Given three integers N, M and K. The task is to find the number of ways to fill N positions using M colors such that there are exactly K pairs of different adjacent colors.
Examples:
Input: N = 3, M = 2, K = 1
Output: 4
Let the colors be 1 and 2, so the ways are:
1, 1, 2
1, 2, 2
2, 2, 1
2, 1, 1
The above 4 ways have exactly one pair of adjacent elements with different colors.Input: N = 3, M = 3, K = 2
Output: 12
Approach: We can use Dynamic Programming with memoization to solve the above problem. There are N positions to fill, hence the recursive function will be composed of two calls, one if the next position is filled with the same color and the other if it is filled with a different color. Hence, the recursive calls will be:
- countWays(index + 1, cnt), if the next index is filled with the same color.
- (m – 1) * countWays(index + 1, cnt + 1), if the next index is filled with a different color. The number of ways is multiplied by (m – 1).
The basic cases will be:
- If index = n, then a check for the value of cnt is done. If cnt = K then it is a possible way, hence return 1, else return 0.
- To avoid repetitive calls, memoize the returned value in a 2-D array and return this value if the recursive call with the same parameters is done again.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define max 4// Recursive function to find the required number of waysint countWays(int index, int cnt, int dp[][max], int n, int m, int k){ // When all positions are filled if (index == n) { // If adjacent pairs are exactly K if (cnt == k) return 1; else return 0; } // If already calculated if (dp[index][cnt] != -1) return dp[index][cnt]; int ans = 0; // Next position filled with same color ans += countWays(index + 1, cnt, dp, n, m, k); // Next position filled with different color // So there can be m-1 different colors ans += (m - 1) * countWays(index + 1, cnt + 1, dp, n, m, k); return dp[index][cnt] = ans;}// Driver Codeint main(){ int n = 3, m = 3, k = 2; int dp[n + 1][max]; memset(dp, -1, sizeof dp); cout << m * countWays(1, 0, dp, n, m, k);} |
Java
//Java implementation of the approachclass solution{static final int max=4; // Recursive function to find the required number of waysstatic int countWays(int index, int cnt, int dp[][], int n, int m, int k){ // When all positions are filled if (index == n) { // If adjacent pairs are exactly K if (cnt == k) return 1; else return 0; } // If already calculated if (dp[index][cnt] != -1) return dp[index][cnt]; int ans = 0; // Next position filled with same color ans += countWays(index + 1, cnt, dp, n, m, k); // Next position filled with different color // So there can be m-1 different colors ans += (m - 1) * countWays(index + 1, cnt + 1, dp, n, m, k); return dp[index][cnt] = ans;} // Driver Codepublic static void main(String args[]){ int n = 3, m = 3, k = 2; int dp[][]= new int [n + 1][max]; for(int i=0;i<n+1;i++) for(int j=0;j<max;j++) dp[i][j]=-1; System.out.println(m * countWays(1, 0, dp, n, m, k));}}//contributed by Arnab Kundu |
Python 3
# Python 3 implementation of the approachmax = 4# Recursive function to find the # required number of waysdef countWays(index, cnt, dp, n, m, k): # When all positions are filled if (index == n) : # If adjacent pairs are exactly K if (cnt == k): return 1 else: return 0 # If already calculated if (dp[index][cnt] != -1): return dp[index][cnt] ans = 0 # Next position filled with same color ans += countWays(index + 1, cnt, dp, n, m, k) # Next position filled with different color # So there can be m-1 different colors ans += (m - 1) * countWays(index + 1, cnt + 1, dp, n, m, k) dp[index][cnt] = ans return dp[index][cnt]# Driver Codeif __name__ == "__main__": n = 3 m = 3 k = 2 dp = [[-1 for x in range(n + 1)] for y in range(max)] print(m * countWays(1, 0, dp, n, m, k))# This code is contributed by ita_c |
C#
