What is the time complexity of Can I Win?

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

What is the space complexity of Can I Win?

The space complexity is O(2^n).

What data structures are used to solve Can I Win?

This problem uses Math, Dynamic Programming, Bit Manipulation, Memoization, Game Theory, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Can I Win?

Insight 1: The problem can be approached using a recursive strategy with memoization. Since the choices made in each turn affect the subsequent turns, and we need to determine if there's a guaranteed win for the first player, a recursive exploration of possible game states is suitable. Insight 2: Memoization is crucial for efficiency. Without it, the same game states would be repeatedly evaluated, leading to exponential time complexity. By storing the results of previously computed states, we can significantly reduce the computation time.

Which companies ask LeetCode 464. Can I Win in interviews?

This problem is frequently asked at LinkedIn.

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Can I Win - LeetCode 464 Solution

Can I Win - Complete Solution Guide

Can I Win is LeetCode problem 464, 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

In the "100 game" two players take turns adding, to a running total, any integer from 1 to 10 . The player who first causes the running total to reach or exceed 100 wins. What if we change the game so that players cannot re-use integers? For example, two players might take turns drawing from a common pool of numbers from 1 to 15 without replacement until they reach a total >= 100. Given two integers maxChoosableInteger and desiredTotal , return true if the first player to move can force a win, o

Detailed Explanation

The "Can I Win" problem presents a game where two players take turns selecting integers from 1 to `maxChoosableInteger` without replacement. They add these integers to a running total. The first player to make the total reach or exceed `desiredTotal` wins. The problem asks whether the first player can force a win, assuming both players play optimally. The inputs are `maxChoosableInteger`, which determines the range of numbers available, and `desiredTotal`, which is the target sum to reach or exceed. The output is a boolean value indicating whether the first player has a winning strategy.

Solution Approach

The solution uses a recursive function with memoization to determine if the first player can win. The recursive function explores all possible moves the first player can make. For each move, it checks if that move immediately leads to a win (i.e., the selected number is greater than or equal to the remaining desired total). If not, it recursively checks if the second player can win given the current state. If the second player cannot win (i.e., the second player will lose), then the first player wins in this scenario. The `memo` dictionary stores the results of previously computed game states to avoid redundant calculations.

Step-by-Step Algorithm

Step 1: Handle base cases: If `desiredTotal` is less than or equal to 0, the first player automatically wins, so return `true`. If the sum of all numbers from 1 to `maxChoosableInteger` is less than `desiredTotal`, no one can win, so return `false`.
Step 2: Initialize a memoization dictionary (or hash map) to store the results of calculated game states. The key is a bitmask representing which numbers have been chosen, and the value is a boolean indicating whether the first player can win from that state.
Step 3: Define a recursive function `can_win_recursive(mask, remaining)` that takes the current bitmask (`mask`) of chosen numbers and the remaining `desiredTotal` as input. The `mask` keeps track of which numbers from 1 to `maxChoosableInteger` have already been used.
Step 4: In the recursive function, first check if the current state (`mask`) is already in the `memo`. If so, return the stored result.
Step 5: Iterate through each number from 1 to `maxChoosableInteger`. For each number `i`, check if it has not been chosen (i.e., the `i`-th bit in the `mask` is 0).
Step 6: If `i` has not been chosen, check if choosing `i` leads to an immediate win (i.e., `i >= remaining`). If so, the first player wins, so store `True` in `memo[mask]` and return `True`.
Step 7: If choosing `i` does not lead to an immediate win, recursively call `can_win_recursive(mask | (1 << (i - 1)), remaining - i)` to determine if the second player can win after the first player chooses `i`. The `mask | (1 << (i - 1))` updates the bitmask to indicate that `i` has been chosen, and `remaining - i` updates the remaining `desiredTotal`.
Step 8: If the second player cannot win (i.e., the recursive call returns `False`), then the first player wins in this scenario. Store `True` in `memo[mask]` and return `True`.
Step 9: If the first player cannot win by choosing any number, it means the second player can always force a win. Store `False` in `memo[mask]` and return `False`.
Step 10: Call the recursive function `can_win_recursive(0, desiredTotal)` with an initial mask of 0 (no numbers chosen) and the original `desiredTotal`. Return the result.

Key Insights

Insight 1: The problem can be approached using a recursive strategy with memoization. Since the choices made in each turn affect the subsequent turns, and we need to determine if there's a guaranteed win for the first player, a recursive exploration of possible game states is suitable.
Insight 2: Memoization is crucial for efficiency. Without it, the same game states would be repeatedly evaluated, leading to exponential time complexity. By storing the results of previously computed states, we can significantly reduce the computation time.
Insight 3: Bit manipulation offers an efficient way to represent the set of used numbers. Each bit in an integer can represent whether a particular number has been chosen or not. This allows us to quickly check if a number is available and update the set of chosen numbers.

Complexity Analysis

Time Complexity: O(2^n * n)

Space Complexity: O(2^n)

Topics

This problem involves: Math, Dynamic Programming, Bit Manipulation, Memoization, Game Theory, Bitmask.

Companies

Asked at: LinkedIn.

