Leetcode #2875: Minimum Size Subarray in Infinite Array
In this guide, we solve Leetcode #2875 Minimum Size Subarray in Infinite Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Prefix Sum, Sliding Window
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is , where n is the length of the array . The space complexity is , where n is the length of the array .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.