https://leetcode.com/problems/minimum-size-subarray-sum/description/
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)).
- code brute force, TLE
class Solution {
public int minSubArrayLen(int target, int[] nums) {
if (Arrays.stream(nums).sum() < target) return 0;
for (int len = 1; len < nums.length + 1; len++){
if (subArrEnough(nums, len, target)) return len;
}
return 0;
}
public boolean subArrEnough(int[] arr, int len, int target){
int sum = 0;
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < len; i++) {
q.offerLast(arr[i]);
sum += arr[i];
}
if (sum >= target) return true;
for (int i = len; i < arr.length; i++){
sum -= (q.pollFirst());
q.offerLast(arr[i]);
sum += arr[i];
if (sum >= target) return true;
}
return false;
}
}
- code #slidingwindow
class Solution {
public int minSubArrayLen(int target, int[] nums) {
int res = nums.length + 1;
int sum = 0;
int left = 0;
for (int i = 0; i< nums.length; i++){
sum += nums[i];
while (sum >= target){
res = Math.min(res, i - left + 1);
sum -= nums[left++];
}
}
return res == nums.length + 1 ? 0 : res;
}
}