# 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[k].

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).

Approach:

• We can use a brute-force approach to iterate over all possible triplets of elements in the array and check if they form an increasing sequence.
• If a triplet forms an increasing sequence, we calculate its cost by adding the costs of its elements.
• We keep track of the minimum cost seen so far and return it as a result.

The array arr, the related costs cost, and the length of the arrays n are passed to the min_cost_triplet method. It iterates through all potential triplets of elements, tests to see if they form an increasing sequence then computes the triplet’s cost. It maintains track of the lowest cost encountered thus far and returns it as the result.

We define the input arrays, determine their length, call the min_cost_triplet function, then print the result in the main function.

Implementation:

## C++

 `#include ``using` `namespace` `std;` `int` `min_cost_triplet(``int` `arr[], ``int` `cost[], ``int` `n) {``    ``int` `min_cost = INT_MAX;``    ``// iterate over all possible triplets of elements``    ``for` `(``int` `i = 0; i < n - 2; i++) {``        ``for` `(``int` `j = i + 1; j < n - 1; j++) {``            ``for` `(``int` `k = j + 1; k < n; k++) {``                ``// check if triplet forms an increasing sequence``                ``if` `(arr[i] < arr[j] && arr[j] < arr[k]) {``                    ``int` `curr_cost = cost[i] + cost[j] + cost[k];``                    ``// update minimum cost seen so far``                    ``if` `(curr_cost < min_cost) {``                        ``min_cost = curr_cost;``                    ``}``                ``}``            ``}``        ``}``    ``}``    ``// return minimum cost of valid triplets``    ``return` `min_cost;``}` `int` `main() {``    ``int` `arr[] = {2, 4, 5, 4, 10};``    ``int` `cost[] = {40, 30, 20, 10, 40};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << min_cost_triplet(arr, cost, n) << endl;``    ``return` `0;``}`

## Java

 `import` `java.io.*;` `import` `java.util.Arrays;` `public` `class` `GFG {``    ``public` `static` `int` `minCostTriplet(``int``[] arr, ``int``[] cost, ``int` `n) {``        ``int` `minCost = Integer.MAX_VALUE;``        ``// iterate over all possible triplets of elements``        ``for` `(``int` `i = ``0``; i < n - ``2``; i++) {``            ``for` `(``int` `j = i + ``1``; j < n - ``1``; j++) {``                ``for` `(``int` `k = j + ``1``; k < n; k++) {``                    ``// check if triplet forms an increasing sequence``                    ``if` `(arr[i] < arr[j] && arr[j] < arr[k]) {``                        ``int` `currCost = cost[i] + cost[j] + cost[k];``                        ``// update minimum cost seen so far``                        ``if` `(currCost < minCost) {``                            ``minCost = currCost;``                        ``}``                    ``}``                ``}``            ``}``        ``}``        ``// return minimum cost of valid triplets``        ``return` `minCost;``    ``}` `    ``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(minCostTriplet(arr, cost, n));``    ``}``}`

## Python

 `def` `min_cost_triplet(arr, cost, n):``    ``min_cost ``=` `float``(``'inf'``)``    ``# iterate over all possible triplets of elements``    ``for` `i ``in` `range``(n ``-` `2``):``        ``for` `j ``in` `range``(i ``+` `1``, n ``-` `1``):``            ``for` `k ``in` `range``(j ``+` `1``, n):``                ``# check if triplet forms an increasing sequence``                ``if` `arr[i] < arr[j] < arr[k]:``                    ``curr_cost ``=` `cost[i] ``+` `cost[j] ``+` `cost[k]``                    ``# update minimum cost seen so far``                    ``if` `curr_cost < min_cost:``                        ``min_cost ``=` `curr_cost``    ``# return minimum cost of valid triplets``    ``return` `min_cost` `arr ``=` `[``2``, ``4``, ``5``, ``4``, ``10``]``cost ``=` `[``40``, ``30``, ``20``, ``10``, ``40``]``n ``=` `len``(arr)``print``(min_cost_triplet(arr, cost, n))`

## C#

 `using` `System;` `public` `class` `GFG``{``    ``public` `static` `int` `MinCostTriplet(``int``[] arr, ``int``[] cost, ``int` `n)``    ``{``        ``int` `minCost = ``int``.MaxValue;``        ``// iterate over all possible triplets of elements``        ``for` `(``int` `i = 0; i < n - 2; i++)``        ``{``            ``for` `(``int` `j = i + 1; j < n - 1; j++)``            ``{``                ``for` `(``int` `k = j + 1; k < n; k++)``                ``{``                    ``// check if triplet forms an increasing sequence``                    ``if` `(arr[i] < arr[j] && arr[j] < arr[k])``                    ``{``                        ``int` `currCost = cost[i] + cost[j] + cost[k];``                        ``// update minimum cost seen so far``                        ``if` `(currCost < minCost)``                        ``{``                            ``minCost = currCost;``                        ``}``                    ``}``                ``}``            ``}``        ``}``        ``// return minimum cost of valid triplets``        ``return` `minCost;``    ``}` `    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int``[] arr = { 2, 4, 5, 4, 10 };``        ``int``[] cost = { 40, 30, 20, 10, 40 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(MinCostTriplet(arr, cost, n));``    ``}``}` `//This code is contributed by aeroabrar_31`

## Javascript

 ``

Output
```90

```

Time Complexity: O(n^3), where n is the size of the given arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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++) {` `        ``// Initialize 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;``}`

## 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++)``    ``{ ` `        ``// Initialize 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):` `        ``# Initialize 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++)``    ``{ ` `        ``// Initialize 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..`

## Javascript

 ``

Output
```90

```

Time Complexity: O(n2), where n is the size of the given arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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