What is the time complexity of Build Array Where You Can Find The Maximum Exactly K Comparisons?

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

What is the space complexity of Build Array Where You Can Find The Maximum Exactly K Comparisons?

The space complexity is O(m*k).

What data structures are used to solve Build Array Where You Can Find The Maximum Exactly K Comparisons?

This problem uses Dynamic Programming, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Build Array Where You Can Find The Maximum Exactly K Comparisons?

Insight 1: Dynamic programming is suitable for this problem because the number of arrays with specific properties can be built incrementally based on smaller subarrays. Insight 2: The `search_cost` (k) and maximum value encountered so far are crucial state variables for defining subproblems.

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Build Array Where You Can Find The Maximum Exactly K Comparisons - LeetCode 1420 Solution

Q: How to solve Build Array Where You Can Find The Maximum Exactly K Comparisons?

The solution uses dynamic programming. `dp[i][c][val]` represents the number of arrays of length `i` with a maximum value of `val` and search cost equal to `c`. We build up the `dp` table from `i = 1` to `n`. A key optimization is using prefix sums to efficiently calculate the number of ways to extend the array when the maximum value increases. The prefix sum array `prefix[c][val]` stores the sum of `dp[c][1]`, `dp[c][2]`, ..., `dp[c][val]`. This allows to compute the number of transitions with `search_cost` incremented efficiently.

Q: Which companies ask LeetCode 1420. Build Array Where You Can Find The Maximum Exactly K Comparisons in interviews?

This problem is frequently asked at Amdocs, Dunzo.

Build Array Where You Can Find The Maximum Exactly K Comparisons - Complete Solution Guide

Build Array Where You Can Find The Maximum Exactly K Comparisons is LeetCode problem 1420, a Hard 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 three integers n , m and k . Consider the following algorithm to find the maximum element of an array of positive integers: You should build the array arr which has the following properties: arr has exactly n integers. 1 <= arr[i] <= m where (0 <= i < n) . After applying the mentioned algorithm to arr , the value search_cost is equal to k . Return the number of ways to build the array arr under the mentioned conditions. As the answer may grow large, the answer must be computed modu

Detailed Explanation

The problem asks us to find the number of arrays of length `n` where each element is between 1 and `m` (inclusive), such that a specific algorithm to find the maximum element of the array makes exactly `k` comparisons (search cost). The provided algorithm iterates through the array, updating the maximum and incrementing a counter (`search_cost`) each time a new maximum is found. We need to count how many such arrays exist, modulo 10^9 + 7.

Solution Approach

The solution uses dynamic programming. `dp[i][c][val]` represents the number of arrays of length `i` with a maximum value of `val` and search cost equal to `c`. We build up the `dp` table from `i = 1` to `n`. A key optimization is using prefix sums to efficiently calculate the number of ways to extend the array when the maximum value increases. The prefix sum array `prefix[c][val]` stores the sum of `dp[c][1]`, `dp[c][2]`, ..., `dp[c][val]`. This allows to compute the number of transitions with `search_cost` incremented efficiently.

Step-by-Step Algorithm

Step 1: Initialize a DP table `dp[k+1][m+1]` and prefix sum table `prefix[k+1][m+1]`. `dp[c][val]` will store the number of arrays such that the search cost is `c` and the maximum value in the array is `val`.
Step 2: Base case: For `i = 1` (arrays of length 1), `dp[1][val] = 1` for all `val` from 1 to `m`. Also, initialize the prefix sum: `prefix[1][val] = prefix[1][val-1] + dp[1][val]`.
Step 3: Iterate from `i = 2` to `n` (array lengths). For each length `i`, iterate `c` from 1 to `k` (search cost) and `val` from 1 to `m` (maximum value).
Step 4: Calculate `dp[c][val]`. It consists of two terms: 1) The number of arrays of length `i-1` with maximum `val` and cost `c`, extended by `val`, which doesn't change the search cost multiplied by the number of array elements not exceeding `val`, which is `dp[c][val]` * `val`. 2) The number of arrays of length `i-1` with max < `val` and cost `c-1`. The contribution of this is `prefix[c - 1][val - 1]`.
Step 5: Calculate `prefix[c][val] = prefix[c][val-1] + dp[c][val]`.
Step 6: After iterating through all array lengths, return `prefix[k][m]`, which represents the total number of arrays of length `n`, with elements between 1 and `m`, and a search cost equal to `k`.

Key Insights

Insight 1: Dynamic programming is suitable for this problem because the number of arrays with specific properties can be built incrementally based on smaller subarrays.
Insight 2: The `search_cost` (k) and maximum value encountered so far are crucial state variables for defining subproblems.
Insight 3: Prefix sums can optimize the calculation of the number of ways to build arrays when the new maximum is introduced.

Complexity Analysis

Time Complexity: O(n*m*k)

Space Complexity: O(m*k)

Topics

This problem involves: Dynamic Programming, Prefix Sum.

Companies

Asked at: Amdocs, Dunzo.

