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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.
Test Cases
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 <= 10<sup>9</sup>1 <= nums.length <= 10<sup>5</sup>1 <= nums[i] <= 10<sup>4</sup>
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
class Solution {
public:
int minSubArrayLen(int target, vector<int>& nums) {
int start = 0, current = 0, end = 0, minlen = INT_MAX;
while(end < nums.size()) {
current += nums[end];
while(current >= target) {
minlen = min(minlen, end-start+1);
current -= nums[start++];
}
end++;
}
return minlen == INT_MAX ? 0: minlen;
}
};class Solution {
public int minSubArrayLen(int target, int[] nums) {
int ml = Integer.MAX_VALUE;
int start = 0, end = 0, curr = 0;
while(end < nums.length) {
curr += nums[end];
while(curr >= target) {
ml = Math.min(ml, end-start+1);
curr -= nums[start++];
}
end++;
}
return ml == Integer.MAX_VALUE ? 0 : ml;
}
}