LeetCode #2682 — EASY

Find the Losers of the Circular Game

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

There are n friends that are playing a game. The friends are sitting in a circle and are numbered from 1 to n in clockwise order. More formally, moving clockwise from the i^th friend brings you to the (i+1)^th friend for 1 <= i < n, and moving clockwise from the n^th friend brings you to the 1^st friend.

The rules of the game are as follows:

1^st friend receives the ball.

After that, 1^st friend passes it to the friend who is k steps away from them in the clockwise direction.
After that, the friend who receives the ball should pass it to the friend who is 2 * k steps away from them in the clockwise direction.
After that, the friend who receives the ball should pass it to the friend who is 3 * k steps away from them in the clockwise direction, and so on and so forth.

In other words, on the i^th turn, the friend holding the ball should pass it to the friend who is i * k steps away from them in the clockwise direction.

The game is finished when some friend receives the ball for the second time.

The losers of the game are friends who did not receive the ball in the entire game.

Given the number of friends, n, and an integer k, return the array answer, which contains the losers of the game in the ascending order.

Example 1:

Input: n = 5, k = 2
Output: [4,5]
Explanation: The game goes as follows:
1) Start at 1^st friend and pass the ball to the friend who is 2 steps away from them - 3^rd friend.
2) 3^rd friend passes the ball to the friend who is 4 steps away from them - 2^nd friend.
3) 2^nd friend passes the ball to the friend who is 6 steps away from them  - 3^rd friend.
4) The game ends as 3^rd friend receives the ball for the second time.

Example 2:

Input: n = 4, k = 4
Output: [2,3,4]
Explanation: The game goes as follows:
1) Start at the 1^st friend and pass the ball to the friend who is 4 steps away from them - 1^st friend.
2) The game ends as 1^st friend receives the ball for the second time.

Constraints:

1 <= k <= n <= 50

Step 01

Brute Force Baseline

Problem summary: There are n friends that are playing a game. The friends are sitting in a circle and are numbered from 1 to n in clockwise order. More formally, moving clockwise from the ith friend brings you to the (i+1)th friend for 1 <= i < n, and moving clockwise from the nth friend brings you to the 1st friend. The rules of the game are as follows: 1st friend receives the ball. After that, 1st friend passes it to the friend who is k steps away from them in the clockwise direction. After that, the friend who receives the ball should pass it to the friend who is 2 * k steps away from them in the clockwise direction. After that, the friend who receives the ball should pass it to the friend who is 3 * k steps away from them in the clockwise direction, and so on and so forth. In other words, on the ith turn, the friend holding the ball should pass it to the friend who is i * k steps away from them in

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

5
2

Example 2

4
4

Core Insight

What unlocks the optimal approach

Simulate the whole game until a player receives the ball for the second time.

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 #2682: Find the Losers of the Circular Game
class Solution {
    public int[] circularGameLosers(int n, int k) {
        boolean[] vis = new boolean[n];
        int cnt = 0;
        for (int i = 0, p = 1; !vis[i]; ++p) {
            vis[i] = true;
            ++cnt;
            i = (i + p * k) % n;
        }
        int[] ans = new int[n - cnt];
        for (int i = 0, j = 0; i < n; ++i) {
            if (!vis[i]) {
                ans[j++] = i + 1;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2682: Find the Losers of the Circular Game
func circularGameLosers(n int, k int) (ans []int) {
	vis := make([]bool, n)
	for i, p := 0, 1; !vis[i]; p++ {
		vis[i] = true
		i = (i + p*k) % n
	}
	for i, x := range vis {
		if !x {
			ans = append(ans, i+1)
		}
	}
	return
}

# Accepted solution for LeetCode #2682: Find the Losers of the Circular Game
class Solution:
    def circularGameLosers(self, n: int, k: int) -> List[int]:
        vis = [False] * n
        i, p = 0, 1
        while not vis[i]:
            vis[i] = True
            i = (i + p * k) % n
            p += 1
        return [i + 1 for i in range(n) if not vis[i]]

// Accepted solution for LeetCode #2682: Find the Losers of the Circular Game
impl Solution {
    pub fn circular_game_losers(n: i32, k: i32) -> Vec<i32> {
        let mut vis: Vec<bool> = vec![false; n as usize];

        let mut i = 0;
        let mut p = 1;
        while !vis[i] {
            vis[i] = true;
            i = (i + p * (k as usize)) % (n as usize);
            p += 1;
        }

        let mut ans = Vec::new();
        for i in 0..vis.len() {
            if !vis[i] {
                ans.push((i + 1) as i32);
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2682: Find the Losers of the Circular Game
function circularGameLosers(n: number, k: number): number[] {
    const vis = new Array(n).fill(false);
    const ans: number[] = [];
    for (let i = 0, p = 1; !vis[i]; p++) {
        vis[i] = true;
        i = (i + p * k) % n;
    }
    for (let i = 0; i < vis.length; i++) {
        if (!vis[i]) {
            ans.push(i + 1);
        }
    }
    return ans;
}

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(n)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.