Count ways to arrange integers in circle such that no two adjacent are same

Given 2 integers N and M, find the number of ways to arrange N integers in a circle such that every integer is in the range 0 to M-1, and no two adjacent integers are same. As the answer could be large, find the number, modulo 998244353.

Examples:

Input: N=3, M=3
Output: 6
Explanation: There are six ways as follows (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0).

Input: N =4, M=2
Output: 2
Explanation: There are two ways as follows (0, 1, 0, 1), (1, 0, 1, 0).

Approach: To solve the problem, follow the below idea:

The approach effectively uses dynamic programming to build up the count of valid arrangements.

State Representation:

• dp[i][0]: stores the number of ways to arrange first i integers such that the ith integer is different from the first integer.
• dp[i][1]: stores the number of ways to arrange first i integers such that the ith integer is same as the first integer.

Base Case:

• dp[1][1] = m: There is only one element, so we can have numbers 0 to m-1 i.e. total m ways

Transitions:

• dp[i][0] += (dp[i-1][0] * (m – 2) + dp[i-1][1] * (m – 1)): If the first and the ith integer are different, then we have to take the sum of both the cases:
  • The (i – 1)th integer is same as the first integer, so the number of ways will be dp[i-1][1] * (m – 1).
  • The (i – 1)th integer is different from the first integer, so the number of ways will be dp[i-1][0] * (m – 2)
• dp[i][1] += dp[i – 1][0]: If the first and the ith integer are same, then the ith element should be same as the last element.

Step-by-step algorithm:

• The dynamic programming array dp[N+1][2] is used to store the number of ways for each integer.
• The base case dp[1][1] = m initializes the number of ways to assign integers to the first person.
• Construct the dp array according to the above state transitions.
• Return dp[N][0] as the final result.

Below is the implementation of the algorithm:

C++

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define MOD 998244353

int main()
{
    int n = 3, m = 3;

    vector<vector<long long>> dp(n + 1, vector<long long>(2, 0));

    dp[1][0] = 0;
    // Initialize base case
    dp[1][1] = m;

    // Bottom-up dynamic programming to fill DP table
    for (int i = 2; i <= n; i++) {
        // No consecutive numbers are the same
        dp[i][0] = (dp[i - 1][0] * (m - 2) + dp[i-1][1] * (m - 1));
        dp[i][1] = dp[i - 1][0];
        dp[i][0] %= MOD;
        dp[1][1] %= MOD;
    }

    // Output the result
    cout << dp[n][0];

    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class Main {
    public static void main(String[] args) {
        int n = 3, m = 3;

        List<List<Long>> dp = new ArrayList<>(n + 1);
        for (int i = 0; i <= n; i++) {
            dp.add(new ArrayList<>(List.of(0L, 0L)));
        }

        dp.get(1).set(0, 0L);
        // Initialize base case
        dp.get(1).set(1, (long) m);

        // Bottom-up dynamic programming to fill DP table
        for (int i = 2; i <= n; i++) {
            // No consecutive numbers are the same
            dp.get(i).set(0, (dp.get(i - 1).get(0) * (m - 2)
                    + dp.get(i - 1).get(1) * (m - 1))
                    % 998244353);
            dp.get(i).set(1, dp.get(i - 1).get(0) % 998244353);
            dp.get(1).set(1, dp.get(1).get(1) % 998244353);
        }

        // Output the result
        System.out.println(dp.get(n).get(0));
    }
}

Python3

MOD = 998244353

def main():
    n = 3
    m = 3

    # Initialize DP table
    dp = [[0 for _ in range(2)] for _ in range(n + 1)]

    # Initialize base case
    dp[1][0] = 0
    dp[1][1] = m

    # Bottom-up dynamic programming to fill DP table
    for i in range(2, n + 1):
        # No consecutive numbers are the same
        dp[i][0] = (dp[i - 1][0] * (m - 2) + dp[i - 1][1] * (m - 1)) % MOD
        dp[i][1] = dp[i - 1][0] % MOD

    # Output the result
    print(dp[n][0])

if __name__ == "__main__":
    main()

C#

using System;
using System.Collections.Generic;

class Program {
    static void Main()
    {
        int n = 3, m = 3;

        List<List<long> > dp = new List<List<long> >(n + 1);
        for (int i = 0; i <= n; i++) {
            dp.Add(new List<long>(new long[2]));
        }

        dp[1][0] = 0;
        // Initialize base case
        dp[1][1] = m;

        // Bottom-up dynamic programming to fill DP table
        for (int i = 2; i <= n; i++) {
            // No consecutive numbers are the same
            dp[i][0] = (dp[i - 1][0] * (m - 2)
                        + dp[i - 1][1] * (m - 1))
                       % 998244353;
            dp[i][1] = dp[i - 1][0] % 998244353;
            dp[1][1] %= 998244353;
        }

        // Output the result
        Console.WriteLine(dp[n][0]);
    }
}

Javascript

const MOD = 998244353;

function main() {
    const n = 3, m = 3;

    // Initialize dp array with dimensions (n+1) x 2 filled with 0s
    let dp = new Array(n + 1).fill(0).map(() => new Array(2).fill(0));

    dp[1][0] = 0;
    // Initialize base case
    dp[1][1] = m;

    // Bottom-up dynamic programming to fill DP table
    for (let i = 2; i <= n; i++) {
        // No consecutive numbers are the same
        dp[i][0] = (dp[i - 1][0] * (m - 2) + dp[i-1][1] * (m - 1)) % MOD;
        dp[i][1] = dp[i - 1][0];
        dp[i][0] %= MOD;
        dp[1][1] %= MOD;
    }

    // Output the result
    console.log(dp[n][0]);
}

main();

Output

6

Time complexity: O(N), where N is total number of integers in the input.
Space complexity: O(N)