Minimum Size Subarray Sum


Problem Statement :


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)).


Solution :



title-img 


                            Solution in C :

int minSubArrayLen(int target, int* nums, int numsSize){
    
    int i = 0;
    int j = 0;
    int a = INT_MAX;
    int sum = 0;
    
    while (i < numsSize) {
        sum += nums[i];
        if (sum >= target) {
            if (i-j+1 < a)
                a = i-j+1;
            sum -= nums[j];
            sum -= nums[i];
            j++;
        } else
            i++;
    }
    return a == INT_MAX ? 0 : a;
}
                        


                        Solution in C++ :

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

        for(right = 0; right < nums.size(); right++) {
            sumOfCurrentWindow += nums[right];

            while (sumOfCurrentWindow >= target) {
                res = min(res, right - left + 1);
                sumOfCurrentWindow -= nums[left];
                left++;
            }
        }

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


                        Solution in Java :

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int left = 0, right = 0, sumOfCurrentWindow = 0;
        int res = Integer.MAX_VALUE;

        for(right = 0; right < nums.length; right++) {
            sumOfCurrentWindow += nums[right];

            while (sumOfCurrentWindow >= target) {
                res = Math.min(res, right - left + 1);
                sumOfCurrentWindow -= nums[left++];
            }
        }

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


                        Solution in Python : 
                            
class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        s,e,sm = 0,0,0
        ans = math.inf
        n = len(nums)
        for e in range(n):
            sm += nums[e]
            while sm>=target:
                ans = min(ans,e-s+1)
                sm -= nums[s]
                s += 1
        return ans if ans!=math.inf else 0


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