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

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: 4
Operation 1: arr[2] = arr[2] – 1
Operation 2: arr[3] = arr[3] – 3
The new array becomes arr[] = {1, 2, 4, 6}
which satisfies the given condition.

Input: arr[] = {-2, 0, 1, 4}, K = 5
Output: 0

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

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[0]);     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

Output:

4

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Improved By : AnkitRai01, Rajput-Ji

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