Question Definition
Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.
For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.
More practice:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
Java Solution(O(n))
public int minSubArrayLen(int s, int[] nums) {
int min = nums.length + 1;
int sum = 0;
int start = 0;
for(int i = 0; i < nums.length; i++){
sum += nums[i];
while(sum - nums[start] >= s)
sum -= nums[start++];
if(sum >= s && min > i - start)
min = i - start + 1;
}
return min == nums.length + 1 ? 0 : min;
}
Java Solution(O(nlogn))
public int minSubArrayLen(int s, int[] nums) {
int[] sum = new int[nums.length + 1];
for(int i = 1; i <= nums.length; i++)
sum[i] = sum[i - 1] + nums[i - 1];
int min = Integer.MAX_VALUE;
for(int i = 0; i < nums.length; i++){
int left = i + 1, right = nums.length, t = sum[i] + s;
while (left <= right) {
int mid = left + (right - left) / 2;
if (sum[mid] < t) left = mid + 1;
else right = mid - 1;
}
if (left == nums.length + 1) break;
min = Math.min(min, left - i);
}
return min == Integer.MAX_VALUE ? 0 : min;
}
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