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Treasure and Cities

  • Difficulty Level : Hard
  • Last Updated : 23 Jul, 2021

Given n cities: x1, x2, …… xn: each associated with T[i] (treasure) and C[i] (color). You can choose to visit a city or skip it. Only moving in the forward direction is allowed.When you visit a city you receive the following amount: 

  1. A*T[i] if the color of visited city is the same as color of previously visited city
  2. B*T[i] if this is the first city visited or if the color of the visited city is different from the color of the previously visited city.The values of T[i], A and B can be negative while C[i] ranges from 1 to n.

We have to compute the maximum profit possible.

Examples:  

Input :  A = -5, B = 7
Treasure = {4, 8, 2, 9}
color = {2, 2, 3, 2}
Output : 133
Visit city 2, 3 and 4. Profit = 8*7+2*7+9*7 = 133

Input : A = 5, B = -7 
Treasure = {4, 8, 2, 9}
color = {2, 2, 3, 2}
Output: 57
Visit city 1, 2, 4. Profit = (-7)*4+8*5+9*5 = 57

Source : Oracle Interview Experience Set 61.

This is a variation of standard 0/1 Knapsack problem. The idea is to either visit a city or skip it and return the maximum of both the cases.

Below is the solution of above problem. 

C++




#include <bits/stdc++.h>
using namespace std;
 
// k is current index and col is previous color.
int MaxProfit(int treasure[], int color[], int n,
              int k, int col, int A, int B)
{
    int sum = 0;
 
    if (k == n) // base case
        return 0;
 
    // we have two options
    // either visit current city or skip that
 
    // check if color of this city is equal
    // to prev visited city
    if (col == color[k])
        sum += max(A * treasure[k] +
                MaxProfit(treasure, color, n,
                       k + 1, color[k], A, B),
                MaxProfit(treasure, color, n,
                          k + 1, col, A, B));
    else
        sum += max(B * treasure[k] +                                         
                MaxProfit(treasure, color, n,
                       k + 1, color[k], A, B),
               MaxProfit(treasure, color, n,
                          k + 1, col, A, B));
 
    // return max of both options
    return sum;
}
 
int main()
{
 
    int A = -5, B = 7;
    int treasure[] = { 4, 8, 2, 9 };
    int color[] = { 2, 2, 6, 2 };
    int n = sizeof(color) / sizeof(color[0]);
 
    // Initially begin with color 0
    cout << MaxProfit(treasure, color, n, 0, 0, A, B);
    return 0;
}

Java




class GFG{
 
// k is current index and col is previous color.
static int MaxProfit(int treasure[], int color[], int n,
            int k, int col, int A, int B)
{
    int sum = 0;
 
    if (k == n) // base case
        return 0;
 
    // we have two options
    // either visit current city or skip that
 
    // check if color of this city is equal
    // to prev visited city
    if (col == color[k])
        sum += Math.max(A * treasure[k] +
                MaxProfit(treasure, color, n,
                    k + 1, color[k], A, B),
                MaxProfit(treasure, color, n,
                        k + 1, col, A, B));
    else
        sum += Math.max(B * treasure[k] +                                        
                MaxProfit(treasure, color, n,
                    k + 1, color[k], A, B),
            MaxProfit(treasure, color, n,
                        k + 1, col, A, B));
 
    // return max of both options
    return sum;
}
 
public static void main(String[] args)
{
 
    int A = -5, B = 7;
    int treasure[] = { 4, 8, 2, 9 };
    int color[] = { 2, 2, 6, 2 };
    int n = color.length;
 
    // Initially begin with color 0
    System.out.print(MaxProfit(treasure, color, n, 0, 0, A, B));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# k is current index and col
# is previous color.
def MaxProfit(treasure, color, n,
                    k, col, A, B):
    sum = 0
    if k == n:
        return 0
 
    # we have two options either
    # visit current city or skip that
 
    # check if color of this city
    # is equal to prev visited city
    if col== color[k]:
        sum += max(A * treasure[k] +
                MaxProfit(treasure, color, n,
                       k + 1, color[k], A, B),
                MaxProfit(treasure, color, n,
                            k + 1, col, A, B))
    else:
        sum += max(B * treasure[k] +                                       
               MaxProfit(treasure, color, n,
                        k + 1, color[k], A, B),
               MaxProfit(treasure, color, n,
                           k + 1, col, A, B))
 
