Number of ways to color N-K blocks using given operation

Given N blocks, out of which K is colored. These K-colored blocks are denoted by an array arr[]. The task is to count the number of ways to color the remaining uncolored blocks such that only any one of the adjacent blocks, of a colored block, can be colored in one step. Print the answer with modulo 10⁹+7.

Examples:

Input: N = 6, K = 3, arr[] = {1, 2, 6}
Output: 4
Explanation:
The following are the 4 ways to color the blocks(each set represents the order in which blocks are colored):
1. {3, 4, 5}
2. {3, 5, 4}
3. {5, 3, 4}
4. {5, 4, 3}

Input: N = 9, K = 3, A = [3, 6, 7]
Output: 180

Naive Approach: The idea is to use recursion. Below are the steps:

Traverse each block from 1 to N.
If the current block(say b) is not colored, then check whether one of the adjacent blocks is colored or not.
If the adjacent block is colored, then color the current block and recursively iterate to find the next uncolored block.
After the above recursive call ends, then, uncolored the block for the blockquotevious recursive call and repeat the above steps for the next uncolored block.
The count of coloring the blocks in all the above recursive calls gives the number of ways to color the uncolored block.

Below is the implementation of the above approach:

C++

 

// C++ program for the above approach

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

const int mod = 1000000007;

// Recursive function to count the ways
int countWays(int colored[], int count,
              int n)
{

    // Base case
    if (count == n) {
        return 1;
    }

    // Initialise answer to 0
    int answer = 0;

    // Color each uncolored block according
    // to the given condition
    for (int i = 1; i < n + 1; i++) {

        // If any block is uncolored
        if (colored[i] == 0) {

            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1
                || colored[i + 1] == 1) {

                // Color the block
                colored[i] = 1;

                // recursively iterate for
                // next uncolored block
                answer = (answer
                          + countWays(colored,
                                      count + 1,
                                      n))
                         % mod;

                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }

    // Return the final count
    return answer;
}

// Function to count the ways to color
// block
int waysToColor(int arr[], int n, int k)
{

    // Mark which blocks are colored in
    // each recursive step
    int colored[n + 2] = { 0 };

    for (int i = 0; i < k; i++) {
        colored[arr[i]] = 1;
    }

    // Function call to count the ways
    return countWays(colored, k, n);
}

// Driver Code
int main()
{
    // Number of blocks
    int N = 6;

    // Number of colored blocks
    int K = 3;
    int arr[K] = { 1, 2, 6 };

    // Function call
    cout << waysToColor(arr, N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

static int mod = 1000000007;

// Recursive function to count the ways
static int countWays(int colored[], 
                     int count, int n)
{

    // Base case
    if (count == n)
    {
        return 1;
    }

    // Initialise answer to 0
    int answer = 0;

    // Color each uncolored block according
    // to the given condition
    for (int i = 1; i < n + 1; i++) 
    {

        // If any block is uncolored
        if (colored[i] == 0) 
        {

            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1 || 
                colored[i + 1] == 1) 
            {

                // Color the block
                colored[i] = 1;

                // recursively iterate for
                // next uncolored block
                answer = (answer + 
                          countWays(colored, 
                                    count + 1, 
                                    n)) % mod;

                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }

    // Return the final count
    return answer;
}

// Function to count the ways to color
// block
static int waysToColor(int arr[], 
                       int n, int k)
{

    // Mark which blocks are colored in
    // each recursive step
    int colored[] = new int[n + 2];

    for (int i = 0; i < k; i++) 
    {
        colored[arr[i]] = 1;
    }

    // Function call to count the ways
    return countWays(colored, k, n);
}

// Driver Code
public static void main(String[] args)
{
    // Number of blocks
    int N = 6;

    // Number of colored blocks
    int K = 3;
    int arr[] = { 1, 2, 6 };

    // Function call
    System.out.print(waysToColor(arr, N, K));
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
mod = 1000000007

# Recursive function to count the ways
def countWays(colored, count, n):

    # Base case
    if (count == n):
        return 1

    # Initialise answer to 0
    answer = 0

    # Color each uncolored block according
    # to the given condition
    for i in range(1, n + 1):

        # If any block is uncolored
        if (colored[i] == 0):

            # Check if adjacent blocks
            # are colored or not
            if (colored[i - 1] == 1 or 
                colored[i + 1] == 1):

                # Color the block
                colored[i] = 1

                # recursively iterate for
                # next uncolored block
                answer = ((answer + 
                           countWays(colored,
                                     count + 1, 
                                       n)) % mod)