// C# implementation of the approach using System;class solution { static int max=4; // Recursive function to find the required number of ways static int countWays(int index, int cnt, int [,]dp, int n, int m, int k) { // When all positions are filled if (index == n) { // If adjacent pairs are exactly K if (cnt == k) return 1; else return 0; } // If already calculated if (dp[index,cnt] != -1) return dp[index,cnt]; int ans = 0; // Next position filled with same color ans += countWays(index + 1, cnt, dp, n, m, k); // Next position filled with different color // So there can be m-1 different colors ans += (m - 1) * countWays(index + 1, cnt + 1, dp, n, m, k); return dp[index,cnt] = ans; } // Driver Code public static void Main() { int n = 3, m = 3, k = 2; int [,]dp= new int [n + 1,max]; for(int i=0;i<n+1;i++) for(int j=0;j<max;j++) dp[i,j]=-1; Console.WriteLine(m * countWays(1, 0, dp, n, m, k)); } // This code is contributed by Ryuga} |
PHP
<?php// PHP implementation of the approach $GLOBALS['max'] = 4; // Recursive function to find the // required number of ways function countWays($index, $cnt, $dp, $n, $m, $k) { // When all positions are filled if ($index == $n) { // If adjacent pairs are exactly K if ($cnt == $k) return 1; else return 0; } // If already calculated if ($dp[$index][$cnt] != -1) return $dp[$index][$cnt]; $ans = 0; // Next position filled with same color $ans += countWays($index + 1, $cnt, $dp, $n, $m, $k); // Next position filled with different color // So there can be m-1 different colors $ans += ($m - 1) * countWays($index + 1, $cnt + 1, $dp, $n, $m, $k); $dp[$index][$cnt] = $ans; return $dp[$index][$cnt];} // Driver Code $n = 3;$m = 3;$k = 2; $dp = array(array());for($i = 0; $i < $n + 1; $i++) for($j = 0; $j < $GLOBALS['max']; $j++) $dp[$i][$j] = -1;echo $m * countWays(1, 0, $dp, $n, $m, $k);// This code is contributed by aishwarya.27?> |
Javascript
<script>//Javascript implementation of the approachlet max=4;// Recursive function to find the required number of waysfunction countWays(index,cnt,dp,n,m,k){ // When all positions are filled if (index == n) { // If adjacent pairs are exactly K if (cnt == k) return 1; else return 0; } // If already calculated if (dp[index][cnt] != -1) return dp[index][cnt]; let ans = 0; // Next position filled with same color ans += countWays(index + 1, cnt, dp, n, m, k); // Next position filled with different color // So there can be m-1 different colors ans += (m - 1) * countWays(index + 1, cnt + 1, dp, n, m, k); return dp[index][cnt] = ans;}// Driver Codelet n = 3, m = 3, k = 2;let dp=new Array(n+1);for(let i=0;i<n+1;i++){ dp[i]=new Array(max); for(let j=0;j<max;j++) dp[i][j]=-1;} document.write(m * countWays(1, 0, dp, n, m, k)); // This code is contributed by rag2127</script> |
Output:
12
Time Complexity: O(n*4), as we are using recursion and the function will be called N times, where N is the number of positions to be filled.
Auxiliary Space: O(n*4), as we are using extra space for the dp matrix, where N is the number of positions to be filled.
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a DP to store the solution of the subproblems and initialize it with 0.
- Initialize the DP with base cases
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in m * dp[1][0].
Implementation :
C++
#include <bits/stdc++.h>using namespace std;#define max 4 int countWays(int n, int m, int k) { int dp[n + 1][max]; memset(dp, 0, sizeof dp); // base case for(int cnt=0; cnt<=k; cnt++) { if(cnt == k) { dp[n][cnt] = 1; } else { dp[n][cnt] = 0; } } // fill the table using bottom-up approach for(int index=n-1; index>=1; index--) { for(int cnt=0; cnt<=k; cnt++) { int ans = 0; // if the current index is not selected ans += dp[index+1][cnt]; // if the current index is selected if(cnt+1 <= k) { ans += (m-1) * dp[index+1][cnt+1]; } dp[index][cnt] = ans; } } return m * dp[1][0];} // Driver Codeint main(){ int n = 3, m = 3, k = 2; cout << countWays(n, m, k);} |
Python
def countWays(n, m, k): MAX = 4 dp = [[0] * MAX for i in range(n + 1)] # base case for cnt in range(k + 1): if cnt == k: dp[n][cnt] = 1 else: dp[n][cnt] = 0 # fill the table using bottom-up approach for index in range(n - 1, 0, -1): for cnt in range(k + 1): ans = 0 # if the current index is not selected ans += dp[index + 1][cnt] # if the current index is selected if cnt + 1 <= k: ans += (m - 1) * dp[index + 1][cnt + 1] dp[index][cnt] = ans return m * dp[1][0] # Driver Codeif __name__ == "__main__": n, m, k = 3, 3, 2 print(countWays(n, m, k)) |
Output:
12
Time Complexity: O(n*4)
Auxiliary Space: O(n*4)
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