What is the time complexity of Predict the Winner?

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

What is the space complexity of Predict the Winner?

The space complexity is O(n^2).

What data structures are used to solve Predict the Winner?

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

What are the key insights for solving Predict the Winner?

Insight 1: Dynamic Programming is essential because both players are making optimal decisions based on the current state of the game. Recursive solutions would recalculate the same subproblems multiple times. Insight 2: The optimal strategy involves maximizing the difference between Player 1's score and Player 2's score. A positive difference means Player 1 wins (or ties).

Which companies ask LeetCode 486. Predict the Winner in interviews?

This problem is frequently asked at Cisco, Flipkart, Salesforce.

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Predict the Winner - LeetCode 486 Solution

Predict the Winner - Complete Solution Guide

Predict the Winner is LeetCode problem 486, 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 integer array nums . Two players are playing a game with this array: player 1 and player 2. Player 1 and player 2 take turns, with player 1 starting first. Both players start the game with a score of 0 . At each turn, the player takes one of the numbers from either end of the array (i.e., nums[0] or nums[nums.length - 1] ) which reduces the size of the array by 1 . The player adds the chosen number to their score. The game ends when there are no more elements in the array. Retur

Detailed Explanation

The 'Predict the Winner' problem involves two players taking turns to select numbers from either end of an array. Each player aims to maximize their score by strategically choosing numbers. Player 1 goes first, and the game continues until the array is empty. The goal is to determine if Player 1 can achieve a score greater than or equal to Player 2's score, assuming both players play optimally. The input is an integer array 'nums', and the output is a boolean value indicating whether Player 1 can win or not. The constraints are that the array length is between 1 and 20, and the numbers in the array are non-negative and less than or equal to 10^7.

Solution Approach

The provided solutions use dynamic programming to solve the problem. A 2D array 'dp' is used to store the maximum difference in scores Player 1 can achieve for each possible subarray. The array is filled in a bottom-up manner, starting with subarrays of length 1 and gradually increasing the length. The value of dp[i][j] is calculated based on whether Player 1 chooses nums[i] or nums[j]. When Player 1 chooses nums[i], Player 2's optimal choice on nums[i+1...j] would minimize Player 1's potential gain, which corresponds to dp[i+1][j]. When Player 1 chooses nums[j], Player 2's optimal choice on nums[i...j-1] would minimize Player 1's potential gain, which corresponds to dp[i][j-1]. Player 1 chooses the option that maximizes the difference between their score and Player 2's. The final result, dp[0][n-1], represents the maximum difference Player 1 can achieve on the entire array. If this value is greater than or equal to 0, Player 1 wins.

Step-by-Step Algorithm

Step 1: Initialize a 2D array 'dp' of size n x n with all elements set to 0. dp[i][j] will store the maximum difference Player 1 can achieve for the subarray nums[i...j].
Step 2: Iterate diagonally through the 'dp' array, starting from the main diagonal (where i == j) and moving towards the top-right corner. This is done by looping through the rows in reverse order (from n-1 to 0) and, for each row, looping through the columns from i+1 to n-1.
Step 3: For each cell dp[i][j], calculate its value based on the following recurrence relation: dp[i][j] = max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1]). This means Player 1 can either choose nums[i] or nums[j]. If Player 1 chooses nums[i], the remaining subarray is nums[i+1...j], and Player 2 will play optimally to minimize Player 1's gain (represented by dp[i+1][j]). Similarly, if Player 1 chooses nums[j], the remaining subarray is nums[i...j-1], and Player 2 will play optimally to minimize Player 1's gain (represented by dp[i][j-1]). Player 1 chooses the option that maximizes the difference.
Step 4: After filling the 'dp' array, return true if dp[0][n-1] >= 0, indicating that Player 1 can achieve a score greater than or equal to Player 2's score; otherwise, return false.

Key Insights

Insight 1: Dynamic Programming is essential because both players are making optimal decisions based on the current state of the game. Recursive solutions would recalculate the same subproblems multiple times.
Insight 2: The optimal strategy involves maximizing the difference between Player 1's score and Player 2's score. A positive difference means Player 1 wins (or ties).
Insight 3: The problem can be solved using a 2D DP array where dp[i][j] represents the maximum difference in score that Player 1 can achieve when playing optimally on the subarray nums[i...j].

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Math, Dynamic Programming, Recursion, Game Theory.

Companies

Asked at: Cisco, Flipkart, Salesforce.

