LeetCode #464 — MEDIUM

Can I Win

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

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, otherwise, return false. Assume both players play optimally.

Example 1:

Input: maxChoosableInteger = 10, desiredTotal = 11
Output: false
Explanation:
No matter which integer the first player choose, the first player will lose.
The first player can choose an integer from 1 up to 10.
If the first player choose 1, the second player can only choose integers from 2 up to 10.
The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.

Example 2:

Input: maxChoosableInteger = 10, desiredTotal = 0
Output: true

Example 3:

Input: maxChoosableInteger = 10, desiredTotal = 1
Output: true

Constraints:

1 <= maxChoosableInteger <= 20
0 <= desiredTotal <= 300

Step 01

Brute Force Baseline

Problem summary: 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, otherwise, return false. Assume both players play optimally.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming · Bit Manipulation

Example 1

10
11

Example 2

10
0

Example 3

10
1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #464: Can I Win
class Solution {
    private Map<Integer, Boolean> f = new HashMap<>();
    private int maxChoosableInteger;
    private int desiredTotal;

    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if ((1 + maxChoosableInteger) * maxChoosableInteger / 2 < desiredTotal) {
            return false;
        }
        this.maxChoosableInteger = maxChoosableInteger;
        this.desiredTotal = desiredTotal;
        return dfs(0, 0);
    }

    private boolean dfs(int mask, int s) {
        if (f.containsKey(mask)) {
            return f.get(mask);
        }
        for (int i = 0; i < maxChoosableInteger; ++i) {
            if ((mask >> i & 1) == 0) {
                if (s + i + 1 >= desiredTotal || !dfs(mask | 1 << i, s + i + 1)) {
                    f.put(mask, true);
                    return true;
                }
            }
        }
        f.put(mask, false);
        return false;
    }
}

// Accepted solution for LeetCode #464: Can I Win
func canIWin(maxChoosableInteger int, desiredTotal int) bool {
	if (1+maxChoosableInteger)*maxChoosableInteger/2 < desiredTotal {
		return false
	}
	f := map[int]bool{}
	var dfs func(int, int) bool
	dfs = func(mask, s int) bool {
		if v, ok := f[mask]; ok {
			return v
		}
		for i := 1; i <= maxChoosableInteger; i++ {
			if mask>>i&1 == 0 {
				if s+i >= desiredTotal || !dfs(mask|1<<i, s+i) {
					f[mask] = true
					return true
				}
			}
		}
		f[mask] = false
		return false
	}
	return dfs(0, 0)
}

# Accepted solution for LeetCode #464: Can I Win
class Solution:
    def canIWin(self, maxChoosableInteger: int, desiredTotal: int) -> bool:
        @cache
        def dfs(mask: int, s: int) -> bool:
            for i in range(1, maxChoosableInteger + 1):
                if mask >> i & 1 ^ 1:
                    if s + i >= desiredTotal or not dfs(mask | 1 << i, s + i):
                        return True
            return False

        if (1 + maxChoosableInteger) * maxChoosableInteger // 2 < desiredTotal:
            return False
        return dfs(0, 0)

// Accepted solution for LeetCode #464: Can I Win
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn can_i_win(max_choosable_integer: i32, desired_total: i32) -> bool {
        let nums: Vec<i32> = (1..=max_choosable_integer).collect();
        let n = nums.len();
        if desired_total > nums.iter().sum::<i32>() {
            return false;
        }
        let mut memo: HashMap<u32, bool> = HashMap::new();
        Self::dp(desired_total, (1 << n) - 1, &mut memo, &nums, n)
    }

    fn dp(total: i32, bitset: u32, memo: &mut HashMap<u32, bool>, nums: &[i32], n: usize) -> bool {
        if let Some(&res) = memo.get(&bitset) {
            return res;
        }
        let mut res = false;
        for i in 0..n {
            if res {
                break;
            }
            if 1 << i & bitset > 0 {
                if total - nums[i] <= 0 {
                    res = true;
                }
                res |= !Self::dp(total - nums[i], bitset & !(1 << i), memo, nums, n);
            }
        }
        memo.insert(bitset, res);
        res
    }
}
#[test]
fn test() {
    let max_choosable_integer = 10;
    let desired_total = 11;
    let res = false;
    assert_eq!(
        Solution::can_i_win(max_choosable_integer, desired_total),
        res
    );
    let max_choosable_integer = 4;
    let desired_total = 6;
    let res = true;
    assert_eq!(
        Solution::can_i_win(max_choosable_integer, desired_total),
        res
    );
    let max_choosable_integer = 5;
    let desired_total = 50;
    let res = false;
    assert_eq!(
        Solution::can_i_win(max_choosable_integer, desired_total),
        res
    );
    let max_choosable_integer = 18;
    let desired_total = 188;
    let res = false;
    assert_eq!(
        Solution::can_i_win(max_choosable_integer, desired_total),
        res
    );
}

// Accepted solution for LeetCode #464: Can I Win
function canIWin(maxChoosableInteger: number, desiredTotal: number): boolean {
    if (((1 + maxChoosableInteger) * maxChoosableInteger) / 2 < desiredTotal) {
        return false;
    }
    const f: Record<string, boolean> = {};
    const dfs = (mask: number, s: number): boolean => {
        if (f.hasOwnProperty(mask)) {
            return f[mask];
        }
        for (let i = 1; i <= maxChoosableInteger; ++i) {
            if (((mask >> i) & 1) ^ 1) {
                if (s + i >= desiredTotal || !dfs(mask ^ (1 << i), s + i)) {
                    return (f[mask] = true);
                }
            }
        }
        return (f[mask] = false);
    };
    return dfs(0, 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(2^n)

Space

O(2^n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.