What is the time complexity of Maximize Score After N Operations?

The optimal solution has a time complexity of O(n^2 * 2^n).

What is the space complexity of Maximize Score After N Operations?

The space complexity is O(2^n).

What data structures are used to solve Maximize Score After N Operations?

This problem uses Array, Math, Dynamic Programming, Backtracking, Bit Manipulation, Number Theory, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximize Score After N Operations?

Insight 1: Since 'n' is small (<= 7), we can use a bitmask to represent which numbers have been used. Insight 2: Dynamic programming with memoization is essential to avoid recomputing results for overlapping subproblems. The state is defined by the bitmask representing the used numbers.

Which companies ask LeetCode 1799. Maximize Score After N Operations in interviews?

This problem is frequently asked at Sprinklr.

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Maximize Score After N Operations - LeetCode 1799 Solution

Q: How to solve Maximize Score After N Operations?

The solution utilizes a dynamic programming approach with bitmasking to represent the state of the array. Each bit in the mask corresponds to an element in the array; if the bit is set to 1, the element is available; otherwise, it's used. The algorithm recursively explores all possible pairings of available numbers, calculates the score for the current operation, and adds it to the maximum score obtained from the remaining subproblem. Memoization is used to store the results of previously computed subproblems, significantly reducing the time complexity.

Q: Which companies ask LeetCode 1799. Maximize Score After N Operations in interviews?

This problem is frequently asked at Sprinklr.

Maximize Score After N Operations - Complete Solution Guide

Maximize Score After N Operations is LeetCode problem 1799, 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 nums , an array of positive integers of size 2 * n . You must perform n operations on this array. In the i th operation (1-indexed) , you will: Choose two elements, x and y . Receive a score of i * gcd(x, y) . Remove x and y from nums . Return the maximum score you can receive after performing n operations. The function gcd(x, y) is the greatest common divisor of x and y . Example 1: Input: nums = [1,2] Output: 1 Explanation: The optimal choice of operations is: (1 * gcd(1, 2)) = 1

Detailed Explanation

The problem requires you to maximize the total score obtained by performing 'n' operations on an array 'nums' of size 2*n. In each operation, you select two numbers, x and y, from the array, calculate the score as i * gcd(x, y) (where 'i' is the operation number, starting from 1), and remove x and y from the array. The goal is to find the optimal sequence of operations that yields the highest possible total score after 'n' operations.

Solution Approach

The solution utilizes a dynamic programming approach with bitmasking to represent the state of the array. Each bit in the mask corresponds to an element in the array; if the bit is set to 1, the element is available; otherwise, it's used. The algorithm recursively explores all possible pairings of available numbers, calculates the score for the current operation, and adds it to the maximum score obtained from the remaining subproblem. Memoization is used to store the results of previously computed subproblems, significantly reducing the time complexity.

Step-by-Step Algorithm

Step 1: Initialize a memoization table (memo) to store the maximum scores for each bitmask configuration. This table will prevent redundant computations.
Step 2: Define a recursive function, solve(mask), that takes a bitmask representing the available numbers as input.
Step 3: Base Case: If the mask is 0 (all numbers are used), return 0 (no score).
Step 4: Check if the result for the current mask is already stored in the memo table. If so, return the stored value.
Step 5: Determine the current operation number ('op') based on the number of used elements. 'op' is equal to (number of used elements / 2) + 1.
Step 6: Iterate through all possible pairs of available numbers (x and y) in the mask.
Step 7: For each pair (x, y), calculate the score for the current operation: op * gcd(x, y).
Step 8: Create a new mask (new_mask) by removing x and y from the current mask.
Step 9: Recursively call solve(new_mask) to obtain the maximum score for the remaining subproblem.
Step 10: Calculate the current total score as the sum of the current operation score and the result of solve(new_mask).
Step 11: Update the maximum score (res) found so far.
Step 12: Store the maximum score (res) in the memo table for the current mask.
Step 13: Return the maximum score (res).

Key Insights

Insight 1: Since 'n' is small (<= 7), we can use a bitmask to represent which numbers have been used.
Insight 2: Dynamic programming with memoization is essential to avoid recomputing results for overlapping subproblems. The state is defined by the bitmask representing the used numbers.
Insight 3: The order in which the operations are performed affects the score. We must explore all possible pairings within the available numbers at each operation.

Complexity Analysis

Time Complexity: O(n^2 * 2^n)

Space Complexity: O(2^n)

Topics

This problem involves: Array, Math, Dynamic Programming, Backtracking, Bit Manipulation, Number Theory, Bitmask.

Companies

Asked at: Sprinklr.

