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209. Minimum Size Subarray Sum 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int minSubArrayLen(int target, 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 >= target) {
        ans = min(ans, r - l + 1);
        sum -= nums[l++];
      }
    }

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

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class Solution {
  public int minSubArrayLen(int target, 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 >= target) {
        ans = Math.min(ans, r - l + 1);
        sum -= nums[l++];
      }
    }

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

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class Solution:
  def minSubArrayLen(self, target: int, nums: list[int]) -> int:
    ans = math.inf
    summ = 0
    j = 0

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

    return 0 if ans == math.inf else ans