Minimum Size Subarray Sum

Description

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.

Example: 

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

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(javascript)

/** 暴力解法 O(n^2)
 * @param {number} s
 * @param {number[]} nums
 * @return {number}
 */
// const minSubArrayLen = function (s, nums = []) {
//   let min = Infinity
//   for (let i = 0; i < nums.length; i++) {
//     let sum = 0
//     for (let j = i; j < nums.length; j++) {
//       sum += nums[j]
//       if (sum >= s) {
//         min = Math.min(min, j - i + 1)
//         break
//       }
//     }
//   }
//   return min === Infinity ? 0 : min
// }

// O(n) two pointers
const minSubArrayLen = function (s, nums) {
  let sum = 0
  let min = Infinity
  let left = 0
  for (let i = 0; i < nums.length; i++) {
    sum += nums[i]
    while (sum >= s) {
      min = Math.min(i - left + 1, min)
      sum -= nums[left]
      left += 1
    }
  }
  return min === Infinity ? 0 : min
}