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# Calculate cost of visiting all array elements in increasing order

• Difficulty Level : Hard
• Last Updated : 10 Jun, 2021

Given an array arr[] consisting of N integers, the task is to find the total cost of visiting all the array elements in ascending order, starting from 0, if the cost of a move from index i to the index j is the absolute difference between i and j.

Examples:

Input: arr[ ] = { 4, 3, 2, 5, 1 }
Output: 11
Explanation:
Jump from index 0 to index 4. Cost = abs(4 – 0) = 4.
Jump from index 4 to index 2. Cost = abs(4 – 2) = 2.
Jump from index 2 to index1. Cost = abs(2 – 1) = 1.
Jump from index 1 to index 0. Cost = abs(1 – 0) = 1.
Jump from index 0 to index 3. Cost = abs(0 – 3) = 3.
Therefore, the total cost of visiting all array elements in ascending order = (4 + 2 + 1 + 1 + 3 = 11).

Input: arr[ ] = { 1, 2, 3 }
Output: 2

Approach: The idea is to use the concept of sorting of the vector of pairs. Follow the steps below to solve the problem:

• Initialize a pair of vector<pair<int, int> >, say v, to store the pairs of elements and their respective positions.
• Traverse the array arr[] and push the pair {arr[i], i} in the vector v.
• Initialize two variables, say ans = 0 and last = 0, to store the total cost required and the index of the last visited element.
• Sort the vector of pairs in ascending order.
• Traverse the vector v and increment ans by abs(v[i].second – last). Update last as last = arr[i].second.
• After completing the above steps, print the answer obtained as ans.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to calculate total``// cost of visiting array``// elements in increasing order``int` `calculateDistance(``int` `arr[], ``int` `N)``{``    ``// Stores the pair of element``    ``// and their positions``    ``vector > v;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < N; i++)` `        ``// Push the pair {arr[i], i} in v``        ``v.push_back({ arr[i], i });` `    ``// Sort the vector in ascending order.``    ``sort(v.begin(), v.end());` `    ``// Stores the total cost``    ``int` `ans = 0;` `    ``// Stores the index of last element visited``    ``int` `last = 0;` `    ``// Traverse the vector v``    ``for` `(``auto` `j : v) {` `        ``// Increment ans``        ``ans += ``abs``(j.second - last);` `        ``// Assign``        ``last = j.second;``    ``}` `    ``// Return ans``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 4, 3, 2, 5, 1 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << calculateDistance(arr, N);``}`

## Java

 `// java program for the above approach``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG {` `  ``// Pair class``  ``static` `class` `Pair {` `    ``int` `first;``    ``int` `second;` `    ``Pair(``int` `first, ``int` `second)``    ``{``      ``this``.first = first;``      ``this``.second = second;``    ``}``  ``}` `  ``// Function to calculate total``  ``// cost of visiting array``  ``// elements in increasing order``  ``static` `int` `calculateDistance(``int` `arr[], ``int` `N)``  ``{``    ``// Stores the pair of element``    ``// and their positions``    ``Pair v[] = ``new` `Pair[N];` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = ``0``; i < N; i++)` `      ``// Push the pair {arr[i], i} in v``      ``v[i] = ``new` `Pair(arr[i], i);` `    ``// Sort the vector in ascending order.``    ``Arrays.sort(v, (p1, p2) -> {``      ``if` `(p1.first != p2.first)``        ``return` `p1.first - p2.first;``      ``return` `p1.second - p2.second;``    ``});` `    ``// Stores the total cost``    ``int` `ans = ``0``;` `    ``// Stores the index of last element visited``    ``int` `last = ``0``;` `    ``// Traverse the vector v``    ``for` `(Pair j : v) {` `      ``// Increment ans``      ``ans += Math.abs(j.second - last);` `      ``// Assign``      ``last = j.second;``    ``}` `    ``// Return ans``    ``return` `ans;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``4``, ``3``, ``2``, ``5``, ``1` `};``    ``int` `N = arr.length;` `    ``// Function call``    ``System.out.println(calculateDistance(arr, N));``  ``}``}` `// This code is contributed by Kingash.`

## Python3

 `# Python3 implementation of the above approach` `# Function to calculate total``# cost of visiting array``# elements in increasing order``def` `calculateDistance(arr, N):` `    ``# Stores the pair of element``    ``# and their positions``    ``v ``=` `[]` `    ``# Traverse the array arr[]``    ``for` `i ``in` `range``(N):` `        ``# Push the pair {arr[i], i} in v``        ``v.append([arr[i], i])` `    ``# Sort the vector in ascending order.``    ``v.sort()` `    ``# Stores the total cost``    ``ans ``=` `0` `    ``# Stores the index of last element visited``    ``last ``=` `0` `    ``# Traverse the vector v``    ``for` `j ``in` `v:` `        ``# Increment ans``        ``ans ``+``=` `abs``(j[``1``] ``-` `last)` `        ``# Assign``        ``last ``=` `j[``1``]` `    ``# Return ans``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``4``, ``3``, ``2``, ``5``, ``1` `]``    ``N ``=` `len``(arr)``    ` `    ``print``(calculateDistance(arr, N))``    ` `# This code is contributed by AnkThon`

## Javascript

 ``
Output:
`11`

Time Complexity: O(N)
Auxiliary Space: O(N)

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