What is the time complexity of Minimum Size Subarray in Infinite Array?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Size Subarray in Infinite Array?

This problem uses Array, Hash Table, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Size Subarray in Infinite Array?

Insight 1: The infinite nature of the array can be handled by using the modulo operator or duplicating the original array. Duplicating the array once is sufficient because any subarray longer than the original array's length can be expressed as a multiple of the array's sum plus a remainder. Insight 2: Prefix sums are crucial for efficiently calculating subarray sums. Storing prefix sums in a hash map (or a similar data structure) allows for quick lookups to determine if a subarray with a specific sum has been encountered previously.

Which companies ask LeetCode 2875. Minimum Size Subarray in Infinite Array in interviews?

This problem is frequently asked at DE Shaw.

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Minimum Size Subarray in Infinite Array - LeetCode 2875 Solution

Minimum Size Subarray in Infinite Array - Complete Solution Guide

Minimum Size Subarray in Infinite Array is LeetCode problem 2875, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. 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

Detailed Explanation

The problem asks us to find the minimum size subarray in an infinitely repeating array 'infinite_nums' that sums up to a given 'target'. The 'infinite_nums' array is created by repeatedly concatenating the input array 'nums' to itself. We need to return the length of the shortest subarray that meets the target sum, or -1 if no such subarray exists. The input consists of an integer array 'nums' and an integer 'target'. The output is an integer representing the minimum length of the subarray. The constraints specify the size and range of values in the input array and the target.

Solution Approach

The solution uses the prefix sum technique and a hash map to efficiently find the minimum-length subarray with the desired sum. It first calculates the total sum of the 'nums' array. Then, it determines how many times 'nums' can be fully added to reach the target and calculates the remaining target value. It then iterates through a doubled version of 'nums', computing prefix sums and storing them in a hash map. For each prefix sum, it checks if there's a prefix sum difference equal to either the original target or the remainder target. The algorithm keeps track of the minimum length found so far and returns it. If no such subarray exists, it returns -1.

Step-by-Step Algorithm

Step 1: Calculate the total sum of the 'nums' array.
Step 2: Calculate 'num_wraps', the number of full copies of 'nums' that can be added to reach the target. This is 'target // total_sum'.
Step 3: Calculate 'remainder_target', the remaining target value after accounting for the full copies. This is 'target % total_sum'.
Step 4: Initialize 'ans' to infinity (or a very large number).
Step 5: Create a doubled array 'arr' by concatenating 'nums' with itself.
Step 6: Initialize a prefix sum 'prefix_sum' to 0 and a hash map 'prefix_map' to store prefix sums and their indices. Insert {0: -1} into 'prefix_map' as the base case.
Step 7: Iterate through the 'arr' array. For each element:
a. Update 'prefix_sum' by adding the current element.
b. Check if 'prefix_sum - remainder_target' exists in 'prefix_map'. If it does, calculate the length of the subarray using `i - prefix_map[prefix_sum - remainder_target]` and update 'ans' by finding the minimum between the current 'ans' and 'num_wraps * n + len_rem'.
c. Check if 'prefix_sum - target' exists in 'prefix_map'. If it does, calculate the length of the subarray using `i - prefix_map[prefix_sum - target]` and update 'ans' by finding the minimum between the current 'ans' and this length.
d. Add 'prefix_sum' and its index 'i' to 'prefix_map'.
Step 8: If 'ans' is still infinity (or the initialized very large number), return -1. Otherwise, return 'ans'.

Key Insights

Insight 1: The infinite nature of the array can be handled by using the modulo operator or duplicating the original array. Duplicating the array once is sufficient because any subarray longer than the original array's length can be expressed as a multiple of the array's sum plus a remainder.
Insight 2: Prefix sums are crucial for efficiently calculating subarray sums. Storing prefix sums in a hash map (or a similar data structure) allows for quick lookups to determine if a subarray with a specific sum has been encountered previously.
Insight 3: The solution involves two cases: a 'short' subarray found within the duplicated array and a 'long' subarray that consists of a whole number of repetitions of 'nums' plus a remainder. We need to consider both scenarios to determine the minimum length.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sliding Window, Prefix Sum.

Companies

Asked at: DE Shaw.

Minimum Size Subarray in Infinite Array | LeetCode 2875

Medium

Array Hash Table Sliding Window Prefix Sum

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minSizeSubarray(int[] nums, int target) {
        int n = nums.length;
        long total_sum = 0;
        for (int num : nums) {
            total_sum += num;
        }

        // The solution is either:
        // 1. A short subarray found within nums + nums.
        // 2. A long subarray composed of k full copies of nums + a remainder part.
        
        long num_wraps = target / total_sum;
        int remainder_target = (int)(target % total_sum);

        int ans = Integer.MAX_VALUE;
        
        // We can find the min length for both the direct target and the remainder target
        // in a single pass through the doubled array.
        long prefix_sum = 0;
        Map<Long, Integer> prefix_map = new HashMap<>();  // map: prefix_sum -> index
        prefix_map.put((long)0, -1);

        int[] arr = new int[2 * n];
        for (int i = 0; i < n; i++) {
            arr[i] = nums[i];
            arr[i + n] = nums[i];
        }
        
        for (int i = 0; i < arr.length; i++) {
            prefix_sum += arr[i];

