209. Minimum Size Subarray Sum

Problem statement:

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.

Solution:

It looks like 53. Maximum Subarray. But they are different solutions. I solve this problem by two pointers, left and right. The return value is the min length which is greater than or equal to the target. I can control the movement of these two pointers to update the min length.

Basic idea:

If sum >= target, left moves to the end and minus the value of left, meanwhile update the min length.

One terminating condition is left > right, that means we already find the min length is 1 and return. Otherwise, I find until the end.

If sum < target, right moves to the end and plus the value of right.

Since left and right move at most n steps. Time complexity is O(2 * n) ---> O(n).

class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
if(nums.empty()){
return 0;
}
int left = 0;
int right = 0;
int sum = nums[left];
int len = INT_MAX;
while(left <= right && left < nums.size() && right < nums.size()){
// if current sum is a solution
if(sum >= s){
// update the len
len = min(len, right - left + 1);
// update the sum
// minus first and move left towards end
sum -= nums[left++];
} else {
// the sum is still less than target
// move right to the end and update the sum
sum += nums[++right];
}
}
// return 0 if no answer
return len == INT_MAX ? 0 : len;
}
};