# 209. 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)).

## Solutions

Solution 1: Prefix Sum + Binary Search

First, we preprocess the prefix sum array $s$ of the array $nums$, where $s[i]$ represents the sum of the first $i$ elements of the array $nums$. Since all elements in the array $nums$ are positive integers, the array $s$ is also monotonically increasing. Also, we initialize the answer $ans = n + 1$, where $n$ is the length of the array $nums$.

Next, we traverse the prefix sum array $s$. For each element $s[i]$, we can find the smallest index $j$ that satisfies $s[j] \geq s[i] + target$ by binary search. If $j \leq n$, it means that there exists a subarray that satisfies the condition, and we can update the answer, i.e., $ans = min(ans, j - i)$.

Finally, if $ans \leq n$, it means that there exists a subarray that satisfies the condition, return $ans$, otherwise return $0$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

Solution 2: Two Pointers

We can use two pointers $j$ and $i$ to maintain a window, where the sum of all elements in the window is less than $target$. Initially, $j = 0$, and the answer $ans = n + 1$, where $n$ is the length of the array $nums$.

Next, the pointer $i$ starts to move to the right from $0$, moving one step each time. We add the element corresponding to the pointer $i$ to the window and update the sum of the elements in the window. If the sum of the elements in the window is greater than or equal to $target$, it means that the current subarray satisfies the condition, and we can update the answer, i.e., $ans = \min(ans, i - j + 1)$. Then we continuously remove the element $nums[j]$ from the window until the sum of the elements in the window is less than $target$, and then repeat the above process.

Finally, if $ans \leq n$, it means that there exists a subarray that satisfies the condition, return $ans$, otherwise return $0$.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public int minSubArrayLen(int target, int[] nums) {
int n = nums.length;
long s = 0;
int ans = n + 1;
for (int i = 0, j = 0; i < n; ++i) {
s += nums[i];
while (j < n && s >= target) {
ans = Math.min(ans, i - j + 1);
s -= nums[j++];
}
}
return ans <= n ? ans : 0;
}
}

• class Solution {
public:
int minSubArrayLen(int target, vector<int>& nums) {
int n = nums.size();
long long s = 0;
int ans = n + 1;
for (int i = 0, j = 0; i < n; ++i) {
s += nums[i];
while (j < n && s >= target) {
ans = min(ans, i - j + 1);
s -= nums[j++];
}
}
return ans == n + 1 ? 0 : ans;
}
};

• class Solution:
def minSubArrayLen(self, target: int, nums: List[int]) -> int:
n = len(nums)
ans = n + 1
s = j = 0
for i, x in enumerate(nums):
s += x
while j < n and s >= target:
ans = min(ans, i - j + 1)
s -= nums[j]
j += 1
return ans if ans <= n else 0


• func minSubArrayLen(target int, nums []int) int {
n := len(nums)
s := 0
ans := n + 1
for i, j := 0, 0; i < n; i++ {
s += nums[i]
for s >= target {
ans = min(ans, i-j+1)
s -= nums[j]
j++
}
}
if ans == n+1 {
return 0
}
return ans
}

• function minSubArrayLen(target: number, nums: number[]): number {
const n = nums.length;
let s = 0;
let ans = n + 1;
for (let i = 0, j = 0; i < n; ++i) {
s += nums[i];
while (s >= target) {
ans = Math.min(ans, i - j + 1);
s -= nums[j++];
}
}
return ans === n + 1 ? 0 : ans;
}


• public class Solution {
public int MinSubArrayLen(int target, int[] nums) {
int n = nums.Length;
long s = 0;
int ans = n + 1;
for (int i = 0, j = 0; i < n; ++i) {
s += nums[i];
while (s >= target) {
ans = Math.Min(ans, i - j + 1);
s -= nums[j++];
}
}
return ans == n + 1 ? 0 : ans;
}
}

• impl Solution {
pub fn min_sub_array_len(target: i32, nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut res = n + 1;
let mut sum = 0;
let mut i = 0;
for j in 0..n {
sum += nums[j];

while sum >= target {
res = res.min(j - i + 1);
sum -= nums[i];
i += 1;
}
}
if res == n + 1 {
return 0;
}
res as i32
}
}