Color N boxes using M colors such that K boxes have different color from the box on its left

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Given N number of boxes arranged in a row and M number of colors. The task is to find the number of ways to paint those N boxes using M colors such that there are exactly K boxes with a color different from the color of the box on its left. Print this answer modulo 998244353.
Examples:

Input: N = 3, M = 3, K = 0
Output: 3
Since the value of K is zero, no box can have a different color from color of the box on its left. Thus, all boxes should be painted with same color and since there are 3 types of colors, so there are total 3 ways.
Input: N = 3, M = 2, K = 1
Output: 4
Let’s number the colors as 1 and 2. Four possible sequences of painting 3 boxes with 1 box having different color from color of box on its left are (1 2 2), (1 1 2), (2 1 1) (2 2 1)

Prerequisites : Dynamic Programming

Approach: This problem can be solved using dynamic programming where dp[i][j] will denote the number of ways to paint i boxes using M colors such that there are exactly j boxes with a color different from the color of the box on its left. For every current box except 1^st, either we can paint the same color as painted on its left box and solve for dp[i – 1][j] or we can paint it with remaining M – 1 color and solve for dp[i – 1][j – 1] recursively.

Steps to follow to according to above approach:

If idx is greater than N, then check if diff is equal to K.
- *If diff is equal to K, then return 1.
  *Else, return 0.
If the result for the current value of idx and diff is not equal to -1 then return the precomputed result dp[idx].
Otherwise, recursively call solve() function with idx+1, diff, N, M, and K and store the result in the variable ans.
recursively call solve() function with idx+1, diff+1, N, M, and K, and multiply the result with (M-1).
add the value obtained in step 5 to the result obtained in step 4, and take its modulo with MOD.
store the result obtained in step 6 in the dp array for the current value of idx and diff, and return the same value.

Below is the code to implement the above approach:

C++

// CPP Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
#include <bits/stdc++.h>
using namespace std;
 
const int MOD = 998244353;
 
vector<vector<int>> dp;
 
// This function returns the required number
// of ways where idx is the current index and
// diff is number of boxes having different
// color from box on its left
int solve(int idx, int diff, int N, int M, int K)
{
    // Base Case
    if (idx > N) {
        if (diff == K)
            return 1;
        return 0;
    }
 
    // If already computed
    if (dp[idx][ diff] != -1)
        return dp[idx][ diff];
 
    // Either paint with same color as
    // previous one
    int ans = solve(idx + 1, diff, N, M, K);
 
    // Or paint with remaining (M - 1)
    // colors
    ans = ans % MOD + ((M - 1) % MOD * solve(idx + 1, diff + 1, N, M, K) % MOD) % MOD;
 
    return dp[idx][ diff] = ans;
}
 
// Driver code
int main()
{
    int N = 3, M = 3, K = 0;
    dp = vector<vector<int>>(N+1,vector<int>(N+1,-1));
 
    // Multiply M since first box can be
    // painted with any of the M colors and
    // start solving from 2nd box
    cout << (M * solve(2, 0, N, M, K)) << endl;
 
    return 0;
}

Java

// Java Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
 
class GFG
{
     
    static int M = 1001;
    static int MOD = 998244353;
 
    static int[][] dp = new int[M][M];
 
    // This function returns the required number
    // of ways where idx is the current index and
    // diff is number of boxes having different
    // color from box on its left
    static int solve(int idx, int diff,
                        int N, int M, int K)
    {
        // Base Case
        if (idx > N) 
        {
            if (diff == K)
                return 1;
            return 0;
        }
 
        // If already computed
        if (dp[idx][ diff] != -1)
            return dp[idx][ diff];
 
        // Either paint with same color as
        // previous one
        int ans = solve(idx + 1, diff, N, M, K);
 
        // Or paint with remaining (M - 1)
        // colors
        ans += (M - 1) * solve(idx + 1, 
                diff + 1, N, M, K);
 
        return dp[idx][ diff] = ans % MOD;
    }
 
    // Driver code
    public static void main (String[] args) 
    {
        int N = 3, M = 3, K = 0;
        for(int i = 0; i <= M; i++)
            for(int j = 0; j <= M; j++)
                dp[i][j] = -1;
     
        // Multiply M since first box can be
        // painted with any of the M colors and
        // start solving from 2nd box
        System.out.println((M * solve(2, 0, N, M, K)));
    }
}
 
// This code is contributed by mits

Python3

# Python3 Program to Paint N boxes using M 
# colors such that K boxes have color 
# different from color of box on its left 
 
M = 1001; 
MOD = 998244353; 
 
dp = [[-1]* M ] * M
 
# This function returns the required number 
# of ways where idx is the current index and 
# diff is number of boxes having different 
# color from box on its left 
def solve(idx, diff, N, M, K) :
     
