Modify Array by modifying adjacent equal elements

Given an array arr[] of size N of positive integers. The task is to rearrange the array after applying the conditions given below: 

• If arr[i] and arr[i+1] are equal then multiply the i^th (current) element with 2 and set the (i +1)^th element to 0. 
• After applying all the conditions move all zeros at the end of the array.

Examples:

Input: N = 6,  arr[] = [1, 2, 2, 1, 1, 0]
Output: [1, 4, 2, 0, 0, 0]
Explanation: At i = 0: arr[0] and arr[1] are not the same, so we do nothing.
At i = 1: arr[1] and arr[2] are the same, so we multiply arr[1] with 2 and change its next element to 0.
array = [1, 4, 0, 1, 1, 0]
 At i = 2: arr[2] and arr[3] are not the same, so we do nothing.
At i = 3: arr[3] and arr[4] are the same, so we multiply arr[3] with 2 and change its next element to 
arr[] = [1, 4, 0, 2, 0, 0]
At i = 4: arr[4] and arr[5] are the same, so we multiply arr[4] with 2 and change its next element to 0.
arr[] = [1, 4, 0, 2, 0, 0]
After applying the above 2 conditions, shift all the 0's to the right side of the array.
arr[]= [1, 4, 2, 0, 0, 0]

Input: N =2, arr[] = [0, 1]
Output: [1, 0]
Explanation: At i = 0: arr[0] and arr[1] are not same, so we do nothing.
No conditions can be applied further, so we shift all 0's to right.
arr[] = [1, 0]

Approach: Linear Iteration

The basic idea is to linearly iterate the array and check whether the conditions are satisfied or not and perform operations according to the conditions.

Illustration:

Consider an array arr[] = {1, 2, 2, 1, 1, 0};

At i = 0: 
arr[0] and arr[1] are not the same, so we do nothing.

At i = 1:
arr[1] and arr[2] are the same, so we multiply arr[1] with 2 and change its next element to 0.
arr[] = [1, 4, 0, 1, 1, 0]

At i = 2:
arr[2] and arr[3] are not the same, so we do nothing.

At i = 3:
arr[3] and arr[4] are the same, so we multiply arr[3] with 2 and change its next element to 0.
arr[]= [1, 4, 0, 2, 0, 0]

At i = 4:
arr[4] and arr[5] are the same, so we multiply arr[4] with 2 and change it's next element to 0.
arr[] = [1, 4, 0, 2, 0, 0]

After applying the above 2 conditions, shift all the 0's to right side of the array.
arr[] = [1, 4, 2, 0, 0, 0]

Follow the steps below to implement the above idea:

• Traverse through the given array from i = 0 to N-2.
• If the i^th element is equal to the next element then,
  • Multiply the current element with 2 arr[i]*2.
  • Set next element arr[i]+1 with 0.
• Now right shift all the 0's.
  • Create a counter variable count to count the number of non-zero elements of the array.
  • If arr[i] is not zero then just update the array with counter variable arr[count++] = arr[i].
• set all zeroes at the end of the array.

Below is the implementation of the above approach.

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to shift all zeros to right side of array
void rightShift(int arr[], int n)
{
    int count = 0;

    for (int i = 0; i < n; i++) {
        if (arr[i] != 0) {
            arr[count++] = arr[i];
        }
    }

    while (count < n) {
        arr[count++] = 0;
    }
}

// Function to apply the given conditions
void applyConditions(int arr[], int n)
{
    for (int i = 0; i < n - 1; i++) {

        // Condition 1
        if (arr[i] == arr[i + 1]) {
            arr[i] = arr[i] * 2;
            arr[i + 1] = 0;
        }

        // Condition 2
        else {
            continue;
        }
    }

    // Function Call
    rightShift(arr, n);
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 1, 1, 0 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    applyConditions(arr, N);

    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }

    return 0;
}

// This code is contributed by Tapesh(tapeshdua420)

// Java code to implement the approach

import java.io.*;

class GFG {

    // Function to apply the given conditions
    static void applyConditions(int[] arr, int n)
    {
        for (int i = 0; i < n - 1; i++) {

