What is the time complexity of Find X Value of Array I?

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

What data structures are used to solve Find X Value of Array I?

This problem uses Array, Math, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find X Value of Array I?

Insight 1: Since k is small (up to 5), we can use dynamic programming to keep track of the counts of subarrays that result in different remainders when divided by k. Insight 2: The crucial observation is that the remainder of the product modulo k depends only on the remainders of the individual numbers modulo k. This allows us to efficiently calculate the remainders of subarrays as we extend them.

Which companies ask LeetCode 3524. Find X Value of Array I in interviews?

This problem is frequently asked at Rubrik.

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Find X Value of Array I - LeetCode 3524 Solution

Find X Value of Array I - Complete Solution Guide

Find X Value of Array I is LeetCode problem 3524, 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 an array of positive integers nums , and a positive integer k . You are allowed to perform an operation once on nums , where in each operation you can remove any non-overlapping prefix and suffix from nums such that nums remains non-empty . You need to find the x-value of nums , which is the number of ways to perform this operation so that the product of the remaining elements leaves a remainder of x when divided by k . Return an array result of size k where result[x] is the x-valu

Detailed Explanation

The problem asks us to find the number of ways to perform a specific operation on an array `nums` such that the product of the remaining elements leaves a remainder of `x` when divided by `k`. The operation involves removing a non-overlapping prefix and suffix from `nums`. We need to return an array `result` of size `k` where `result[x]` represents the count of such operations that result in a remainder of `x` when the product of the remaining subarray is divided by `k`, for each `x` from 0 to `k-1`. The key is to iterate through all possible subarrays, calculate their product modulo k, and increment the corresponding count in the `result` array. Both prefix and suffix can be empty, meaning we can keep the entire array or remove elements from the beginning or end. Non-overlapping means that the prefix and suffix cannot share any common elements.

Solution Approach

The solution uses dynamic programming to efficiently count the number of subarrays that result in a specific remainder when their product is divided by `k`. The main idea is to iterate through the input array `nums` and maintain a `dp` array of size `k`, where `dp[v]` stores the number of subarrays ending at the current index that have a product with a remainder of `v` when divided by `k`. At each iteration, we update `dp` based on the current number in `nums` and the previous values in `dp`. We also update a `result` array to accumulate the counts for each remainder. The algorithm then returns the `result` array.

Step-by-Step Algorithm

Step 1: Initialize `result` and `dp` arrays of size `k` with all elements set to 0. `result` will store the final counts for each remainder, and `dp` will store the counts of subarrays ending at the current index for each remainder.
Step 2: Iterate through the `nums` array from left to right.
Step 3: For each element `nums[i]`, calculate `num_mod = nums[i] % k` to get the remainder of the current number when divided by `k`.
Step 4: Create a new `new_dp` array of size `k` initialized with all elements set to 0. This array will hold the new counts after processing the current number.
Step 5: Update `new_dp[num_mod]` by adding 1, representing the subarray of length 1 consisting of only the current number.
Step 6: Iterate through the previous `dp` array from `v_prev = 0` to `k - 1`. If `dp[v_prev]` is greater than 0, it means there were subarrays ending at the previous index with a remainder of `v_prev` when their product was divided by `k`.
Step 7: For each such `v_prev`, calculate the new remainder `v_new = (v_prev * num_mod) % k`, which is the remainder of the product of the previous subarray and the current number.
Step 8: Update `new_dp[v_new]` by adding `dp[v_prev]`, which means we are extending the previous subarrays by the current number.
Step 9: Update the `dp` array by copying all the values from `new_dp` to `dp`
Step 10: Update `result` array. For each `v` from 0 to `k-1`, add the `dp[v]` value to the `result[v]` value.
Step 11: After iterating through all numbers in `nums`, return the `result` array.

Key Insights

Insight 1: Since k is small (up to 5), we can use dynamic programming to keep track of the counts of subarrays that result in different remainders when divided by k.
Insight 2: The crucial observation is that the remainder of the product modulo k depends only on the remainders of the individual numbers modulo k. This allows us to efficiently calculate the remainders of subarrays as we extend them.
Insight 3: The order of iteration is essential to correctly accumulating the result. We should process the numbers one by one and update a temporary array for calculating the remainders.

Complexity Analysis

Time Complexity: O(nk)

Space Complexity: O(k)

Topics

This problem involves: Array, Math, Dynamic Programming.

Companies

Asked at: Rubrik.

