In this guide, you will get 209. Minimum Size Subarray Sum LeetCode Solution with the best time and space complexity. The solution to Minimum Size Subarray Sum problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
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)).
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
209. Minimum Size Subarray Sum LeetCode Solution in C++
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;
}
}; /* code provided by PROGIEZ */
209. Minimum Size Subarray Sum LeetCode Solution in Java
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;
}
} // code provided by PROGIEZ
209. Minimum Size Subarray Sum LeetCode Solution in Python
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 # code by PROGIEZ
