LeetCode #1535 — MEDIUM

Find the Winner of an Array Game

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

The Problem

Problem Statement

Given an integer array arr of distinct integers and an integer k.

A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0, and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds.

Return the integer which will win the game.

It is guaranteed that there will be a winner of the game.

Example 1:

Input: arr = [2,1,3,5,4,6,7], k = 2
Output: 5
Explanation: Let's see the rounds of the game:
Round |       arr       | winner | win_count
  1   | [2,1,3,5,4,6,7] | 2      | 1
  2   | [2,3,5,4,6,7,1] | 3      | 1
  3   | [3,5,4,6,7,1,2] | 5      | 1
  4   | [5,4,6,7,1,2,3] | 5      | 2
So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games.

Example 2:

Input: arr = [3,2,1], k = 10
Output: 3
Explanation: 3 will win the first 10 rounds consecutively.

Constraints:

2 <= arr.length <= 10⁵
1 <= arr[i] <= 10⁶
arr contains distinct integers.
1 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given an integer array arr of distinct integers and an integer k. A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0, and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds. Return the integer which will win the game. It is guaranteed that there will be a winner of the game.

Baseline thinking

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

Pattern signal: Array

Example 1

[2,1,3,5,4,6,7]
2

Example 2

[3,2,1]
10

Step 02

Core Insight

What unlocks the optimal approach

If k ≥ arr.length return the max element of the array.
If k < arr.length simulate the game until a number wins k consecutive games.

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 #1535: Find the Winner of an Array Game
class Solution {
    public int getWinner(int[] arr, int k) {
        int mx = arr[0];
        for (int i = 1, cnt = 0; i < arr.length; ++i) {
            if (mx < arr[i]) {
                mx = arr[i];
                cnt = 1;
            } else {
                ++cnt;
            }
            if (cnt == k) {
                break;
            }
        }
        return mx;
    }
}

// Accepted solution for LeetCode #1535: Find the Winner of an Array Game
func getWinner(arr []int, k int) int {
	mx, cnt := arr[0], 0
	for _, x := range arr[1:] {
		if mx < x {
			mx = x
			cnt = 1
		} else {
			cnt++
		}
		if cnt == k {
			break
		}
	}
	return mx
}

# Accepted solution for LeetCode #1535: Find the Winner of an Array Game
class Solution:
    def getWinner(self, arr: List[int], k: int) -> int:
        mx = arr[0]
        cnt = 0
        for x in arr[1:]:
            if mx < x:
                mx = x
                cnt = 1
            else:
                cnt += 1
            if cnt == k:
                break
        return mx

// Accepted solution for LeetCode #1535: Find the Winner of an Array Game
struct Solution;
use std::collections::VecDeque;

impl Solution {
    fn get_winner(arr: Vec<i32>, k: i32) -> i32 {
        let mut queue = VecDeque::new();
        let k = k as usize;
        let k = k.min(arr.len());
        for x in arr {
            queue.push_back(x);
        }
        loop {
            let first = queue.pop_front().unwrap();
            let second = queue.pop_front().unwrap();
            if second > first {
                queue.push_front(first);
                queue.push_front(second);
            } else {
                queue.push_front(second);
                queue.push_front(first);
            }
            let mut round = 0;
            let winner = queue.pop_front().unwrap();
            while let Some(first) = queue.pop_front() {
                if first < winner {
                    queue.push_back(first);
                    round += 1;
                    if round == k {
                        return winner;
                    }
                } else {
                    queue.push_front(first);
                    queue.push_front(winner);
                    break;
                }
            }
        }
    }
}

#[test]
fn test() {
    let arr = vec![2, 1, 3, 5, 4, 6, 7];
    let k = 2;
    let res = 5;
    assert_eq!(Solution::get_winner(arr, k), res);
    let arr = vec![3, 2, 1];
    let k = 10;
    let res = 3;
    assert_eq!(Solution::get_winner(arr, k), res);
    let arr = vec![1, 9, 8, 2, 3, 7, 6, 4, 5];
    let k = 7;
    let res = 9;
    assert_eq!(Solution::get_winner(arr, k), res);
    let arr = vec![1, 11, 22, 33, 44, 55, 66, 77, 88, 99];
    let k = 1000000000;
    let res = 99;
    assert_eq!(Solution::get_winner(arr, k), res);
}

// Accepted solution for LeetCode #1535: Find the Winner of an Array Game
function getWinner(arr: number[], k: number): number {
    let mx = arr[0];
    let cnt = 0;
    for (const x of arr.slice(1)) {
        if (mx < x) {
            mx = x;
            cnt = 1;
        } else {
            ++cnt;
        }
        if (cnt === k) {
            break;
        }
    }
    return mx;
}

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.