Build Array Where You Can Find The Maximum Exactly K Comparisons | LeetCode 1420

Hard

Dynamic Programming Prefix Sum

Time: O(n*m*k)Space: O(m*k)

Solution Code

Available in:

class Solution {
    public int numOfArrays(int n, int m, int k) {
        int MOD = 1_000_000_007;

        int[][] prev_dp = new int[k + 1][m + 1];
        int[][] prev_prefix = new int[k + 1][m + 1];

        for (int val = 1; val <= m; val++) {
            prev_dp[1][val] = 1;
            prev_prefix[1][val] = (prev_prefix[1][val - 1] + prev_dp[1][val]) % MOD;
        }

        for (int i = 2; i <= n; i++) {
            int[][] curr_dp = new int[k + 1][m + 1];
            int[][] curr_prefix = new int[k + 1][m + 1];
            for (int c = 1; c <= k; c++) {
                for (int val = 1; val <= m; val++) {
                    long term1 = (long)prev_dp[c][val] * val % MOD;
                    long term2 = prev_prefix[c - 1][val - 1];

                    curr_dp[c][val] = (int)((term1 + term2) % MOD);
                    curr_prefix[c][val] = (int)((curr_prefix[c][val - 1] + curr_dp[c][val]) % MOD);
                }
            }
            
            prev_dp = curr_dp;
            prev_prefix = curr_prefix;
        }
        
        return prev_prefix[k][m];
    }
}

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Problem Info

Description

Problem 1420: Build Array Where You Can Find The Maximum Exactly K Comparisons - Detailed explanation and solution approaches.

Topics

Dynamic Programming Prefix Sum

Companies

AmdocsDunzo

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1420

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

6 steps

1
Step 1: Initialize a DP table `dp[k+1][m+1]` and prefix sum table `prefix[k+1][m+1]`. `dp[c][val]` will store the number of arrays such that the search cost is `c` and the maximum value in the array is `val`.
2
Step 2: Base case: For `i = 1` (arrays of length 1), `dp[1][val] = 1` for all `val` from 1 to `m`. Also, initialize the prefix sum: `prefix[1][val] = prefix[1][val-1] + dp[1][val]`.
3
Step 3: Iterate from `i = 2` to `n` (array lengths). For each length `i`, iterate `c` from 1 to `k` (search cost) and `val` from 1 to `m` (maximum value).
4
Step 4: Calculate `dp[c][val]`. It consists of two terms: 1) The number of arrays of length `i-1` with maximum `val` and cost `c`, extended by `val`, which doesn't change the search cost multiplied by the number of array elements not exceeding `val`, which is `dp[c][val]` * `val`. 2) The number of arrays of length `i-1` with max < `val` and cost `c-1`. The contribution of this is `prefix[c - 1][val - 1]`.
5
Step 5: Calculate `prefix[c][val] = prefix[c][val-1] + dp[c][val]`.
6
Step 6: After iterating through all array lengths, return `prefix[k][m]`, which represents the total number of arrays of length `n`, with elements between 1 and `m`, and a search cost equal to `k`.

💡

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*m*k)

💾

Space Complexity

O(m*k)

📝

Why?

The time complexity arises directly from the three nested loops that iterate through all possible values of n, m, and k. The space complexity is determined by the size of the dp and prefix arrays, which are both proportional to m*k.

📖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

Recursion with Memoization

A recursive approach with memoization can also solve this problem. The recursive function would explore the possible values for each array element and cache the results of subproblems to avoid redundant calculations.

Time: O(n*m*k)Space: O(n*m*k) due to the memoization table (similar to the DP table) and the recursion stack.

Trade-offs: Recursion with memoization might be slightly easier to conceptualize for some, but the iterative DP solution is generally more efficient due to avoiding the overhead of function calls.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for this problem because the number of arrays with specific properties can be built incrementally based on smaller subarrays.
2Insight 2: The `search_cost` (k) and maximum value encountered so far are crucial state variables for defining subproblems.
3Insight 3: Prefix sums can optimize the calculation of the number of ways to build arrays when the new maximum is introduced.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `k > n`. In this case, the function should return 0 because the search cost cannot exceed the array length. The algorithm implicitly handles this because the dp states are never computed beyond k = n, so the final result is 0.
Edge case 2: `m = 0` or `n = 0`. If either is zero, function returns zero, due to the nature of dp calculation loop and the initial value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of ways to extend the array when the maximum value increases. Failing to use prefix sums efficiently can lead to a much higher time complexity.

✗Common mistake 2: Forgetting to apply the modulo operator (%) at each step of the calculation, which can lead to integer overflow and incorrect results.

✅

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: Algorithm optimization where you want to minimize a comparison operation (similar to the search cost) when finding the maximum value or performing a search.

🏦Real-world scenario 2: Resource allocation problems where you have a limited number of resources (k) and you need to find the number of ways to allocate them among different tasks (n), subject to certain constraints (m).

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 8, 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

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Build Array Where You Can Find The Maximum Exactly K Comparisons"?

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

🤖 Need help with this problem?