Find minimum cost to buy all books

Given an array of ratings for n books. Find the minimum cost to buy all books with below conditions :

  1. Cost of every book would be at-least 1 dollar.
  2. A book has higher cost than an adjacent (left or right) if rating is more than the adjacent.

Examples :

Input : Ratings[] = {1, 3, 4, 3, 7, 1}
Output : 10
Exp :- 1 + 2 + 3 + 1 + 2 + 1 = 10

Input : ratings[] = {1, 6, 8, 3, 4, 1, 5, 7}
Output : 15
Exp :- 1 + 2 + 3 + 1 + 2 + 1 + 2 + 3 = 15 

  1. Make two arrays left2right and right2left and fill 1 in both of them.
  2. Traverse left to right and fill left2right array and update it by seeing previous rating of given array. Do not care about next rating of given array.
  3. Traverse right to left and fill right2left array and update it by seeing next rating of given array. Do not care about previous rating of given array.
  4. Find maximum value of ith position in both array (left2right and right2left) and add it to result

min_cost

C++

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// C++ program to find minimum cost to buy
// n books.
#include <bits/stdc++.h>
using namespace std;
  
int minCost(int ratings[], int n)
{
    int res = 0;
    int left2right[n];
    int right2left[n];
  
    // fill 1 in both array
    fill_n(left2right, n, 1);
    fill_n(right2left, n, 1);
  
    // Traverse from left to right and assign
    // minimum possible rating considering only
    // left adjacent
    for (int i = 1; i < n; i++) 
        if (ratings[i] > ratings[i - 1]) 
            left2right[i] = left2right[i - 1] + 1;     
  
    // Traverse from right to left and assign
    // minimum possible rating considering only
    // right adjacent
    for (int i = n - 2; i >= 0; i--) 
        if (ratings[i] > ratings[i + 1]) 
            right2left[i] = right2left[i + 1] + 1;       
      
    // Since we need to follow rating rule for
    // both adjacent, we pick maximum of two
    for (int i = 0; i < n; i++) 
        res += max(left2right[i], right2left[i]);
      
    return res;
}
  
// Driver function
int main()
{
    int ratings[] = { 1, 6, 8, 3, 4, 1, 5, 7 };
    int n = sizeof(ratings) / sizeof(ratings[0]);
    cout << minCost(ratings, n);
    return 0;
}

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Java

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// JAVA Code For Find minimum cost to
// buy all books
import java.util.*;
  
class GFG {
      
    public static int minCost(int ratings[], int n)
    {
        int res = 0;
        int left2right[] = new int[n];
        int right2left[] = new int[n];;
       
        // fill 1 in both array
        Arrays.fill(left2right,  1);
        Arrays.fill(right2left, 1);
       
        // Traverse from left to right and assign
        // minimum possible rating considering 
        // only left adjacent
        for (int i = 1; i < n; i++) 
            if (ratings[i] > ratings[i - 1]) 
                left2right[i] = left2right[i - 1] + 1;     
       
        // Traverse from right to left and assign
        // minimum possible rating considering only
        // right adjacent
        for (int i = n - 2; i >= 0; i--) 
            if (ratings[i] > ratings[i + 1]) 
                right2left[i] = right2left[i + 1] + 1;       
           
        // Since we need to follow rating rule for
        // both adjacent, we pick maximum of two
        for (int i = 0; i < n; i++) 
            res += Math.max(left2right[i], 
                            right2left[i]);
           
        return res;
    }
      
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        int ratings[] = { 1, 6, 8, 3, 4, 1, 5, 7 };
        int n = ratings.length;
        System.out.print(minCost(ratings, n));
          
    }
}
    
// This code is contributed by Arnav Kr. Mandal.

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Python

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# Python program to find minimum cost to buy
# n books.
  
def minCost(ratings, n):
    res = 0
     
    # fill 1 in both array
    left2right = [1 for i in range(n)]
    right2left = [1 for i in range(n)]
   
    # Traverse from left to right and assign
    # minimum possible rating considering only
    # left adjacent
    for i in range(1, n):
        if (ratings[i] > ratings[i - 1]):
            left2right[i] = left2right[i - 1] + 1
   
    # Traverse from right to left and assign
    # minimum possible rating considering only
    # right adjacent
    i = n - 2
    while(i >= 0):
        if (ratings[i] > ratings[i + 1]):
            right2left[i] = right2left[i + 1] + 1
        i -= 1
       
    # Since we need to follow rating rule for
    # both adjacent, we pick maximum of two
    for i in range(n):
          
        res += max(left2right[i], right2left[i])
       
    return res
   
# Driver function
ratings = [ 1, 6, 8, 3, 4, 1, 5, 7 ]
n = len(ratings)
print minCost(ratings, n)
  
# This code is contributed by Sachin Bisht

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C#

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// C# code For Finding minimum
// cost to buy all books
using System;
  
class GFG {
      
    public static int minCost(int []ratings, int n)
    {
        int res = 0;
        int []left2right = new int[n];
        int []right2left = new int[n];
      
        // fill 1 in both array
        for(int i = 0; i < n; i++)
        left2right[i] = 1;
          
        for(int i = 0; i < n; i++)
        right2left[i] = 1;
      
        // Traverse from left to right and assign
        // minimum possible rating considering 
        // only left adjacent
        for (int i = 1; i < n; i++) 
            if (ratings[i] > ratings[i - 1]) 
                left2right[i] = left2right[i - 1] + 1;     
      
        // Traverse from right to left and assign
        // minimum possible rating considering only
        // right adjacent
        for (int i = n - 2; i >= 0; i--) 
            if (ratings[i] > ratings[i + 1]) 
                right2left[i] = right2left[i + 1] + 1;     
          
        // Since we need to follow rating rule for
        // both adjacent, we pick maximum of two
        for (int i = 0; i < n; i++) 
            res += Math.Max(left2right[i], 
                            right2left[i]);
          
        return res;
    }
      
    /* Driver program to test above function */
    public static void Main() 
    {
        int []ratings = {1, 6, 8, 3, 4, 1, 5, 7};
        int n = ratings.Length;
        Console.Write(minCost(ratings, n));
    }
}
      
// This code is contributed by nitin mittal.

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Output:

15

Time complexity – O(n)
Auxiliary Space – O(n)

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Improved By : nitin mittal



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