Find X Value of Array I | LeetCode 3524

Medium

Array Math Dynamic Programming

Time: O(nk)Space: O(k)

Solution Code

Available in:

class Solution {
    public long[] resultArray(int[] nums, int k) {
        int n = nums.length;
        long[] dp = new long[k];
        long[] result = new long[k];

        for (int i = 0; i < n; i++) {
            int num_mod = nums[i] % k;
            
            long[] new_dp = new long[k];
            
            // Subarray of length 1: [nums[i]]
            new_dp[num_mod] += 1;
            
            // Extending subarrays ending at nums[i-1]
            for (int v_prev = 0; v_prev < k; v_prev++) {
                if (dp[v_prev] > 0) {
                    int v_new = (int)(((long)v_prev * num_mod) % k);
                    new_dp[v_new] += dp[v_prev];
                }
            }
            
            dp = new_dp;
            
            for (int v = 0; v < k; v++) {
                result[v] += dp[v];
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 3524: Find X Value of Array I - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming

Companies

Rubrik

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3524

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

11 steps

1
Step 1: Initialize `result` and `dp` arrays of size `k` with all elements set to 0. `result` will store the final counts for each remainder, and `dp` will store the counts of subarrays ending at the current index for each remainder.
2
Step 2: Iterate through the `nums` array from left to right.
3
Step 3: For each element `nums[i]`, calculate `num_mod = nums[i] % k` to get the remainder of the current number when divided by `k`.
4
Step 4: Create a new `new_dp` array of size `k` initialized with all elements set to 0. This array will hold the new counts after processing the current number.
5
Step 5: Update `new_dp[num_mod]` by adding 1, representing the subarray of length 1 consisting of only the current number.
6
Step 6: Iterate through the previous `dp` array from `v_prev = 0` to `k - 1`. If `dp[v_prev]` is greater than 0, it means there were subarrays ending at the previous index with a remainder of `v_prev` when their product was divided by `k`.
7
Step 7: For each such `v_prev`, calculate the new remainder `v_new = (v_prev * num_mod) % k`, which is the remainder of the product of the previous subarray and the current number.
8
Step 8: Update `new_dp[v_new]` by adding `dp[v_prev]`, which means we are extending the previous subarrays by the current number.
9
Step 9: Update the `dp` array by copying all the values from `new_dp` to `dp`
10
Step 10: Update `result` array. For each `v` from 0 to `k-1`, add the `dp[v]` value to the `result[v]` value.
11
Step 11: After iterating through all numbers in `nums`, return the `result` array.

💡

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(nk)

💾

Space Complexity

O(k)

📝

Why?

The time complexity is determined by the nested loops in the algorithm. The space complexity is primarily determined by the auxiliary `dp`, `result` and `new_dp` arrays.

📖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

Brute-Force Approach

Iterate through all possible subarrays and calculate the product modulo k for each subarray. Count the occurrences of each remainder.

Time: O(n^2), where n is the length of nums array. Generating all subarrays takes O(n^2) time and each subarray's product modulo k takes O(n) in the worst case, assuming product calculation doesn't overflow. However, since k <= 5, the total calculation is not expensive.Space: O(k), to store the `result` array.

Trade-offs: The brute-force approach is simpler to understand but less efficient than the dynamic programming approach, especially when n is large. Dynamic programming solution is much more efficient. Brute force solution could lead to overflows if the product of the numbers exceeds the integer limits. The given dynamic programming solution avoids the product overflow problem by calculating modulo at each step.

✨

Key Insights

1Insight 1: Since k is small (up to 5), we can use dynamic programming to keep track of the counts of subarrays that result in different remainders when divided by k.
2Insight 2: The crucial observation is that the remainder of the product modulo k depends only on the remainders of the individual numbers modulo k. This allows us to efficiently calculate the remainders of subarrays as we extend them.
3Insight 3: The order of iteration is essential to correctly accumulating the result. We should process the numbers one by one and update a temporary array for calculating the remainders.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty prefix or suffix - The algorithm implicitly handles empty prefixes and suffixes by considering all possible subarrays. The initial `new_dp[num_mod] += 1` accounts for the subarray containing only the current element.
Edge case 2: `k = 1` - If k is 1, then the remainder will always be 0, so the correct answer would be number of all subarrays. Since every number modulo 1 is 0, the code still functions correctly for this case.
Edge case 3: `nums` contains a 0 (if the constraints allowed 0). If any `nums[i]` is 0, `nums[i] % k` will be 0. This case is correctly handled by the provided solution. It means all subarrays containing this zero element will lead to zero remainder in the end.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to take the modulo k at each step. This can lead to incorrect results or potential integer overflows if the product of elements becomes very large.

✗Common mistake 2: Not handling the base case correctly. The single-element subarrays are a crucial starting point for the dynamic programming process.

✗Common mistake 3: Incorrectly updating the DP table, leading to incorrect accumulation of counts. For example, not using a temporary `new_dp` array can result in overwriting values prematurely and messing up the counts.

✅

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: Cryptography - Analyzing the distribution of remainders when processing encrypted data might reveal patterns or weaknesses in the encryption algorithm.

🏦Real-world scenario 2: Data analysis - When dealing with large datasets, calculating remainders modulo a specific value can be used for data binning or hash table implementations.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

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: September 13, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find X Value of Array I"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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