                # Uncolored for the next
                # recursive call
                colored[i] = 0

    # Return the final count
    return answer

# Function to count the ways to color
# block
def waysToColor( arr, n, k):

    # Mark which blocks are colored in
    # each recursive step
    colored = [0] * (n + 2)
    
    for i in range(k):
        colored[arr[i]] = 1

    # Function call to count the ways
    return countWays(colored, k, n)

# Driver Code
if __name__ == "__main__":
    
    # Number of blocks
    N = 6

    # Number of colored blocks
    K = 3
    arr = [ 1, 2, 6 ]

    # Function call
    print(waysToColor(arr, N, K))

# This code is contributed by chitranayal

// C# program for the above approach
using System;
class GFG{

static int mod = 1000000007;

// Recursive function to count the ways
static int countWays(int []colored, 
                     int count, int n)
{

    // Base case
    if (count == n)
    {
        return 1;
    }

    // Initialise answer to 0
    int answer = 0;

    // Color each uncolored block according
    // to the given condition
    for (int i = 1; i < n + 1; i++) 
    {

        // If any block is uncolored
        if (colored[i] == 0) 
        {

            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1 || 
                colored[i + 1] == 1) 
            {

                // Color the block
                colored[i] = 1;

                // recursively iterate for
                // next uncolored block
                answer = (answer + 
                          countWays(colored, 
                                    count + 1, 
                                    n)) % mod;

                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }

    // Return the final count
    return answer;
}

// Function to count the ways to color
// block
static int waysToColor(int []arr, 
                    int n, int k)
{

    // Mark which blocks are colored in
    // each recursive step
    int []colored = new int[n + 2];

    for (int i = 0; i < k; i++) 
    {
        colored[arr[i]] = 1;
    }

    // Function call to count the ways
    return countWays(colored, k, n);
}

// Driver Code
public static void Main()
{
    // Number of blocks
    int N = 6;

    // Number of colored blocks
    int K = 3;
    int []arr = { 1, 2, 6 };

    // Function call
    Console.Write(waysToColor(arr, N, K));
}
}

// This code is contributed by Code_Mech

JavaScript

<script>

// Javascript program for the above approach

let mod = 1000000007;
 
// Recursive function to count the ways
function countWays(colored,
                     count, n)
{
 
    // Base case
    if (count == n)
    {
        return 1;
    }
 
    // Let initialise answer to 0
    let answer = 0;
 
    // Color each uncolored block according
    // to the given condition
    for (let i = 1; i < n + 1; i++)
    {
 
        // If any block is uncolored
        if (colored[i] == 0)
        {
 
            // Check if adjacent blocks
            // are colored or not
            if (colored[i - 1] == 1 ||
                colored[i + 1] == 1)
            {
 
                // Color the block
                colored[i] = 1;
 
                // recursively iterate for
                // next uncolored block
                answer = (answer +
                          countWays(colored,
                                    count + 1,
                                    n)) % mod;
 
                // Uncolored for the next
                // recursive call
                colored[i] = 0;
            }
        }
    }
 
    // Return the final count
    return answer;
}
 
// Function to count the ways to color
// block
function waysToColor(arr, n, k)
{
 
    // Mark which blocks are colored in
    // each recursive step
    let colored = Array.from({length: n+2}, (_, i) => 0);
 
    for (let i = 0; i < k; i++)
    {
        colored[arr[i]] = 1;
    }
 
    // Function call to count the ways
    return countWays(colored, k, n);
}

// Driver Code
    
    // Number of blocks
    let N = 6;
 
    // Number of colored blocks
    let K = 3;
    let arr = [ 1, 2, 6 ];
 
    // Function call
    document.write(waysToColor(arr, N, K));
        
</script>

Output:

Time Complexity: O(N^N-K)

Auxiliary Space: O(N)

Efficient Approach: For solving this problem efficiently we will use the concept of Permutation and Combination. Below are the steps:

1. If the number of blocks between two consecutive colored blocks is x, then the number of ways to color these set of blocks is given by:

ways = 2^x-1

2. Coloring each set of uncolored blocks is independent of the other. Suppose there are x blocks in one section and y blocks in the other section. To find the total combination when the two sections are merged is given by:

total combinations = {n \choose x}*2^{x-1}*2^{y-1}

3. Sort the colored block indices to find the length of each uncolored block section and iterate and find the combination of each two-section using the above formula.

