LeetCode #1510 — HARD

Stone Game IV

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there are n stones in a pile. On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile.

Also, if a player cannot make a move, he/she loses the game.

Given a positive integer n, return true if and only if Alice wins the game otherwise return false, assuming both players play optimally.

Example 1:

Input: n = 1
Output: true
Explanation: Alice can remove 1 stone winning the game because Bob doesn't have any moves.

Example 2:

Input: n = 2
Output: false
Explanation: Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).

Example 3:

Input: n = 4
Output: true
Explanation: n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).

Constraints:

1 <= n <= 10⁵

Step 01

Brute Force Baseline

Problem summary: Alice and Bob take turns playing a game, with Alice starting first. Initially, there are n stones in a pile. On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile. Also, if a player cannot make a move, he/she loses the game. Given a positive integer n, return true if and only if Alice wins the game otherwise return false, assuming 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

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

Use dynamic programming to keep track of winning and losing states. Given some number of stones, Alice can win if she can force Bob onto a losing 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

Largest constraint values

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 #1510: Stone Game IV
class Solution {
    private Boolean[] f;

    public boolean winnerSquareGame(int n) {
        f = new Boolean[n + 1];
        return dfs(n);
    }

    private boolean dfs(int i) {
        if (i <= 0) {
            return false;
        }
        if (f[i] != null) {
            return f[i];
        }
        for (int j = 1; j <= i / j; ++j) {
            if (!dfs(i - j * j)) {
                return f[i] = true;
            }
        }
        return f[i] = false;
    }
}

// Accepted solution for LeetCode #1510: Stone Game IV
func winnerSquareGame(n int) bool {
	f := make([]int, n+1)
	var dfs func(int) bool
	dfs = func(i int) bool {
		if i <= 0 {
			return false
		}
		if f[i] != 0 {
			return f[i] == 1
		}
		for j := 1; j <= i/j; j++ {
			if !dfs(i - j*j) {
				f[i] = 1
				return true
			}
		}
		f[i] = -1
		return false
	}
	return dfs(n)
}

# Accepted solution for LeetCode #1510: Stone Game IV
class Solution:
    def winnerSquareGame(self, n: int) -> bool:
        @cache
        def dfs(i: int) -> bool:
            if i == 0:
                return False
            j = 1
            while j * j <= i:
                if not dfs(i - j * j):
                    return True
                j += 1
            return False

        return dfs(n)

// Accepted solution for LeetCode #1510: Stone Game IV
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn winner_square_game(n: i32) -> bool {
        let mut memo: HashMap<i32, bool> = HashMap::new();
        Self::dp(n, &mut memo)
    }

    fn dp(n: i32, memo: &mut HashMap<i32, bool>) -> bool {
        if let Some(&res) = memo.get(&n) {
            return res;
        }
        let x = Self::sqrt(n);
        let res = if x * x == n {
            true
        } else {
            let mut res = false;
            for i in 1..=x {
                let y = n - i * i;
                if !Self::dp(y, memo) {
                    res = true;
                    break;
                }
            }
            res
        };
        memo.insert(n, res);
        res
    }

    fn sqrt(x: i32) -> i32 {
        (x as f64).sqrt() as i32
    }
}

#[test]
fn test() {
    let n = 1;
    let res = true;
    assert_eq!(Solution::winner_square_game(n), res);
    let n = 2;
    let res = false;
    assert_eq!(Solution::winner_square_game(n), res);
    let n = 4;
    let res = true;
    assert_eq!(Solution::winner_square_game(n), res);
    let n = 7;
    let res = false;
    assert_eq!(Solution::winner_square_game(n), res);
    let n = 17;
    let res = false;
    assert_eq!(Solution::winner_square_game(n), res);
}

// Accepted solution for LeetCode #1510: Stone Game IV
function winnerSquareGame(n: number): boolean {
    const f: number[] = new Array(n + 1).fill(0);
    const dfs = (i: number): boolean => {
        if (i <= 0) {
            return false;
        }
        if (f[i] !== 0) {
            return f[i] === 1;
        }
        for (let j = 1; j * j <= i; ++j) {
            if (!dfs(i - j * j)) {
                f[i] = 1;
                return true;
            }
        }
        f[i] = -1;
        return false;
    };
    return dfs(n);
}

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(n × m)

Space

O(n × m)

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.