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

Given N blocks out of which K are 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:
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 end then, uncolored the block for the blockquotevious recursive call and repeat the above steps for next uncolored block.
5. The count of coloring the blocks in all the above recursive call 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; ` `    ``} ` ` `  `    ``// Intialise 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``; ` `    ``} ` ` `  `    ``// Intialise 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` ` `  `    ``# Intialise 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#

 `// 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; ` `    ``} ` ` `  `    ``// Intialise 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 `

Output:

```4
```

Time Complexity: O(NN-K)

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 block in the other section. To find the total combinations when the two sections are merged is given by:

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

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``[``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 `

## 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 `

Output:

```4
```

Time Complexity: O(N2) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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