            // Condition 1
            if (arr[i] == arr[i + 1]) {
                arr[i] = arr[i] * 2;
                arr[i + 1] = 0;
            }

            // Condition 2
            else {
                continue;
            }
        }

        // Function Call
        rightShift(arr, n);
    }

    // Function to shift all zeros to right side of array
    static void rightShift(int[] arr, int n)
    {
        int count = 0;

        for (int i = 0; i < n; i++) {
            if (arr[i] != 0) {
                arr[count++] = arr[i];
            }
        }

        while (count < n) {
            arr[count++] = 0;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 2, 1, 1, 0 };
        int N = arr.length;

        // Function Call
        applyConditions(arr, N);

        for (int i = 0; i < N; i++) {
            System.out.print(arr[i] + " ");
        }
    }
}

# Python code to implement the approach

# Function to shift all zeros to right side of array


def rightShift(arr, n):

    count = 0

    for i in range(n):
        if (arr[i] != 0):
            arr[count] = arr[i]
            count += 1

    while (count < n):
        arr[count] = 0
        count += 1

# Function to apply the given conditions


def applyConditions(arr, n):

    for i in range(n - 1):

        # Condition 1
        if (arr[i] == arr[i + 1]):
            arr[i] = arr[i] * 2
            arr[i + 1] = 0

        # Condition 2
        else:
            continue

    # Function Call
    rightShift(arr, n)


# Driver Code
if __name__ == '__main__':

    arr = [1, 2, 2, 1, 1, 0]
    N = len(arr)

    # Function Call
    applyConditions(arr, N)

    for i in range(N):
        print(arr[i], end=" ")

# This code is contributed by Tapesh(tapeshdua420)

// C# code to implement the approach

using System;

public class GFG {

    // Function to apply the given conditions
    static void applyConditions(int[] arr, int n)
    {
        for (int i = 0; i < n - 1; i++) {

            // Condition 1
            if (arr[i] == arr[i + 1]) {
                arr[i] = arr[i] * 2;
                arr[i + 1] = 0;
            }

            // Condition 2
            else {
                continue;
            }
        }

        // Function Call
        rightShift(arr, n);
    }

    // Function to shift all zeros to right side of array
    static void rightShift(int[] arr, int n)
    {
        int count = 0;

        for (int i = 0; i < n; i++) {
            if (arr[i] != 0) {
                arr[count++] = arr[i];
            }
        }

        while (count < n) {
            arr[count++] = 0;
        }
    }

    static public void Main()
    {

        // Code
        int[] arr = { 1, 2, 2, 1, 1, 0 };
        int N = arr.Length;

        // Function Call
        applyConditions(arr, N);

        for (int i = 0; i < N; i++) {
            Console.Write(arr[i] + " ");
        }
    }
}

// This code is contributed by lokeshmvs21.

// JS code to implement the approach


    // Function to apply the given conditions
function applyConditions(arr, n)
    {
        for (let i = 0; i < n - 1; i++) {

            // Condition 1
            if (arr[i] == arr[i + 1]) {
                arr[i] = arr[i] * 2;
                arr[i + 1] = 0;
            }

            // Condition 2
            else {
                continue;
            }
        }

        // Function Call
        rightShift(arr, n);
    }

    // Function to shift all zeros to right side of array
function rightShift(arr, n)
    {
        let count = 0;

        for (let i = 0; i < n; i++) {
            if (arr[i] != 0) {
                arr[count++] = arr[i];
            }
        }

        while (count < n) {
            arr[count++] = 0;
        }
    }


        let arr = [ 1, 2, 2, 1, 1, 0 ];
        let N = arr.length;

        // Function Call
        applyConditions(arr, N);

        for (let i = 0; i < N; i++) {
            console.log(arr[i] + " ");
        }

// This code is contributed by ksam24000.

1 4 2 0 0 0 

Time Complexity: O(N), because we are iterating the array two times.
Auxiliary Space: O(1)

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• Introduction to Arrays - Data Structures and Algorithms Tutorials