Given an array of ratings for n books. Find the minimum cost to buy all books with below conditions :
- Cost of every book would be at-least 1 dollar.
- 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
- Make two arrays left2right and right2left and fill 1 in both of them.
- 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.
- 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.
- Find maximum value of ith position in both array (left2right and right2left) and add it to result
Implementation:
C++
// 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 functionint main(){ int ratings[] = { 1, 6, 8, 3, 4, 1, 5, 7 }; int n = sizeof(ratings) / sizeof(ratings[0]); cout << minCost(ratings, n); return 0;} |
Java
// JAVA Code For Find minimum cost to// buy all booksimport 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. |
Python
# 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 functionratings = [ 1, 6, 8, 3, 4, 1, 5, 7 ]n = len(ratings)print minCost(ratings, n)# This code is contributed by Sachin Bisht |
C#
// C# code For Finding minimum// cost to buy all booksusing 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. |
Javascript
<script>// JavaScript program to find minimum cost to buy// n books.function minCost(ratings, n) { let res = 0; let left2right = new Array(n).fill(1); let right2left = new Array(n).fill(1); // Traverse from left to right and assign // minimum possible rating considering only // left adjacent for (let 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 (let 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 (let i = 0; i < n; i++) res += Math.max(left2right[i], right2left[i]); return res;}// Driver functionlet ratings = [1, 6, 8, 3, 4, 1, 5, 7];let n = ratings.length;document.write(minCost(ratings, n));</script> |
Output
15
Time complexity – O(n)
Auxiliary Space – O(n)
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