4. Find the Binomial Coefficient using the approach discussed in this article.

Below is the implementation of the above approach:

C++

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

const int mod = 1000000007;

// Function to count the ways to color
// block
int waysToColor(int arr[], int n, int k)
{
    // For storing powers of 2
    int powOf2[500] = { 0 };

    // For storing binomial coefficient
    // values
    int c[500][500];

    // Calculating binomial coefficient
    // using DP
    for (int i = 0; i <= n; i++) {

        c[i][0] = 1;
        for (int j = 1; j <= i; j++) {
            c[i][j] = (c[i - 1][j]
                       + c[i - 1][j - 1])
                      % mod;
        }
    }

    powOf2[0] = powOf2[1] = 1;

    // Calculating powers of 2
    for (int i = 2; i <= n; i++) {

        powOf2[i] = powOf2[i - 1] * 2 % mod;
    }

    int rem = n - k;
    arr[k++] = n + 1;

    // Sort the indices to calculate
    // length of each section
    sort(arr, arr + k);

    // Initialise answer to 1
    int answer = 1;

    for (int i = 0; i < k; i++) {

        // Find the length of each section
        int x = arr[i] - (i - 1 >= 0
                              ? arr[i - 1]
                              : 0)
                - 1;

        // Merge this section
        answer *= c[rem][x] % mod * (i != 0
                                             && i != k - 1
                                         ? powOf2[x]
                                         : 1)
                  % mod;
        rem -= x;
    }

    // Return the final count
    return answer;
}

// Driver Code
int main()
{
    // Number of blocks
    int N = 6;

    // Number of colored blocks
    int K = 3;
    int arr[K] = { 1, 2, 6 };

    // Function call
    cout << waysToColor(arr, N, K);
    return 0;
}

Java

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

class GFG{

static int mod = 1000000007;

// Function to count the ways to color
// block
static int waysToColor(int arr[], int n, int k)
{
    
    // For storing powers of 2
    int powOf2[] = new int[500];

    // For storing binomial coefficient
    // values
    int [][]c = new int[500][500];

    // Calculating binomial coefficient
    // using DP
    for(int i = 0; i <= n; i++) 
    {
       c[i][0] = 1;
       for(int j = 1; j <= i; j++)
       {
          c[i][j] = (c[i - 1][j] + 
                     c[i - 1][j - 1]) % mod;
       }
    }

    powOf2[0] = powOf2[1] = 1;

    // Calculating powers of 2
    for(int i = 2; i <= n; i++) 
    {
       powOf2[i] = powOf2[i - 1] * 2 % mod;
    }

    int rem = n - k;
    arr[k++] = n + 1;
    
    // Sort the indices to calculate
    // length of each section
    Arrays.sort(arr);

    // Initialise answer to 1
    int answer = 1;

    for(int i = 0; i < k; i++)
    {
        
       // Find the length of each section
       int x = arr[i] - (i - 1 >= 0 ? 
                     arr[i - 1] : 0) - 1;
       
       // Merge this section
       answer *= c[rem][x] % mod * (i != 0 && 
                                    i != k - 1 ?
                                    powOf2[x] : 1) %
                                    mod;
       rem -= x;
    }
    
    // Return the final count
    return answer;
}

// Driver Code
public static void main(String[] args)
{
    
    // Number of blocks
    int N = 6;

    // Number of colored blocks
    int K = 3;
    int arr[] = { 1, 2, 6 ,0 };

    // Function call
    System.out.print(waysToColor(arr, N, K));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
mod = 1000000007
 
# Function to count the ways to color
# block
def waysToColor(arr, n, k):
    
    global mod

    # For storing powers of 2
    powOf2 = [0 for i in range(500)]
 
    # For storing binomial coefficient
    # values
    c = [[0 for i in range(500)] for j in range(500)]
 
    # Calculating binomial coefficient
    # using DP
    for i in range(n + 1):
 
        c[i][0] = 1;
        
        for j in range(1, i + 1):
        
            c[i][j] = (c[i - 1][j]+ c[i - 1][j - 1])% mod;
 
    powOf2[0] = 1
    powOf2[1] = 1;
 
    # Calculating powers of 2
    for i in range(2, n + 1):
 
        powOf2[i] = (powOf2[i - 1] * 2) % mod;
    
    rem = n - k;
    arr[k] = n + 1;
    k += 1
 
    # Sort the indices to calculate
    # length of each section
    arr.sort()
 
    # Initialise answer to 1
    answer = 1;
    
    for i in range(k):
        
        x = 0 
        
        # Find the length of each section
        if i - 1 >= 0:
            x = arr[i] - arr[i - 1] -1
        else:
            x = arr[i] - 1 
 
