# 2875. Minimum Size Subarray in Infinite Array

## Description

You are given a 0-indexed array nums and an integer target.

A 0-indexed array infinite_nums is generated by infinitely appending the elements of nums to itself.

Return the length of the shortest subarray of the array infinite_nums with a sum equal to target. If there is no such subarray return -1.

Example 1:

Input: nums = [1,2,3], target = 5
Output: 2
Explanation: In this example infinite_nums = [1,2,3,1,2,3,1,2,...].
The subarray in the range [1,2], has the sum equal to target = 5 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 5.


Example 2:

Input: nums = [1,1,1,2,3], target = 4
Output: 2
Explanation: In this example infinite_nums = [1,1,1,2,3,1,1,1,2,3,1,1,...].
The subarray in the range [4,5], has the sum equal to target = 4 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 4.


Example 3:

Input: nums = [2,4,6,8], target = 3
Output: -1
Explanation: In this example infinite_nums = [2,4,6,8,2,4,6,8,...].
It can be proven that there is no subarray with sum equal to target = 3.


Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105
• 1 <= target <= 109

## Solutions

Solution 1: Prefix Sum + Hash Table

First, we calculate the sum of all elements in the array $nums$, denoted as $s$.

If $target \gt s$, we can reduce $target$ to the range $[0, s)$ by subtracting $\lfloor \frac{target}{s} \rfloor \times s$ from it. Then, the length of the subarray is $a = \lfloor \frac{target}{s} \rfloor \times n$, where $n$ is the length of the array $nums$.

Next, we need to find the shortest subarray in $nums$ whose sum equals $target$, or the shortest subarray whose prefix sum plus suffix sum equals $s - target$. We can use prefix sum and a hash table to find such subarrays.

If we find such a subarray, the final answer is $a + b$. Otherwise, the answer is $-1$.

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

• class Solution {
public int minSizeSubarray(int[] nums, int target) {
long s = Arrays.stream(nums).sum();
int n = nums.length;
int a = 0;
if (target > s) {
a = n * (target / (int) s);
target -= target / s * s;
}
if (target == s) {
return n;
}
Map<Long, Integer> pos = new HashMap<>();
pos.put(0L, -1);
long pre = 0;
int b = 1 << 30;
for (int i = 0; i < n; ++i) {
pre += nums[i];
if (pos.containsKey(pre - target)) {
b = Math.min(b, i - pos.get(pre - target));
}
if (pos.containsKey(pre - (s - target))) {
b = Math.min(b, n - (i - pos.get(pre - (s - target))));
}
pos.put(pre, i);
}
return b == 1 << 30 ? -1 : a + b;
}
}

• class Solution {
public:
int minSizeSubarray(vector<int>& nums, int target) {
long long s = accumulate(nums.begin(), nums.end(), 0LL);
int n = nums.size();
int a = 0;
if (target > s) {
a = n * (target / s);
target -= target / s * s;
}
if (target == s) {
return n;
}
unordered_map<int, int> pos{ {0, -1} };
long long pre = 0;
int b = 1 << 30;
for (int i = 0; i < n; ++i) {
pre += nums[i];
if (pos.count(pre - target)) {
b = min(b, i - pos[pre - target]);
}
if (pos.count(pre - (s - target))) {
b = min(b, n - (i - pos[pre - (s - target)]));
}
pos[pre] = i;
}
return b == 1 << 30 ? -1 : a + b;
}
};

• class Solution:
def minSizeSubarray(self, nums: List[int], target: int) -> int:
s = sum(nums)
n = len(nums)
a = 0
if target > s:
a = n * (target // s)
target -= target // s * s
if target == s:
return n
pos = {0: -1}
pre = 0
b = inf
for i, x in enumerate(nums):
pre += x
if (t := pre - target) in pos:
b = min(b, i - pos[t])
if (t := pre - (s - target)) in pos:
b = min(b, n - (i - pos[t]))
pos[pre] = i
return -1 if b == inf else a + b


• func minSizeSubarray(nums []int, target int) int {
s := 0
for _, x := range nums {
s += x
}
n := len(nums)
a := 0
if target > s {
a = n * (target / s)
target -= target / s * s
}
if target == s {
return n
}
pos := map[int]int{0: -1}
pre := 0
b := 1 << 30
for i, x := range nums {
pre += x
if j, ok := pos[pre-target]; ok {
b = min(b, i-j)
}
if j, ok := pos[pre-(s-target)]; ok {
b = min(b, n-(i-j))
}
pos[pre] = i
}
if b == 1<<30 {
return -1
}
return a + b
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

• function minSizeSubarray(nums: number[], target: number): number {
const s = nums.reduce((a, b) => a + b);
const n = nums.length;
let a = 0;
if (target > s) {
a = n * ((target / s) | 0);
target -= ((target / s) | 0) * s;
}
if (target === s) {
return n;
}
const pos: Map<number, number> = new Map();
let pre = 0;
pos.set(0, -1);
let b = Infinity;
for (let i = 0; i < n; ++i) {
pre += nums[i];
if (pos.has(pre - target)) {
b = Math.min(b, i - pos.get(pre - target)!);
}
if (pos.has(pre - (s - target))) {
b = Math.min(b, n - (i - pos.get(pre - (s - target))!));
}
pos.set(pre, i);
}
return b === Infinity ? -1 : a + b;
}