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Ways to fill N positions using M colors such that there are exactly K pairs of adjacent different colors

  • Difficulty Level : Hard
  • Last Updated : 26 May, 2021

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: 

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Input: N = 3, M = 2, K = 1 
Output:
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 ways
int 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 Code
int 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 approach
class solution
{
static final 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(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 approach
 
max = 4
 
# Recursive function to find the
# required number of ways
def 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
if __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 approach
 
let 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];
    
    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 Code
let 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

 




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