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++ 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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