Question

There are N boxes placed in a row and M diamonds are distributed in these boxes such that each box contains at least 1 diamond into it. Each diamond is of certain colour and has a certain value. Formally, diamond i has colour Ci and value Vi. Each diamond is put in a certain box denoted by Bi. Now your task is to take exactly one diamond from each box such that total value of diamonds taken is maximum and diamonds taken from two consecutive boxes are of different colours. If it is impossible to satisfy the conditions, output -1.

Answer 1

Answer:

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

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)

Explanation:

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 1st, 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 colors and solve for dp[i – 1][j – 1] recursively.

Below is the implementation of the above approach:

// 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 M = 1001;

const int MOD = 998244353;

int dp[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

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

int main()

{

int N = 3, M = 3, K = 0;

memset(dp, -1, sizeof(dp));

// 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;