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

• Difficulty Level : Medium
• Last Updated : 15 Nov, 2021

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:
Explanation:
The following are the 4 ways to color the blocks(each set reblockquotesents 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:

1. Traverse each block from 1 to N.
2. If the current block(say b) is not colored, then check whether one of the adjacent blocks is colored or not.
3. If the adjacent block is colored, then color the current block and recursively iterate to find the next uncolored block.
4. 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.
5. 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 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 =  * (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#

 // 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



Output:

4

Time Complexity: O(NN-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 = 2x-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 = 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 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 = { 0 };       // For storing binomial coefficient     // values     int c;       // Calculating binomial coefficient     // using DP     for (int i = 0; i <= n; i++) {           c[i] = 1;         for (int j = 1; j <= i; j++) {             c[i][j] = (c[i - 1][j]                        + c[i - 1][j - 1])                       % mod;         }     }       powOf2 = powOf2 = 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;       // For storing binomial coefficient     // values     int [][]c = new int;       // Calculating binomial coefficient     // using DP     for(int i = 0; i <= n; i++)     {        c[i] = 1;        for(int j = 1; j <= i; j++)        {           c[i][j] = (c[i - 1][j] +                      c[i - 1][j - 1]) % mod;        }     }       powOf2 = powOf2 = 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] = 1;                   for j in range(1, i + 1):                       c[i][j] = (c[i - 1][j]+ c[i - 1][j - 1])% mod;        powOf2 = 1     powOf2 = 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#

 // 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;       // 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 = powOf2 = 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



Output:

4

Time Complexity: O(N2

Auxiliary Space: O(52 * 104)

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