Count ways to generate an array having distinct elements at M consecutive indices

Given an array arr[] consisting of N integers in the range [0, M] and an integer M, the task is to count the number of ways to replace all array elements whose value is 0 with non-zero values from the range [0, M] such that all possible M consecutive elements are distinct.

Examples:

Input: arr[] = { 1, 0, 3, 0, 0 }, M = 4 
Output: 2 
Explanation:
Possible ways to replace arr[1], arr[3] and arr[4] with non-zero values such that no M( = 4) consecutive elements contains any duplicate elements are { { 1, 2, 3, 4, 1 }, { 1, 4, 3, 2, 1 } }. 
Therefore, the required output is 2.

Input: arr[] = {0, 1, 2, 1, 0}, M = 4 
Output: 0 
Explanation:No such arrangements possible.

Approach: The idea is to replace 0s with non-zero elements such that arr[i] must be equal to arr[i % M]. Follow the steps below to solve the problem:

• Initialize an array B[] of size M + 1, to store M consecutive array array elements such that arr[i] is equal to B[i % M].
• Traverse the array and check the following conditions:
  • If arr[i] is not 0 and B[i % M] is 0, then B[i % M] will be equal to arr[i] as this number should be present as it is.
  • If arr[i] is not equal to B[i % M], then print 0 as there exist no such arrangements.
• Calculate the count of 0s in the array B[], say X.
• Then, there exist factorial X possible arrangements, therefore, print the value of factorial X .

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Modular function
// to calculate factorial
long long int Fact(int N)
{
    // Stores factorial of N
    long long int result = 1;

    // Iterate over the range [1, N]
    for (int i = 1; i <= N; i++) {

        // Update result
        result = (result * i);
    }

    return result;
}

// Function to count ways to replace array
// elements having 0s with non-zero elements
// such that any M consecutive elements are distinct
void numberOfWays(int M, int arr[], int N)
{

    // Store m consecutive distinct elements
    // such that arr[i] is equal to B[i % M]
    int B[M] = { 0 };

    // Stores frequency of array elements
    int counter[M + 1] = { 0 };

    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {

        // If arr[i] is non-zero
        if (arr[i] != 0) {

            // If B[i % M] is equal to 0
            if (B[i % M] == 0) {

                // Update B[i % M]
                B[i % M] = arr[i];

                // Update frequency of arr[i]
                counter[arr[i]]++;

                // If a duplicate element found
                // in M consecutive elements
                if (counter[arr[i]] > 1) {
                    cout << 0 << endl;
                    return;
                }
            }

            // Handling the case of
            // inequality
            else if (B[i % M] != arr[i]) {

                cout << 0 << endl;
                return;
            }
        }
    }

    // Stores count of 0s
    // in B[]
    int cnt = 0;

    // Traverse the array, B[]
    for (int i = 0; i < M; i++) {

        // If B[i] is 0
        if (B[i] == 0) {

            // Update cnt
            cnt++;
        }
    }

    // Calculate factorial
    cout << Fact(cnt) << endl;
}

// Driver Code
int main()
{

    // Given M
    int M = 4;

    // Given array
    int arr[] = { 1, 0, 3, 0, 0 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    numberOfWays(M, arr, N);
}

Java

// Java program for the above approach
import java.io.*;

class GFG{

// Modular function
// to calculate factorial
static int Fact(int N)
{

    // Stores factorial of N
    int result = 1;

    // Iterate over the range [1, N]
    for(int i = 1; i <= N; i++)
    {

        // Update result
        result = (result * i);
    }
    return result;
}

// Function to count ways to replace array
// elements having 0s with non-zero elements
// such that any M consecutive elements are distinct
static void numberOfWays(int M, int[] arr, int N)
{

    // Store m consecutive distinct elements
    // such that arr[i] is equal to B[i % M]
    int[] B = new int[M];

    // Stores frequency of array elements
    int[] counter = new int[M + 1];

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // If arr[i] is non-zero
        if (arr[i] != 0)
        {

            // If B[i % M] is equal to 0
            if (B[i % M] == 0)
            {

                // Update B[i % M]
                B[i % M] = arr[i];

                // Update frequency of arr[i]
                counter[arr[i]]++;

                // If a duplicate element found
                // in M consecutive elements
                if (counter[arr[i]] > 1)
                {
                    System.out.println(0);
                    return;
                }
            }

            // Handling the case of
            // inequality
            else if (B[i % M] != arr[i])
            {
                System.out.println(0);
                return;
            }
        }
    }

    // Stores count of 0s
    // in B[]
    int cnt = 0;

    // Traverse the array, B[]
    for(int i = 0; i < M; i++)
    {

        // If B[i] is 0
        if (B[i] == 0)
        {

            // Update cnt
            cnt++;
        }
    }

    // Calculate factorial
    System.out.println(Fact(cnt));
}

// Driver Code
public static void main(String[] args)
{

    // Given M
    int M = 4;

    // Given array
    int[] arr = new int[]{ 1, 0, 3, 0, 0 };

    // Size of the array
    int N = arr.length;

    // Function Call
    numberOfWays(M, arr, N);
}
}

// This code is contributed by Dharanendra L V

Python3

# Python3 program for the above approach

# Modular function
# to calculate factorial
def Fact(N):

    # Stores factorial of N
    result = 1

    # Iterate over the range [1, N]
    for i in range(1, N + 1):

        # Update result
        result = (result * i)

    return result

# Function to count ways to replace array
# elements having 0s with non-zero elements
# such that any M consecutive elements are distinct
def numberOfWays(M, arr, N):

