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 waysint 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// blockint 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 Codeint 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 approachimport java.util.*;class GFG{static int mod = 1000000007;// Recursive function to count the waysstatic 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// blockstatic 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 Codepublic 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 approachmod = 1000000007# Recursive function to count the waysdef 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# blockdef 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 Codeif __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 approachusing System;class GFG{static int mod = 1000000007;// Recursive function to count the waysstatic 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// blockstatic 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 Codepublic 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 approachlet mod = 1000000007; // Recursive function to count the waysfunction 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// blockfunction 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:
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 <bits/stdc++.h>using namespace std;const int mod = 1000000007;// Function to count the ways to color// blockint 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 Codeint 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 approachimport java.util.*;class GFG{static int mod = 1000000007;// Function to count the ways to color// blockstatic 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 Codepublic 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 approachmod = 1000000007 # Function to count the ways to color# blockdef 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 Codeif __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 approachusing System;class GFG{static int mod = 1000000007;// Function to count the ways to color// blockstatic 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 Codepublic 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 approachlet mod = 1000000007;// Function to count the ways to color// blockfunction 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 blockslet N = 6;// Number of colored blockslet K = 3;let arr=[ 1, 2, 6 ,0];// Function calldocument.write(waysToColor(arr, N, K));// This code is contributed by avanitrachhadiya2155</script> |
Output:
4
Time Complexity: O(N2)
Auxiliary Space: O(52 * 104)
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