    # Base Case 
    if (idx > N) : 
        if (diff == K) :
            return 1
        return 0
 
    # If already computed 
    if (dp[idx][ diff] != -1) :
        return dp[idx]; 
 
    # Either paint with same color as 
    # previous one 
    ans = solve(idx + 1, diff, N, M, K); 
 
    # Or paint with remaining (M - 1) 
    # colors 
    ans += (M - 1) * solve(idx + 1, diff + 1, N, M, K); 
 
    dp[idx][ diff] = ans % MOD; 
     
    return dp[idx][ diff]
 
# Driver code 
if __name__ == "__main__" : 
 
    N = 3
    M = 3
    K = 0
 
    # Multiply M since first box can be 
    # painted with any of the M colors and 
    # start solving from 2nd box 
    print(M * solve(2, 0, N, M, K)) 
 
# This code is contributed by Ryuga

C#

// C# Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
using System;
class GFG
{
     
static int M = 1001;
static int MOD = 998244353;
 
static int[,] dp = new int[M, M];
 
// This function returns the required number
// of ways where idx is the current index and
// diff is number of boxes having different
// color from box on its left
static int solve(int idx, int diff,
                 int N, int M, int K)
{
    // Base Case
    if (idx > N) 
    {
        if (diff == K)
            return 1;
        return 0;
    }
 
    // If already computed
    if (dp[idx, diff] != -1)
        return dp[idx, diff];
 
    // Either paint with same color as
    // previous one
    int ans = solve(idx + 1, diff, N, M, K);
 
    // Or paint with remaining (M - 1)
    // colors
    ans += (M - 1) * solve(idx + 1, 
                diff + 1, N, M, K);
 
    return dp[idx, diff] = ans % MOD;
}
 
// Driver code
public static void Main () 
{
    int N = 3, M = 3, K = 0;
    for(int i = 0; i <= M; i++)
        for(int j = 0; j <= M; j++)
            dp[i, j] = -1;
 
    // Multiply M since first box can be
    // painted with any of the M colors and
    // start solving from 2nd box
    Console.WriteLine((M * solve(2, 0, N, M, K)));
}
}
 
// This code is contributed by chandan_jnu

Javascript

<script>
 
    // JavaScript Program to Paint N boxes using M
    // colors such that K boxes have color
    // different from color of box on its left
     
    let m = 1001;
    let MOD = 998244353;
   
    let dp = new Array(m);
    for(let i = 0; i < m; i++)
    {
        dp[i] = new Array(m);
        for(let j = 0; j < m; j++)
        {
            dp[i][j] = 0;
        }
    }
   
    // This function returns the required number
    // of ways where idx is the current index and
    // diff is number of boxes having different
    // color from box on its left
    function solve(idx, diff, N, M, K)
    {
        // Base Case
        if (idx > N) 
        {
            if (diff == K)
                return 1;
            return 0;
        }
   
        // If already computed
        if (dp[idx][ diff] != -1)
            return dp[idx][ diff];
   
        // Either paint with same color as
        // previous one
        let ans = solve(idx + 1, diff, N, M, K);
   
        // Or paint with remaining (M - 1)
        // colors
        ans += (M - 1) * solve(idx + 1, 
                diff + 1, N, M, K);
          dp[idx][ diff] = ans % MOD;
        return dp[idx][ diff];
    }
     
    let N = 3, M = 3, K = 0;
    for(let i = 0; i <= M; i++)
      for(let j = 0; j <= M; j++)
        dp[i][j] = -1;
 
    // Multiply M since first box can be
    // painted with any of the M colors and
    // start solving from 2nd box
    document.write((M * solve(2, 0, N, M, K)));
     
</script>

PHP

<?php
// PHP Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
 
$M = 1001;
$MOD = 998244353;
 
$dp = array_fill(0, $M,
      array_fill(0, $M, -1));
 
// This function returns the required number
// of ways where idx is the current index 
// and diff is number of boxes having 
// different color from box on its left
function solve($idx, $diff, $N, $M, $K)
{
    global $dp, $MOD;
     
    // Base Case
    if ($idx > $N) 
    {
        if ($diff == $K)
            return 1;
        return 0;
    }
 
    // If already computed
    if ($dp[$idx][$diff] != -1)
        return $dp[$idx][$diff];
 
    // Either paint with same color 
    // as previous one
    $ans = solve($idx + 1, $diff, $N, $M, $K);
 
    // Or paint with remaining (M - 1)
    // colors
    $ans += ($M - 1) * solve($idx + 1,
             $diff + 1, $N, $M, $K);
 
    return $dp[$idx][$diff] = $ans % $MOD;
}
 
// Driver code
$N = 3;
$M = 3;
$K = 0;
 
// Multiply M since first box can be
// painted with any of the M colors and
// start solving from 2nd box
echo ($M * solve(2, 0, $N, $M, $K));
 
// This code is contributed by chandan_jnu
?>

Output

Time Complexity: O(M*M)
Auxiliary Space: O(M*M)

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 .
Initialize the DP with base cases by initializing the first row of DP.
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 dp[1][K].