    # Store m consecutive distinct elements
    # such that arr[i] is equal to B[i % M]
    B = [0] * (M)

    # Stores frequency of array elements
    counter = [0] * (M + 1)

    # Traverse the array arr
    for i in range(0, N):

        # If arr[i] is non-zero
        if (arr[i] != 0):

            # If B[i % M] is equal to 0
            if (B[i % M] == 0):

                # Update B[i % M]
                B[i % M] = arr[i]

                # Update frequency of arr[i]
                counter[arr[i]] += 1

                # If a duplicate element found
                # in M consecutive elements
                if (counter[arr[i]] > 1):
                    print(0)
                    return

            # Handling the case of
            # inequality
            elif (B[i % M] != arr[i]):
                print(0)
                return

    # Stores count of 0s
    # in B
    cnt = 0

    # Traverse the array, B
    for i in range(0, M):

        # If B[i] is 0
        if (B[i] == 0):

            # Update cnt
            cnt += 1

    # Calculate factorial
    print(Fact(cnt))

# Driver Code
if __name__ == '__main__':

    # Given M
    M = 4

    # Given array
    arr = [ 1, 0, 3, 0, 0 ]

    # Size of the array
    N = len(arr)

    # Function Call
    numberOfWays(M, arr, N)

# This code is contributed by shikhasingrajput

C#

// C# program for the above approach
using System;

class GFG{

// Modular function
// to calculate factorial
static int Fact(int N)
{

    // Stores factorial of N
    int result = 1;

    // Iterate over the range [1, N]
    for(int i = 1; i <= N; i++)
    {

        // Update result
        result = (result * i);
    }
    return result;
}

// Function to count ways to replace array
// elements having 0s with non-zero elements
// such that any M consecutive elements are distinct
static void numberOfWays(int M, int[] arr, int N)
{

    // Store m consecutive distinct elements
    // such that arr[i] is equal to B[i % M]
    int[] B = new int[M];

    // Stores frequency of array elements
    int[] counter = new int[M + 1];

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // If arr[i] is non-zero
        if (arr[i] != 0)
        {

            // If B[i % M] is equal to 0
            if (B[i % M] == 0)
            {

                // Update B[i % M]
                B[i % M] = arr[i];

                // Update frequency of arr[i]
                counter[arr[i]]++;

                // If a duplicate element found
                // in M consecutive elements
                if (counter[arr[i]] > 1)
                {
                    Console.WriteLine(0);
                    return;
                }
            }

            // Handling the case of
            // inequality
            else if (B[i % M] != arr[i])
            {
                Console.WriteLine(0);
                return;
            }
        }
    }

    // Stores count of 0s
    // in B[]
    int cnt = 0;

    // Traverse the array, B[]
    for(int i = 0; i < M; i++)
    {

        // If B[i] is 0
        if (B[i] == 0)
        {

            // Update cnt
            cnt++;
        }
    }

    // Calculate factorial
    Console.WriteLine(Fact(cnt));
}

// Driver Code
static public void Main()
{

    // Given M
    int M = 4;

    // Given array
    int[] arr = new int[]{ 1, 0, 3, 0, 0 };

    // Size of the array
    int N = arr.Length;

    // Function Call
    numberOfWays(M, arr, N);
}
}

// This code is contributed by Dharanendra L V

Javascript

<script>

// Javascript program of the above approach

// Modular function
// to calculate factorial
function Fact(N)
{

    // Stores factorial of N
    let result = 1;

    // Iterate over the range [1, N]
    for(let i = 1; i <= N; i++)
    {

        // Update result
        result = (result * i);
    }
    return result;
}

// Function to count ways to replace array
// elements having 0s with non-zero elements
// such that any M consecutive elements are distinct
function numberOfWays(M, arr, N)
{

    // Store m consecutive distinct elements
    // such that arr[i] is equal to B[i % M]
    let B = new Array(M).fill(0);

    // Stores frequency of array elements
    let counter = new Array(M+1).fill(0);

    // Traverse the array arr[]
    for(let i = 0; i < N; i++)
    {

        // If arr[i] is non-zero
        if (arr[i] != 0)
        {

            // If B[i % M] is equal to 0
            if (B[i % M] == 0)
            {

                // Update B[i % M]
                B[i % M] = arr[i];

                // Update frequency of arr[i]
                counter[arr[i]]++;

                // If a duplicate element found
                // in M consecutive elements
                if (counter[arr[i]] > 1)
                {
                    document.write(0);
                    return;
                }
            }

            // Handling the case of
            // inequality
            else if (B[i % M] != arr[i])
            {
                document.write(0);
                return;
            }
        }
    }

    // Stores count of 0s
    // in B[]
    let cnt = 0;

    // Traverse the array, B[]
    for(let i = 0; i < M; i++)
    {

        // If B[i] is 0
        if (B[i] == 0)
        {

            // Update cnt
            cnt++;
        }
    }

    // Calculate factorial
    document.write(Fact(cnt));
}

    // Driver Code

          // Given M
    let M = 4;

    // Given array
    let arr = [ 1, 0, 3, 0, 0 ];

    // Size of the array
    let N = arr.length;

    // Function Call
    numberOfWays(M, arr, N);

</script>

Output:

2

Time Complexity: O(N)
Auxiliary Space: O(N)