        # Merge this section
        answer = answer * (c[rem][x] % mod) * ((powOf2[x] if (i != 0 and i != k - 1) else 1))% mod
        rem -= x;
 
    # Return the final count
    return answer;

# Driver Code
if __name__=='__main__':
 
    # Number of blocks
    N = 6;
 
    # Number of colored blocks
    K = 3;
    arr = [ 1, 2, 6, 0]
 
    # Function call
    print(waysToColor(arr, N, K))

# This code is contributed by rutvik_56

// C# program for the above approach
using System;
class GFG{

static int mod = 1000000007;

// Function to count the ways to color
// block
static int waysToColor(int []arr, int n, int k)
{
    
    // For storing powers of 2
    int []powOf2 = new int[500];

    // For storing binomial coefficient
    // values
    int [,]c = new int[500, 500];

    // Calculating binomial coefficient
    // using DP
    for(int i = 0; i <= n; i++) 
    {
        c[i, 0] = 1;
        for(int j = 1; j <= i; j++)
        {
            c[i, j] = (c[i - 1, j] + 
                       c[i - 1, j - 1]) % mod;
        }
    }

    powOf2[0] = powOf2[1] = 1;

    // Calculating powers of 2
    for(int i = 2; i <= n; i++) 
    {
        powOf2[i] = powOf2[i - 1] * 2 % mod;
    }

    int rem = n - k;
    arr[k++] = n + 1;
    
    // Sort the indices to calculate
    // length of each section
    Array.Sort(arr);

    // Initialise answer to 1
    int answer = 1;

    for(int i = 0; i < k; i++)
    {
        
        // Find the length of each section
        int x = arr[i] - (i - 1 >= 0 ? 
                arr[i - 1] : 0) - 1;
            
        // Merge this section
        answer *= c[rem, x] % mod * (i != 0 && 
                                     i != k - 1 ?
                                     powOf2[x] : 1) %
                                     mod;
        rem -= x;
    }
    
    // Return the readonly count
    return answer;
}

// Driver Code
public static void Main(String[] args)
{
    
    // Number of blocks
    int N = 6;

    // Number of colored blocks
    int K = 3;
    int []arr = { 1, 2, 6, 0 };

    // Function call
    Console.Write(waysToColor(arr, N, K));
}
}

// This code is contributed by 29AjayKumar

JavaScript

<script>

// JavaScript program for the above approach

let mod = 1000000007;

// Function to count the ways to color
// block
function waysToColor(arr,n,k)
{
    // For storing powers of 2
    let powOf2 = new Array(500);
 
    // For storing binomial coefficient
    // values
    let c = new Array(500);
    for(let i=0;i<500;i++)
    {
        c[i]=new Array(500);
        for(let j=0;j<500;j++)
        {
            c[i][j]=0;
        }
    }
 
    // Calculating binomial coefficient
    // using DP
    for(let i = 0; i <= n; i++)
    {
       c[i][0] = 1;
       for(let j = 1; j <= i; j++)
       {
          c[i][j] = (c[i - 1][j] +
                     c[i - 1][j - 1]) % mod;
       }
    }
 
    powOf2[0] = powOf2[1] = 1;
 
    // Calculating powers of 2
    for(let i = 2; i <= n; i++)
    {
       powOf2[i] = powOf2[i - 1] * 2 % mod;
    }
 
    let rem = n - k;
    arr[k++] = n + 1;
     
    // Sort the indices to calculate
    // length of each section
    arr.sort(function(a,b){return a-b;});
 
    // Initialise answer to 1
    let answer = 1;
 
    for(let i = 0; i < k; i++)
    {
         
       // Find the length of each section
       let x = arr[i] - (i - 1 >= 0 ?
                     arr[i - 1] : 0) - 1;
        
       // Merge this section
       answer *= c[rem][x] % mod * (i != 0 &&
                                    i != k - 1 ?
                                    powOf2[x] : 1) %
                                    mod;
       rem -= x;
    }
     
    // Return the final count
    return answer;
}

// Driver Code
// Number of blocks
let N = 6;

// Number of colored blocks
let K = 3;
let arr=[ 1, 2, 6 ,0];

// Function call
document.write(waysToColor(arr, N, K));

// This code is contributed by avanitrachhadiya2155

</script>

Output:

Time Complexity: O(N²)

Auxiliary Space: O(5² * 10⁴)

Number of ways to color N-K blocks using given operation

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