LeetCode #486 — MEDIUM

Predict the Winner

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

The Problem

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.

Return true if Player 1 can win the game. If the scores of both players are equal, then player 1 is still the winner, and you should also return true. You may assume that both players are playing optimally.

Example 1:

Input: nums = [1,5,2]
Output: false
Explanation: Initially, player 1 can choose between 1 and 2. 
If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. If player 2 chooses 5, then player 1 will be left with 1 (or 2). 
So, final score of player 1 is 1 + 2 = 3, and player 2 is 5. 
Hence, player 1 will never be the winner and you need to return false.

Example 2:

Input: nums = [1,5,233,7]
Output: true
Explanation: Player 1 first chooses 1. Then player 2 has to choose between 5 and 7. No matter which number player 2 choose, player 1 can choose 233.
Finally, player 1 has more score (234) than player 2 (12), so you need to return True representing player1 can win.

Constraints:

1 <= nums.length <= 20
0 <= nums[i] <= 10⁷

Step 01

Brute Force Baseline

Problem summary: 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. Return true if Player 1 can win the game. If the scores of both players are equal, then player 1 is still the winner, and you should also return true. You may assume that both players are playing optimally.

Baseline thinking

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

Pattern signal: Array · Math · Dynamic Programming

Example 1

[1,5,2]

Example 2

[1,5,233,7]

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 #486: Predict the Winner
class Solution {
    private int[] nums;
    private int[][] f;

    public boolean predictTheWinner(int[] nums) {
        this.nums = nums;
        int n = nums.length;
        f = new int[n][n];
        return dfs(0, n - 1) >= 0;
    }

    private int dfs(int i, int j) {
        if (i > j) {
            return 0;
        }
        if (f[i][j] != 0) {
            return f[i][j];
        }
        return f[i][j] = Math.max(nums[i] - dfs(i + 1, j), nums[j] - dfs(i, j - 1));
    }
}

// Accepted solution for LeetCode #486: Predict the Winner
func predictTheWinner(nums []int) bool {
	n := len(nums)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, n)
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i > j {
			return 0
		}
		if f[i][j] == 0 {
			f[i][j] = max(nums[i]-dfs(i+1, j), nums[j]-dfs(i, j-1))
		}
		return f[i][j]
	}
	return dfs(0, n-1) >= 0
}

# Accepted solution for LeetCode #486: Predict the Winner
class Solution:
    def predictTheWinner(self, nums: List[int]) -> bool:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            return max(nums[i] - dfs(i + 1, j), nums[j] - dfs(i, j - 1))

        return dfs(0, len(nums) - 1) >= 0

// Accepted solution for LeetCode #486: Predict the Winner
impl Solution {
    pub fn predict_the_winner(nums: Vec<i32>) -> bool {
        let n = nums.len();
        let mut f = vec![vec![0; n]; n];
        Self::dfs(&nums, &mut f, 0, n - 1) >= 0
    }

    fn dfs(nums: &Vec<i32>, f: &mut Vec<Vec<i32>>, i: usize, j: usize) -> i32 {
        if i == j {
            return nums[i] as i32;
        }
        if f[i][j] != 0 {
            return f[i][j];
        }
        f[i][j] = std::cmp::max(
            nums[i] - Self::dfs(nums, f, i + 1, j),
            nums[j] - Self::dfs(nums, f, i, j - 1),
        );
        f[i][j]
    }
}

// Accepted solution for LeetCode #486: Predict the Winner
function predictTheWinner(nums: number[]): boolean {
    const n = nums.length;
    const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    const dfs = (i: number, j: number): number => {
        if (i > j) {
            return 0;
        }
        if (f[i][j] === 0) {
            f[i][j] = Math.max(nums[i] - dfs(i + 1, j), nums[j] - dfs(i, j - 1));
        }
        return f[i][j];
    };
    return dfs(0, n - 1) >= 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(n^2)

Space

O(n^2)

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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.