Minimum cost of choosing 3 increasing elements in an array of size N
Given two arrays arr[] and cost[] where cost[i] is the cost associated with arr[i], the task is to find the minimum cost of choosing three elements from the array such that arr[i] < arr[j] < arr[j].
Examples:
Input: arr[] = {2, 4, 5, 4, 10}, cost[] = {40, 30, 20, 10, 40}
Output: 90
(2, 4, 5), (2, 4, 10) and (4, 5, 10) are
the only valid triplets with cost 90.Input: arr[] = {1, 2, 3, 4, 5, 6}, cost[] = {10, 13, 11, 14, 15, 12}
Output: 33
Naive approach: A basic approach is two run three nested loops and to check every possible triplet. The time complexity of this approach will be O(n3).
Efficient approach: An efficient approach is to fix the middle element and search for the smaller element with minimum cost on its left and the larger element with minimum cost on its right in the given array. If a valid triplet is found then update the minimum cost far. The time complexity of this approach will be O(n2).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the minimum required cost int minCost(int arr[], int cost[], int n) { // To store the cost of choosing three elements int costThree = INT_MAX; // Fix the middle element for (int j = 0; j < n; j++) { // Initialse cost of the first // and the third element int costI = INT_MAX, costK = INT_MAX; // Search for the first element // in the left subarray for (int i = 0; i < j; i++) { // If smaller element is found // then update the cost if (arr[i] < arr[j]) costI = min(costI, cost[i]); } // Search for the third element // in the right subarray for (int k = j + 1; k < n; k++) { // If greater element is found // then update the cost if (arr[k] > arr[j]) costK = min(costK, cost[k]); } // If a valid triplet was found then // update the minimum cost so far if (costI != INT_MAX && costK != INT_MAX) { costThree = min(costThree, cost[j] + costI + costK); } } // No such triplet found if (costThree == INT_MAX) return -1; return costThree; } // Driver code int main() { int arr[] = { 2, 4, 5, 4, 10 }; int cost[] = { 40, 30, 20, 10, 40 }; int n = sizeof(arr) / sizeof(arr[0]); cout << minCost(arr, cost, n); return 0; } |
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Java
// Java implementation of the approach class GFG { // Function to return the minimum required cost static int minCost(int arr[], int cost[], int n) { // To store the cost of choosing three elements int costThree = Integer.MAX_VALUE; // Fix the middle element for (int j = 0; j < n; j++) { // Initialse cost of the first // and the third element int costI = Integer.MAX_VALUE; int costK = Integer.MAX_VALUE; // Search for the first element // in the left subarray for (int i = 0; i < j; i++) { // If smaller element is found // then update the cost if (arr[i] < arr[j]) costI = Math.min(costI, cost[i]); } // Search for the third element // in the right subarray for (int k = j + 1; k < n; k++) { // If greater element is found // then update the cost if (arr[k] > arr[j]) costK = Math.min(costK, cost[k]); } // If a valid triplet was found then // update the minimum cost so far if (costI != Integer.MAX_VALUE && costK != Integer.MAX_VALUE) { costThree = Math.min(costThree, cost[j] + costI + costK); } } // No such triplet found if (costThree == Integer.MAX_VALUE) return -1; return costThree; } // Driver code public static void main (String[] args) { int arr[] = { 2, 4, 5, 4, 10 }; int cost[] = { 40, 30, 20, 10, 40 }; int n = arr.length; System.out.println(minCost(arr, cost, n)); } } // This code is contributed by AnkitRai01 |
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Python3
# Python3 implementation of the approach # Function to return the minimum required cost def minCost(arr, cost, n): # To store the cost of choosing three elements costThree = 10**9 # Fix the middle element for j in range(n): # Initialse cost of the first # and the third element costI = 10**9 costK = 10**9 # Search for the first element # in the left subarray for i in range(j): # If smaller element is found # then update the cost if (arr[i] < arr[j]): costI = min(costI, cost[i]) # Search for the third element # in the right subarray for k in range(j + 1, n): # If greater element is found # then update the cost if (arr[k] > arr[j]): costK = min(costK, cost[k]) # If a valid triplet was found then # update the minimum cost so far if (costI != 10**9 and costK != 10**9): costThree = min(costThree, cost[j] + costI + costK) # No such triplet found if (costThree == 10**9): return -1 return costThree # Driver code arr = [2, 4, 5, 4, 10] cost = [40, 30, 20, 10, 40] n = len(arr) print(minCost(arr, cost, n)) # This code is contributed by Mohit Kumar |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the // minimum required cost static int minCost(int []arr, int []cost, int n) { // To store the cost of // choosing three elements int costThree = int.MaxValue; // Fix the middle element for (int j = 0; j < n; j++) { // Initialse cost of the first // and the third element int costI = int.MaxValue; int costK = int.MaxValue; // Search for the first element // in the left subarray for (int i = 0; i < j; i++) { // If smaller element is found // then update the cost if (arr[i] < arr[j]) costI = Math.Min(costI, cost[i]); } // Search for the third element // in the right subarray for (int k = j + 1; k < n; k++) { // If greater element is found // then update the cost if (arr[k] > arr[j]) costK = Math.Min(costK, cost[k]); } // If a valid triplet was found then // update the minimum cost so far if (costI != int.MaxValue && costK != int.MaxValue) { costThree = Math.Min(costThree, cost[j] + costI + costK); } } // No such triplet found if (costThree == int.MaxValue) return -1; return costThree; } // Driver code static public void Main () { int []arr = { 2, 4, 5, 4, 10 }; int []cost = { 40, 30, 20, 10, 40 }; int n = arr.Length; Console.Write(minCost(arr, cost, n)); } } // This code is contributed by Sachin.. |
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Output:
90
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