# Count of operations required to update the array such that it satisfies the given conditions

• Difficulty Level : Easy
• Last Updated : 17 Mar, 2021

Given an array arr[] of size N and an integer K. The task is to find the operations required to update the array such that it is possible to move from index 0 to index N – 1 when any index j can be visited from index i if index j is adjacent to index i and abs(arr[i] – arr[j]) ≤ K. In a single operation, any element of the array can be incremented or decremented by 1.
Examples:

Input: arr[] = {1, 2, 5, 9}, K = 2
Output:
Operation 1: arr = arr – 1
Operation 2: arr = arr – 3
The new array becomes arr[] = {1, 2, 4, 6}
which satisfies the given condition.
Input: arr[] = {-2, 0, 1, 4}, K = 5
Output:

Approach:

• Traverse the array starting from the second element and calculate the absolute difference between the current and the previous element.
• If the absolute difference is greater than K then the current element needs to be updated i.e. add the value to the smaller element or subtract the value from the larger element such that the absolute difference becomes K.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of``// operations required to update``// the array such that it is possible``// to move from index 0 to index n - 1``int` `countOp(``int` `arr[], ``int` `n, ``int` `k)``{` `    ``int` `operations = 0;``    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Current element needs to be updated``        ``if` `(``abs``(arr[i] - arr[i - 1]) > k) {` `            ``// Get the absolute difference``            ``int` `absDiff = ``abs``(arr[i] - arr[i - 1]);` `            ``// The value which needs to``            ``// be added or subtracted``            ``int` `currOp = absDiff - k;` `            ``// Add value to arr[i]``            ``if` `(arr[i] < arr[i - 1])``                ``arr[i] += currOp;` `            ``// Subtract value from arr[i]``            ``else``                ``arr[i] -= currOp;` `            ``// Update the operations``            ``operations += currOp;``        ``}``    ``}` `    ``return` `operations;``}` `// Driver code``int` `main()``{` `    ``int` `arr[] = { 1, 2, 5, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `k = 2;` `    ``cout << countOp(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;``    ` `class` `GFG``{` `// Function to return the count of``// operations required to update``// the array such that it is possible``// to move from index 0 to index n - 1``static` `int` `countOp(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `operations = ``0``;``    ``for` `(``int` `i = ``1``; i < n; i++)``    ``{` `        ``// Current element needs to be updated``        ``if` `(Math.abs(arr[i] - arr[i - ``1``]) > k)``        ``{` `            ``// Get the absolute difference``            ``int` `absDiff = Math.abs(arr[i] - arr[i - ``1``]);` `            ``// The value which needs to``            ``// be added or subtracted``            ``int` `currOp = absDiff - k;` `            ``// Add value to arr[i]``            ``if` `(arr[i] < arr[i - ``1``])``                ``arr[i] += currOp;` `            ``// Subtract value from arr[i]``            ``else``                ``arr[i] -= currOp;` `            ``// Update the operations``            ``operations += currOp;``        ``}``    ``}``    ``return` `operations;``}` `// Driver code``static` `public` `void` `main (String []arg)``{``    ``int` `arr[] = { ``1``, ``2``, ``5``, ``9` `};``    ``int` `n = arr.length;``    ``int` `k = ``2``;` `    ``System.out.println(countOp(arr, n, k));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of``# operations required to update``# the array such that it is possible``# to move from index 0 to index n - 1``def` `countOp(arr, n, k) :` `    ``operations ``=` `0``;``    ``for` `i ``in` `range``(``1``, n) :` `        ``# Current element needs to be updated``        ``if` `(``abs``(arr[i] ``-` `arr[i ``-` `1``]) > k) :` `            ``# Get the absolute difference``            ``absDiff ``=` `abs``(arr[i] ``-` `arr[i ``-` `1``]);` `            ``# The value which needs to``            ``# be added or subtracted``            ``currOp ``=` `absDiff ``-` `k;` `            ``# Add value to arr[i]``            ``if` `(arr[i] < arr[i ``-` `1``]) :``                ``arr[i] ``+``=` `currOp;` `            ``# Subtract value from arr[i]``            ``else` `:``                ``arr[i] ``-``=` `currOp;` `            ``# Update the operations``            ``operations ``+``=` `currOp;` `    ``return` `operations;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``1``, ``2``, ``5``, ``9` `];``    ``n ``=` `len``(arr);``    ``k ``=` `2``;` `    ``print``(countOp(arr, n, k));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function to return the count of``// operations required to update``// the array such that it is possible``// to move from index 0 to index n - 1``static` `int` `countOp(``int` `[]arr, ``int` `n, ``int` `k)``{``    ``int` `operations = 0;``    ``for` `(``int` `i = 1; i < n; i++)``    ``{` `        ``// Current element needs to be updated``        ``if` `(Math.Abs(arr[i] - arr[i - 1]) > k)``        ``{` `            ``// Get the absolute difference``            ``int` `absDiff = Math.Abs(arr[i] -``                                   ``arr[i - 1]);` `            ``// The value which needs to``            ``// be added or subtracted``            ``int` `currOp = absDiff - k;` `            ``// Add value to arr[i]``            ``if` `(arr[i] < arr[i - 1])``                ``arr[i] += currOp;` `            ``// Subtract value from arr[i]``            ``else``                ``arr[i] -= currOp;` `            ``// Update the operations``            ``operations += currOp;``        ``}``    ``}``    ``return` `operations;``}` `// Driver code``static` `public` `void` `Main (String []arg)``{``    ``int` `[]arr = { 1, 2, 5, 9 };``    ``int` `n = arr.Length;``    ``int` `k = 2;` `    ``Console.WriteLine(countOp(arr, n, k));``}``}``    ` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`4`

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