + 1);
        sum -= nums[l++];
      }
    }

    return ans != Integer.MAX_VALUE ? ans : 0;
  }
}

Python

class Solution:
  def minSubArrayLen(self, s: int, nums: List[int]) -> int:
    ans = math.inf
    sum = 0
    j = 0

    for i, num in enumerate(nums):
      sum += num
      while sum >= s:
        ans = min(ans, i - j + 1)
        sum -= nums[j]
        j += 1

    return ans if ans != math.inf else 0

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