LeetCode #2550 — MEDIUM

Count Collisions of Monkeys on a Polygon

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

The Problem

Problem Statement

There is a regular convex polygon with n vertices. The vertices are labeled from 0 to n - 1 in a clockwise direction, and each vertex has exactly one monkey. The following figure shows a convex polygon of 6 vertices.

Simultaneously, each monkey moves to a neighboring vertex. A collision happens if at least two monkeys reside on the same vertex after the movement or intersect on an edge.

Return the number of ways the monkeys can move so that at least one collision happens. Since the answer may be very large, return it modulo 10⁹+ 7.

Example 1:

Input: n = 3

Output: 6

Explanation:

There are 8 total possible movements.
Two ways such that they collide at some point are:

Monkey 1 moves in a clockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 2 collide.
Monkey 1 moves in an anticlockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 3 collide.

Example 2:

Input: n = 4

Output: 14

Constraints:

3 <= n <= 10⁹

Step 01

Brute Force Baseline

Problem summary: There is a regular convex polygon with n vertices. The vertices are labeled from 0 to n - 1 in a clockwise direction, and each vertex has exactly one monkey. The following figure shows a convex polygon of 6 vertices. Simultaneously, each monkey moves to a neighboring vertex. A collision happens if at least two monkeys reside on the same vertex after the movement or intersect on an edge. Return the number of ways the monkeys can move so that at least one collision happens. Since the answer may be very large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Try counting the number of ways in which the monkeys will not collide.

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 #2550: Count Collisions of Monkeys on a Polygon
class Solution {
    public int monkeyMove(int n) {
        final int mod = (int) 1e9 + 7;
        return (qpow(2, n, mod) - 2 + mod) % mod;
    }

    private int qpow(long a, int n, int mod) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2550: Count Collisions of Monkeys on a Polygon
func monkeyMove(n int) int {
	const mod = 1e9 + 7
	qpow := func(a, n int) int {
		ans := 1
		for ; n > 0; n >>= 1 {
			if n&1 == 1 {
				ans = ans * a % mod
			}
			a = a * a % mod
		}
		return ans
	}
	return (qpow(2, n) - 2 + mod) % mod
}

# Accepted solution for LeetCode #2550: Count Collisions of Monkeys on a Polygon
class Solution:
    def monkeyMove(self, n: int) -> int:
        mod = 10**9 + 7
        return (pow(2, n, mod) - 2) % mod

// Accepted solution for LeetCode #2550: Count Collisions of Monkeys on a Polygon
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2550: Count Collisions of Monkeys on a Polygon
// class Solution {
//     public int monkeyMove(int n) {
//         final int mod = (int) 1e9 + 7;
//         return (qpow(2, n, mod) - 2 + mod) % mod;
//     }
// 
//     private int qpow(long a, int n, int mod) {
//         long ans = 1;
//         for (; n > 0; n >>= 1) {
//             if ((n & 1) == 1) {
//                 ans = ans * a % mod;
//             }
//             a = a * a % mod;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #2550: Count Collisions of Monkeys on a Polygon
function monkeyMove(n: number): number {
    const mod = 10 ** 9 + 7;
    const qpow = (a: number, n: number): number => {
        let ans = 1n;
        for (; n; n >>>= 1) {
            if (n & 1) {
                ans = (ans * BigInt(a)) % BigInt(mod);
            }
            a = Number((BigInt(a) * BigInt(a)) % BigInt(mod));
        }
        return Number(ans);
    };
    return (qpow(2, n) - 2 + mod) % mod;
}

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(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

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.