Can I Win | LeetCode 464

Medium

Math Dynamic Programming Bit Manipulation Memoization Game Theory Bitmask

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

Solution Code

Available in:

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

class Solution {
    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if (desiredTotal <= 0) {
            return true;
        }
        
        if ((maxChoosableInteger * (maxChoosableInteger + 1)) / 2 < desiredTotal) {
            return false;
        }

        Map<Integer, Boolean> memo = new HashMap<>();

        return canWinRecursive(0, desiredTotal, maxChoosableInteger, memo);
    }

    private boolean canWinRecursive(int mask, int remaining, int maxChoosableInteger, Map<Integer, Boolean> memo) {
        if (memo.containsKey(mask)) {
            return memo.get(mask);
        }

        for (int i = 1; i <= maxChoosableInteger; i++) {
            int currentBit = 1 << (i - 1);
            if ((mask & currentBit) == 0) {
                if (i >= remaining) {
                    memo.put(mask, true);
                    return true;
                }
                
                if (!canWinRecursive(mask | currentBit, remaining - i, maxChoosableInteger, memo)) {
                    memo.put(mask, true);
                    return true;
                }
            }
        }
        
        memo.put(mask, false);
        return false;
    }
}

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Problem Info

Description

Problem 464: Can I Win - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Bit Manipulation Memoization Game Theory Bitmask

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:464

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

10 steps

1
Step 1: Handle base cases: If `desiredTotal` is less than or equal to 0, the first player automatically wins, so return `true`. If the sum of all numbers from 1 to `maxChoosableInteger` is less than `desiredTotal`, no one can win, so return `false`.
2
Step 2: Initialize a memoization dictionary (or hash map) to store the results of calculated game states. The key is a bitmask representing which numbers have been chosen, and the value is a boolean indicating whether the first player can win from that state.
3
Step 3: Define a recursive function `can_win_recursive(mask, remaining)` that takes the current bitmask (`mask`) of chosen numbers and the remaining `desiredTotal` as input. The `mask` keeps track of which numbers from 1 to `maxChoosableInteger` have already been used.
4
Step 4: In the recursive function, first check if the current state (`mask`) is already in the `memo`. If so, return the stored result.
5
Step 5: Iterate through each number from 1 to `maxChoosableInteger`. For each number `i`, check if it has not been chosen (i.e., the `i`-th bit in the `mask` is 0).
6
Step 6: If `i` has not been chosen, check if choosing `i` leads to an immediate win (i.e., `i >= remaining`). If so, the first player wins, so store `True` in `memo[mask]` and return `True`.
7
Step 7: If choosing `i` does not lead to an immediate win, recursively call `can_win_recursive(mask | (1 << (i - 1)), remaining - i)` to determine if the second player can win after the first player chooses `i`. The `mask | (1 << (i - 1))` updates the bitmask to indicate that `i` has been chosen, and `remaining - i` updates the remaining `desiredTotal`.
8
Step 8: If the second player cannot win (i.e., the recursive call returns `False`), then the first player wins in this scenario. Store `True` in `memo[mask]` and return `True`.
9
Step 9: If the first player cannot win by choosing any number, it means the second player can always force a win. Store `False` in `memo[mask]` and return `False`.
10
Step 10: Call the recursive function `can_win_recursive(0, desiredTotal)` with an initial mask of 0 (no numbers chosen) and the original `desiredTotal`. Return the result.

💡

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(2^n * n)

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity arises from exploring each possible combination of numbers being chosen which can be represented by all possible bit masks (2^n). Then for each combination we potentially iterate across all possible numbers to choose (n). The space complexity is dominated by the size of the memoization map which can grow to 2^n if we visit every state.

📖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

Iterative Dynamic Programming

Instead of using recursion with memoization, we can use an iterative dynamic programming approach. This would involve building a table to store the results of subproblems and iteratively filling it in from the base cases.

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

Trade-offs: Iterative DP can avoid recursion overhead, but it might be less intuitive to understand and implement for this problem. The time and space complexity remain the same.

✨

Key Insights

1Insight 1: The problem can be approached using a recursive strategy with memoization. Since the choices made in each turn affect the subsequent turns, and we need to determine if there's a guaranteed win for the first player, a recursive exploration of possible game states is suitable.
2Insight 2: Memoization is crucial for efficiency. Without it, the same game states would be repeatedly evaluated, leading to exponential time complexity. By storing the results of previously computed states, we can significantly reduce the computation time.
3Insight 3: Bit manipulation offers an efficient way to represent the set of used numbers. Each bit in an integer can represent whether a particular number has been chosen or not. This allows us to quickly check if a number is available and update the set of chosen numbers.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `desiredTotal <= 0`. In this case, the first player automatically wins because the target is already met or exceeded, so the function should return `true`. This is handled by the initial check at the beginning of the `canIWin` function.
Edge case 2: The sum of all numbers from 1 to `maxChoosableInteger` is less than `desiredTotal`. In this case, no player can ever win, so the function should return `false`. This is also handled by an initial check within the `canIWin` function.
Edge case 3: `maxChoosableInteger` is 1. If `desiredTotal` is greater than 1, the first player cannot win, return `false`. If `desiredTotal` is 0 or 1, the first player wins, return `true`. This is implicitly handled by the overall logic.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to memoize the results of the recursive calls. This leads to exponential time complexity and can cause the program to run very slowly, especially for larger values of `maxChoosableInteger`.

✗Common mistake 2: Incorrectly updating the bitmask or the remaining `desiredTotal` in the recursive calls. This can lead to incorrect results, as the game state will not be accurately represented.

✗Common mistake 3: Not handling the base cases where `desiredTotal <= 0` or where the sum of all numbers from 1 to `maxChoosableInteger` is less than `desiredTotal`. This will lead to incorrect results or infinite recursion.

✅

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: Game strategy analysis. This algorithm can be adapted to analyze the winning strategies in other turn-based games with a finite number of choices and a defined winning condition.

🏦Real-world scenario 2: Resource allocation. This can model optimal resource allocation where resources are limited (the numbers) and you need to reach a target value (desiredTotal) before another entity.

🎮Real-world scenario 3: Decision-making under constraints. The algorithm's core logic of exploring possible choices and their consequences to achieve a desired outcome can be applied in various decision-making scenarios where there are constraints on available options.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

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 13

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

@DevExpert92Jul 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."

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 "Can I Win"?

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

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