Question

2875. Minimum Size Subarray in Infinite Array

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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:
  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

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;
  }
};

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
}

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;
}

Solution 2

class Solution {
  public int shortestSubarray(int[] nums, int k) {
    int n = nums.length;

    int minLength = n * 2 + 1;
    int l = 0;
    int sum = 0;

    for (int r = 0; r < n * 2; r++) {
      int start = l % n;
      int end = r % n;
      sum += nums[end];

      while (sum > k && l <= r) {
        start = l % n;
        sum -= nums[start];
        l++;
      }

      if (sum == k) {
        minLength = Math.min(minLength, r - l + 1);
        start = l % n;
        sum -= nums[start];
        l++;
      }
    }

    return minLength == n * 2 + 1 ? -1 : minLength;
  }
  public int minSizeSubarray(int[] nums, int target) {
    int n = nums.length;
    int sum = 0;

    for (int num : nums) {
      sum += num;
    }
    int k = target % sum;
    int ans = target / sum * n;
    if (k == 0) {
      return ans;
    }
    int res = shortestSubarray(nums, k);
    return res == -1 ? -1 : ans + res;
  }
}