            // Check for a subarray summing to the remainder_target.
            // This is used to construct the "long" subarray solution.
            if (prefix_map.containsKey(prefix_sum - remainder_target)) {
                int len_rem = i - prefix_map.get(prefix_sum - remainder_target);
                ans = Math.min(ans, (int)(num_wraps * n + len_rem));
            }

            // Check for a subarray summing to the target itself.
            // This is the "short" subarray solution.
            if (prefix_map.containsKey(prefix_sum - target)) {
                ans = Math.min(ans, i - prefix_map.get(prefix_sum - target));
            }
            
            prefix_map.put(prefix_sum, i);
        }
        
        return ans == Integer.MAX_VALUE ? -1 : ans;
    }
}

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Problem Info

Description

Problem 2875: Minimum Size Subarray in Infinite Array - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window Prefix Sum

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2875

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Calculate the total sum of the 'nums' array.
2
Step 2: Calculate 'num_wraps', the number of full copies of 'nums' that can be added to reach the target. This is 'target // total_sum'.
3
Step 3: Calculate 'remainder_target', the remaining target value after accounting for the full copies. This is 'target % total_sum'.
4
Step 4: Initialize 'ans' to infinity (or a very large number).
5
Step 5: Create a doubled array 'arr' by concatenating 'nums' with itself.
6
Step 6: Initialize a prefix sum 'prefix_sum' to 0 and a hash map 'prefix_map' to store prefix sums and their indices. Insert {0: -1} into 'prefix_map' as the base case.
7
Step 7: Iterate through the 'arr' array. For each element:
8
a. Update 'prefix_sum' by adding the current element.
9
b. Check if 'prefix_sum - remainder_target' exists in 'prefix_map'. If it does, calculate the length of the subarray using `i - prefix_map[prefix_sum - remainder_target]` and update 'ans' by finding the minimum between the current 'ans' and 'num_wraps * n + len_rem'.
10
c. Check if 'prefix_sum - target' exists in 'prefix_map'. If it does, calculate the length of the subarray using `i - prefix_map[prefix_sum - target]` and update 'ans' by finding the minimum between the current 'ans' and this length.
11
d. Add 'prefix_sum' and its index 'i' to 'prefix_map'.
12
Step 8: If 'ans' is still infinity (or the initialized very large number), return -1. Otherwise, return 'ans'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operations are the loop iterating through the doubled array (2n) and the hash map lookups (constant time on average). Therefore, the time complexity is O(n). The space complexity is determined by the hash map, which stores prefix sums. In the worst case, all 2n prefix sums can be unique, leading to a space complexity of O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sliding Window

A sliding window approach can be used to find the subarray with the minimum length. However, due to the infinite nature of the array, it's necessary to efficiently manage the sliding window over the repeated array. This is more complicated to implement efficiently compared to the prefix sum method.

Time: O(n^2) in the worst case if not handled efficiently over the infinite array. Optimizations might reduce it.Space: O(1) if implemented with constant extra space. However managing the window over 'infinite_nums' is complicated.

Trade-offs: The sliding window approach might be more intuitive for some, but it's generally less efficient and more difficult to adapt to the infinite array nature of the problem compared to the prefix sum method.

✨

Key Insights

1Insight 1: The infinite nature of the array can be handled by using the modulo operator or duplicating the original array. Duplicating the array once is sufficient because any subarray longer than the original array's length can be expressed as a multiple of the array's sum plus a remainder.
2Insight 2: Prefix sums are crucial for efficiently calculating subarray sums. Storing prefix sums in a hash map (or a similar data structure) allows for quick lookups to determine if a subarray with a specific sum has been encountered previously.
3Insight 3: The solution involves two cases: a 'short' subarray found within the duplicated array and a 'long' subarray that consists of a whole number of repetitions of 'nums' plus a remainder. We need to consider both scenarios to determine the minimum length.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: When the target is smaller than the smallest number in the array, there will be no solution. The algorithm correctly returns -1 because the prefix sum will never reach the target.
Edge case 2: When the target is a multiple of the sum of 'nums'. The algorithm efficiently handles this case by calculating 'num_wraps' correctly. The 'remainder_target' will be 0, and the prefix map will already contain 0 initialized to -1.
Edge case 3: When the array only contains zeros, and the target is not zero. The total sum will be zero, resulting in division by zero error if not handled. The provided solutions do not specifically handle this case, as the problem constraints do not allow zero values within the array.
Edge case 4: When the array contains negative values. The solutions will still function correctly as prefix sums can handle negative numbers. However, ensure that the initial size of the hashmap is sufficient to accomodate for the negative values for C implementation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to account for the infinite nature of the array and only searching within the original 'nums' array. Doubling the array is a simple trick to search for subarrays that wrap around.

✗Common mistake 2: Incorrectly calculating the remainder target after accounting for full repetitions of 'nums'. The correct formula is 'target % total_sum'.

✗Common mistake 3: Not handling the case where no subarray sums to the target, leading to incorrect output (e.g., not returning -1).

✗Common mistake 4: Integer overflow issues with large prefix sums when not using long/long long

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing buffer sizes in circular buffers. The algorithm can be adapted to find the smallest buffer segment that meets a certain data requirement, considering the cyclic nature of circular buffers.

🏦Real-world scenario 2: Finding patterns in recurring event logs. If event logs are treated as a repeating sequence, this algorithm can help identify the shortest sequence of events that contribute to a specific outcome (target).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Size Subarray in Infinite Array"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?