    # return max of both options
    return sum
 
# Driver Code
A = -5
B= 7
treasure = [4, 8, 2, 9]
color = [2, 2, 6, 2]
n = len(color)
 
# Initially begin with color 0
print( MaxProfit(treasure, color,
                 n, 0, 0, A, B))
 
# This code is contributed
# by Shrikant13

C#




using System;
 
class GFG
{
 
// k is current index and col is previous color.
static int MaxProfit(int []treasure, int []color, int n,
            int k, int col, int A, int B)
{
    int sum = 0;
 
    if (k == n) // base case
        return 0;
 
    // we have two options
    // either visit current city or skip that
 
    // check if color of this city is equal
    // to prev visited city
    if (col == color[k])
        sum += Math.Max(A * treasure[k] +
                MaxProfit(treasure, color, n,
                    k + 1, color[k], A, B),
                MaxProfit(treasure, color, n,
                        k + 1, col, A, B));
    else
        sum += Math.Max(B * treasure[k] +                                        
                MaxProfit(treasure, color, n,
                    k + 1, color[k], A, B),
            MaxProfit(treasure, color, n,
                        k + 1, col, A, B));
 
    // return max of both options
    return sum;
}
 
// Driver code
public static void Main(String[] args)
{
 
    int A = -5, B = 7;
    int []treasure = { 4, 8, 2, 9 };
    int []color = { 2, 2, 6, 2 };
    int n = color.Length;
 
    // Initially begin with color 0
    Console.Write(MaxProfit(treasure, color, n, 0, 0, A, B));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// k is current index and col is previous color.
function MaxProfit(treasure,color,n,k,col,A,B)
{
    let sum = 0;
  
    if (k == n) // base case
        return 0;
  
    // we have two options
    // either visit current city or skip that
  
    // check if color of this city is equal
    // to prev visited city
    if (col == color[k])
        sum += Math.max(A * treasure[k] +
                MaxProfit(treasure, color, n,
                    k + 1, color[k], A, B),
                MaxProfit(treasure, color, n,
                        k + 1, col, A, B));
    else
        sum += Math.max(B * treasure[k] +                                       
                MaxProfit(treasure, color, n,
                    k + 1, color[k], A, B),
            MaxProfit(treasure, color, n,
                        k + 1, col, A, B));
  
    // return max of both options
    return sum;
}
 
let A = -5, B = 7;
let treasure = [ 4, 8, 2, 9 ];
let color = [ 2, 2, 6, 2 ];
let n = color.length;
 
// Initially begin with color 0
document.write(MaxProfit(treasure, color, n, 0, 0, A, B));
 
// This code is contributed by rag2127
</script>
Output: 
133

 

Since subproblems are evaluated again, this problem has Overlapping Subproblems property. 

Following is Dynamic Programming based implementation.

C++




// A memoization based program to find maximum
// treasure that can be collected.
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1001;
 
int dp[MAX][MAX];
 
// k is current index and col is previous color.
int MaxProfit(int treasure[], int color[], int n,
              int k, int col, int A, int B)
{
    if (k == n) // base case
        return dp[k][col] = 0;
 
    if (dp[k][col] != -1)
        return dp[k][col];
 
    int sum = 0;
 
    // we have two options
    // either visit current city or skip that
 
    if (col == color[k]) // check if color of this city is equal to prev visited city
        sum += max(A * treasure[k] +
               MaxProfit(treasure, color, n, k + 1,
                                     color[k], A, B), 
               MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
    else
        sum += max(B * treasure[k] +
                MaxProfit(treasure, color, n, k + 1,
                                   color[k], A, B),
                MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
 
    // return max of both options
    return dp[k][col] = sum;
}
 
int main()
{
    int A = -5, B = 7;
    int treasure[] = { 4, 8, 2, 9 };
    int color[] = { 2, 2, 6, 2 };
    int n = sizeof(color) / sizeof(color[0]);
    memset(dp, -1, sizeof(dp));
    cout << MaxProfit(treasure, color, n, 0, 0, A, B);
    return 0;
}

Java




// A memoization based program to find maximum
// treasure that can be collected.
import java.util.*;
 
class GFG
{
 
static int MAX = 1001;
 
static int [][]dp = new int[MAX][MAX];
 
// k is current index and col is previous color.
static int MaxProfit(int treasure[], int color[], int n,
            int k, int col, int A, int B)
{
    if (k == n) // base case
        return dp[k][col] = 0;
 
    if (dp[k][col] != -1)
        return dp[k][col];
 
    int sum = 0;
 
    // we have two options
    // either visit current city or skip that
     
    // check if color of this city
    // is equal to prev visited city
    if (col == color[k])
        sum += Math.max(A * treasure[k] +
            MaxProfit(treasure, color, n, k + 1,
                                    color[k], A, B),
            MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
    else
        sum += Math.max(B * treasure[k] +
                MaxProfit(treasure, color, n, k + 1,
                                color[k], A, B),
                MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
 
