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:
#include <bits/stdc++.h>
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;
}
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));
}
}
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))
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
function minCostTriplet(arr, cost, n) {
let minCost = Number.MAX_VALUE;
// Iterate over all possible triplets of elements
for (let i = 0; i < n - 2; i++) {
for (let j = i + 1; j < n - 1; j++) {
for (let k = j + 1; k < n; k++) {
// Check if triplet forms an increasing sequence
if (arr[i] < arr[j] && arr[j] < arr[k]) {
const 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;
}
function main() {
const arr = [2, 4, 5, 4, 10];
const cost = [40, 30, 20, 10, 40];
const n = arr.length;
console.log(minCostTriplet(arr, cost, n));
}
main();
//This code is contributed by aeroabrar_31
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++ 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++) {
// 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 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 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# 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 implementation of the approach
// Function to return the minimum required cost
function minCost(arr,cost,n)
{
// To store the cost of choosing three elements
let costThree = Number.MAX_VALUE;
// Fix the middle element
for (let j = 0; j < n; j++)
{
// Initialize cost of the first
// and the third element
let costI = Number.MAX_VALUE;
let costK = Number.MAX_VALUE;
// Search for the first element
// in the left subarray
for (let 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 (let 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 != Number.MAX_VALUE &&
costK != Number.MAX_VALUE)
{
costThree = Math.min(costThree, cost[j] +
costI + costK);
}
}
// No such triplet found
if (costThree == Number.MAX_VALUE)
return -1;
return costThree;
}
// Driver code
let arr=[2, 4, 5, 4, 10];
let cost=[40, 30, 20, 10, 40 ];
let n = arr.length;
console.log(minCost(arr, cost, n));
// This code is contributed by unknown2108
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.