Implementation :

C++

// CPP Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
 
#include <bits/stdc++.h>
using namespace std;
 
const int MOD = 998244353;
 
int dp[2][1010];
 
// This function returns the required number
// of ways where idx is the current index and
// diff is number of boxes having different
// color from box on its left
int solve(int N, int M, int K)
{
    // Initialize the first row of dp table
    for (int i = 0; i <= N; i++) {
        dp[0][i] = 1;
    }
 
    // Process the remaining rows
    for (int i = 2; i <= N + 1; i++) {
        for (int j = 0; j <= N; j++) {
            int ans = dp[0][j];
 
            if (j == 0) {
                ans = (M * dp[1][j]) % MOD;
            } else {
                ans = (ans % MOD + ((M - 1) % MOD * dp[1][j - 1]) % MOD) % MOD;
            }
 
            dp[0][j] = ans;
        }
 
        // Swap the arrays
        swap(dp[0], dp[1]);
    }
     
    // return final answer
    return dp[1][K];
}
 
// Driver code
int main()
{
    int N = 3, M = 3, K = 0;
     
    // function call
    cout << (M * solve(N, M, K + 1)) % MOD << endl;
 
    return 0;
}

Java

// Java Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
 
import java.util.*;
 
public class Main {
  static final int MOD = 998244353;
 
static int[][] dp;
 
// This function returns the required number
// of ways where idx is the current index and
// diff is number of boxes having different
// color from box on its left
static int solve(int N, int M, int K) {
    // Initialize the first row of dp table
    for (int i = 0; i <= N; i++) {
        dp[0][i] = 1;
    }
 
    // Process the remaining rows
    for (int i = 2; i <= N + 1; i++) {
        for (int j = 0; j <= N; j++) {
            int ans = dp[0][j];
 
            if (j == 0) {
                ans = (M * dp[1][j]) % MOD;
            } else {
                ans = (ans % MOD + ((M - 1) % MOD * dp[1][j - 1]) % MOD) % MOD;
            }
 
            dp[0][j] = ans;
        }
 
        // Swap the arrays
        int[] temp = dp[0];
        dp[0] = dp[1];
        dp[1] = temp;
    }
 
    // return final answer
    return dp[1][K];
}
 
// Driver code
public static void main(String[] args) {
    int N = 3, M = 3, K = 0;
 
    // Initialize dp array
    dp = new int[2][1010];
 
    // function call
    System.out.println((M * solve(N, M, K + 1)) % MOD);
}
}

Python3

MOD = 998244353
 
# This function returns the required number
# of ways where idx is the current index and
# diff is number of boxes having different
# color from box on its left
def solve(N, M, K):
    global dp
     
    # Initialize the first row of dp table
    for i in range(N+1):
        dp[0][i] = 1
 
    # Process the remaining rows
    for i in range(2, N+2):
        for j in range(N+1):
            ans = dp[0][j]
 
            if j == 0:
                ans = (M * dp[1][j]) % MOD
            else:
                ans = (ans % MOD + ((M - 1) % MOD * dp[1][j - 1]) % MOD) % MOD
 
            dp[0][j] = ans
 
        # Swap the arrays
        dp[0], dp[1] = dp[1], dp[0]
 
    # return final answer
    return dp[1][K]
 
# Driver code
if __name__ == '__main__':
    N, M, K = 3, 3, 0
 
    # Initialize dp array
    dp = [[0] * 1010 for _ in range(2)]
 
    # function call
    print((M * solve(N, M, K+1)) % MOD)

Javascript

// JavaScript Program to Paint N boxes using M
// colors such that K boxes have color
// different from color of box on its left
 
const MOD = 998244353;
 
// This function returns the required number
// of ways where idx is the current index and
// diff is the number of boxes having a different
// color from the box on its left
function solve(N, M, K) {
    // Initialize a 2D array for dynamic programming
    const dp = new Array(2).fill().map(() => new Array(N + 1).fill(0));
 
    // Initialize the first row of dp table
    for (let i = 0; i <= N; i++) {
        dp[0][i] = 1;
    }
 
    // Process the remaining rows
    for (let i = 2; i <= N + 1; i++) {
        for (let j = 0; j <= N; j++) {
            let ans = dp[0][j];
 
            if (j === 0) {
                ans = (M * dp[1][j]) % MOD;
            } else {
                ans = (ans % MOD + ((M - 1) % MOD * dp[1][j - 1]) % MOD) % MOD;
            }
 
            dp[0][j] = ans;
        }
 
        // Swap the arrays
        [dp[0], dp[1]] = [dp[1], dp[0]];
    }
 
    // Return the final answer
    return dp[1][K];
}
 
// Driver code
const N = 3, M = 3, K = 0;
 
// Function call
console.log((M * solve(N, M, K + 1)) % MOD);
 
// This code is contributed by prasad264

Output:

Time Complexity: O(N * N)
Auxiliary Space: O(N)

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