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 ith (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 ith 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++
// 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
// 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] + " ");
}
}
}
Python3
# 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#
// 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.
Javascript
// 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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