Maximize Score After N Operations | LeetCode 1799

Hard

Array Math Dynamic Programming Backtracking Bit Manipulation Number Theory Bitmask

Time: O(n^2 * 2^n)Space: O(2^n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    private int[] nums;
    private int m;
    private Integer[] memo;

    public int maxScore(int[] nums) {
        this.nums = nums;
        this.m = nums.length;
        this.memo = new Integer[1 << m];
        Arrays.fill(memo, null);

        return solve((1 << m) - 1);
    }

    private int solve(int mask) {
        if (mask == 0) {
            return 0;
        }

        if (memo[mask] != null) {
            return memo[mask];
        }

        int numUsed = m - Integer.bitCount(mask);
        int op = numUsed / 2 + 1;

        int res = 0;

        int p1Idx = -1;
        for (int i = 0; i < m; i++) {
            if ((mask >> i) % 2 == 1) {
                p1Idx = i;
                break;
            }
        }

        for (int p2Idx = p1Idx + 1; p2Idx < m; p2Idx++) {
            if ((mask >> p2Idx) % 2 == 1) {
                int newMask = mask ^ (1 << p1Idx) ^ (1 << p2Idx);

                int currentScore = op * gcd(nums[p1Idx], nums[p2Idx]) + solve(newMask);
                res = Math.max(res, currentScore);
            }
        }

        memo[mask] = res;
        return res;
    }

    private int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}

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Problem Info

Description

Problem 1799: Maximize Score After N Operations - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Backtracking Bit Manipulation Number Theory Bitmask

Companies

Sprinklr

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1799

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

13 steps

1
Step 1: Initialize a memoization table (memo) to store the maximum scores for each bitmask configuration. This table will prevent redundant computations.
2
Step 2: Define a recursive function, solve(mask), that takes a bitmask representing the available numbers as input.
3
Step 3: Base Case: If the mask is 0 (all numbers are used), return 0 (no score).
4
Step 4: Check if the result for the current mask is already stored in the memo table. If so, return the stored value.
5
Step 5: Determine the current operation number ('op') based on the number of used elements. 'op' is equal to (number of used elements / 2) + 1.
6
Step 6: Iterate through all possible pairs of available numbers (x and y) in the mask.
7
Step 7: For each pair (x, y), calculate the score for the current operation: op * gcd(x, y).
8
Step 8: Create a new mask (new_mask) by removing x and y from the current mask.
9
Step 9: Recursively call solve(new_mask) to obtain the maximum score for the remaining subproblem.
10
Step 10: Calculate the current total score as the sum of the current operation score and the result of solve(new_mask).
11
Step 11: Update the maximum score (res) found so far.
12
Step 12: Store the maximum score (res) in the memo table for the current mask.
13
Step 13: Return the maximum score (res).

💡

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^2 * 2^n)

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity arises from the nested loops within the recursive function, where we examine all possible pairs for each state (bitmask). The space complexity is dominated by the memoization table, which stores the results for each unique bitmask state. Since the input array 'nums' has a length of 2*n, there are 2^(2n) possible subsets of used numbers.

📖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

Backtracking without Memoization

A naive approach using only recursion to explore all possible combinations without storing intermediate results. This approach would be very slow.

Time: O((2n)!) which is factorial and much slower.Space: O(n) due to recursion depth.

Trade-offs: The backtracking approach is simple to understand but quickly becomes infeasible for larger values of 'n'. The DP approach offers a significant performance improvement at the cost of increased space complexity.

✨

Key Insights

1Insight 1: Since 'n' is small (<= 7), we can use a bitmask to represent which numbers have been used.
2Insight 2: Dynamic programming with memoization is essential to avoid recomputing results for overlapping subproblems. The state is defined by the bitmask representing the used numbers.
3Insight 3: The order in which the operations are performed affects the score. We must explore all possible pairings within the available numbers at each operation.

📚

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: Input array with only two elements (n=1). The algorithm correctly calculates the score as 1 * gcd(nums[0], nums[1]).
Edge case 2: All numbers in the array are relatively prime. The algorithm still explores all possible pairings and selects the combination that maximizes the overall score, even though the gcd values will be small.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to memoize the results of the subproblems, leading to exponential time complexity.

✗Common mistake 2: Incorrectly calculating the operation number, which directly affects the score calculation.

✗Common mistake 3: Not handling the bitmask operations correctly, which could lead to incorrect subproblem results.

✗Common mistake 4: Not considering all possible pairs for each operation. The inner loop should iterate only over unused numbers.

✅

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: Task scheduling where assigning tasks to processors has varying costs (analogous to gcd) and order matters.

🏦Real-world scenario 2: Resource allocation where pairing resources provides different efficiencies and we want to maximize overall efficiency through optimal pairings.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: November 27, 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

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

@AlgoMasterJun 30

❓

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Maximize Score After N Operations"?

For this hard-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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