    // return max of both options
    return dp[k][col] = sum;
}
 
// Driver code
public static void main(String[] args)
{
    int A = -5, B = 7;
    int treasure[] = { 4, 8, 2, 9 };
    int color[] = { 2, 2, 6, 2 };
    int n = color.length;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < MAX; j++)
            dp[i][j] = -1;
    System.out.print(MaxProfit(treasure, color, n, 0, 0, A, B));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# A memoization based program to find maximum
# treasure that can be collected.
MAX = 1001
 
dp = [[-1 for i in range(MAX)] for i in range(MAX)]
 
# k is current index and col is previous color.
def MaxProfit(treasure, color, n,k, col, A, B):
    if (k == n):
         
        # base case
        dp[k][col] = 0
        return dp[k][col]
    if (dp[k][col] != -1):
        return dp[k][col]
     
    summ = 0
     
    # we have two options
    # either visit current city or skip that
    if (col == color[k]):
         
        # check if color of this city is equal to prev visited city
        summ += max(A * treasure[k] + MaxProfit(treasure,
                color, n, k + 1,color[k], A, B),
                MaxProfit(treasure, color, n, k + 1, col, A, B))
    else:
        summ += max(B * treasure[k] + MaxProfit(treasure,
                color, n, k + 1,color[k], A, B),
                MaxProfit(treasure, color, n, k + 1, col, A, B))
    dp[k][col] = summ
     
    # return max of both options
    return dp[k][col]
 
# Driver code
A = -5
B = 7
treasure = [ 4, 8, 2, 9 ]
color = [ 2, 2, 6, 2 ]
n = len(color)
print(MaxProfit(treasure, color, n, 0, 0, A, B))
 
# This code is contributed by shubhamsingh10

C#




// A memoization based program to find maximum
// treasure that can be collected.
using System;
 
class GFG
{
  
static int MAX = 1001;
  
static int [,]dp = new int[MAX, MAX];
  
// k is current index and col is previous color.
static int MaxProfit(int []treasure, int []color, int n,
            int k, int col, int A, int B)
{
    if (k == n) // base case
        return dp[k, col] = 0;
  
    if (dp[k, col] != -1)
        return dp[k, col];
  
    int sum = 0;
  
    // we have two options
    // either visit current city or skip that
      
    // check if color of this city
    // is equal to prev visited city
    if (col == color[k])
        sum += Math.Max(A * treasure[k] +
            MaxProfit(treasure, color, n, k + 1,
                                    color[k], A, B),
            MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
    else
        sum += Math.Max(B * treasure[k] +
                MaxProfit(treasure, color, n, k + 1,
                                color[k], A, B),
                MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
  
    // return max of both options
    return dp[k, col] = sum;
}
  
// Driver code
public static void Main(String[] args)
{
    int A = -5, B = 7;
    int []treasure = { 4, 8, 2, 9 };
    int []color = { 2, 2, 6, 2 };
    int n = color.Length;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < MAX; j++)
            dp[i, j] = -1;
    Console.Write(MaxProfit(treasure, color, n, 0, 0, A, B));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
// A memoization based program to find maximum
// treasure that can be collected.
 
let MAX = 1001;
let dp = new Array(MAX);
 
for(let i = 0; i < MAX; i++)
{
    dp[i] = new Array(MAX);
    for(let j = 0; j < MAX; j++)
        dp[i][j] = -1;
}
 
// k is current index and col is previous color.
function MaxProfit(treasure, color, n, k, col, A, B)
{
    if (k == n) // base case
        return dp[k][col] = 0;
  
    if (dp[k][col] != -1)
        return dp[k][col];
  
    let sum = 0;
  
    // we have two options
    // either visit current city or skip that
      
    // check if color of this city
    // is equal to prev visited city
    if (col == color[k])
        sum += Math.max(A * treasure[k] +
            MaxProfit(treasure, color, n, k + 1,
                                    color[k], A, B),
            MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
    else
        sum += Math.max(B * treasure[k] +
                MaxProfit(treasure, color, n, k + 1,
                                color[k], A, B),
                MaxProfit(treasure, color, n, k + 1,
                                        col, A, B));
  
    // return max of both options
    return dp[k][col] = sum;
}
 
// Driver code
let A = -5, B = 7;
let treasure=[4, 8, 2, 9];
let color=[2, 2, 6, 2];
let n = color.length;
document.write(MaxProfit(treasure, color, n, 0, 0, A, B));
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
133

 


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