# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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); ` ` `  `    ``cout << minCost(arr, cost, n); ` ` `  `    ``return` `0; ` `} `

## 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 `

## 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 `

## 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.. `

Output:

```90
```

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