Predict the Winner | LeetCode 486

Medium

Array Math Dynamic Programming Recursion Game Theory

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

Solution Code

Available in:

class Solution {
    public boolean predictTheWinner(int[] nums) {
        int n = nums.length;
        if (n == 1) {
            return true;
        }
        
        int[][] dp = new int[n][n];
        
        for (int i = n - 1; i >= 0; i--) {
            dp[i][i] = nums[i];
            for (int j = i + 1; j < n; j++) {
                dp[i][j] = Math.max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1]);
            }
        }
        
        return dp[0][n-1] >= 0;
    }
}

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Problem Info

Description

Problem 486: Predict the Winner - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Recursion Game Theory

Companies

CiscoFlipkartSalesforce

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:486

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

4 steps

1
Step 1: Initialize a 2D array 'dp' of size n x n with all elements set to 0. dp[i][j] will store the maximum difference Player 1 can achieve for the subarray nums[i...j].
2
Step 2: Iterate diagonally through the 'dp' array, starting from the main diagonal (where i == j) and moving towards the top-right corner. This is done by looping through the rows in reverse order (from n-1 to 0) and, for each row, looping through the columns from i+1 to n-1.
3
Step 3: For each cell dp[i][j], calculate its value based on the following recurrence relation: dp[i][j] = max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1]). This means Player 1 can either choose nums[i] or nums[j]. If Player 1 chooses nums[i], the remaining subarray is nums[i+1...j], and Player 2 will play optimally to minimize Player 1's gain (represented by dp[i+1][j]). Similarly, if Player 1 chooses nums[j], the remaining subarray is nums[i...j-1], and Player 2 will play optimally to minimize Player 1's gain (represented by dp[i][j-1]). Player 1 chooses the option that maximizes the difference.
4
Step 4: After filling the 'dp' array, return true if dp[0][n-1] >= 0, indicating that Player 1 can achieve a score greater than or equal to Player 2's score; otherwise, return false.

💡

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)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The nested loops each iterate up to 'n' times, resulting in O(n^2) time complexity. Similarly, the 2D DP table stores 'n^2' integers, leading to O(n^2) space complexity.

📖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 take the start and end indices of the subarray as input and return the maximum difference Player 1 can achieve. Memoization is used to store the results of previously computed subproblems, avoiding redundant calculations.

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

Trade-offs: Recursion with memoization is conceptually similar to dynamic programming but may have higher overhead due to function call stack. In many cases, the iterative DP approach is preferred for its clarity and slightly better performance. However, the recursive implementation can be more intuitive for some developers.

✨

Key Insights

1Insight 1: Dynamic Programming is essential because both players are making optimal decisions based on the current state of the game. Recursive solutions would recalculate the same subproblems multiple times.
2Insight 2: The optimal strategy involves maximizing the difference between Player 1's score and Player 2's score. A positive difference means Player 1 wins (or ties).
3Insight 3: The problem can be solved using a 2D DP array where dp[i][j] represents the maximum difference in score that Player 1 can achieve when playing optimally on the subarray nums[i...j].

📚

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.

Recursion

A function that calls itself to solve smaller versions of the same problem.

Edge Cases

Edge case 1: When the input array has only one element (n == 1), Player 1 automatically wins because they get the only number. The code handles this case with a specific check: `if n == 1: return True`.
Edge case 2: An empty array (n == 0) isn't explicitly handled in the main logic, but the problem statement constraints specify that the array length will be between 1 and 20. If an empty array was possible, the code would need to be adjusted to return true as per the problem's definition (player 1 wins if they tie or exceed the score of player 2)

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not recognizing the need for dynamic programming and attempting a greedy approach. A greedy approach will not always yield the optimal solution because a player's immediate choice impacts future choices.

✗Common mistake 2: Incorrectly implementing the recurrence relation or the base cases for the dynamic programming solution. It's crucial to carefully consider how the optimal choice for Player 1 affects the remaining subarray and Player 2's subsequent optimal play.

✅

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: Resource allocation problems where two entities compete for resources, and each entity aims to maximize their own gain. The 'Predict the Winner' problem can be adapted to model scenarios where the resource values are known and the entities take turns to claim them.

🏦Real-world scenario 2: Game theory situations such as negotiations or bargaining, where players have to make strategic decisions to achieve their desired outcome. Understanding optimal strategies, as demonstrated in the 'Predict the Winner' problem, is fundamental to making informed decisions in such scenarios.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: October 5, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 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 "Predict the Winner"?

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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