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

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:
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 ` `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

## 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

## PHP

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## Javascript

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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 ` `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 Code` `int` `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 Code` `if` `__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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