Double the first element and move zero to end
Last Updated :
09 Dec, 2022
For a given array of n integers and assume that '0' is an invalid number and all others a valid number. Convert the array in such a way that if both current and next element is valid and both have same value then double current value and replace the next number with 0. After the modification, rearrange the array such that all 0’s shift to the end.
Examples:
Input : arr[] = {2, 2, 0, 4, 0, 8}
Output : 4 4 8 0 0 0
Input : arr[] = {0, 2, 2, 2, 0, 6, 6, 0, 0, 8}
Output : 4 2 12 8 0 0 0 0 0 0
Source: Microsoft IDC Interview Experience | Set 150.
Approach: First modify the array as mentioned, i.e., if the next valid number is the same as the current number, double its value and replace the next number with 0.
Algorithm for Modification:
1. if n == 1
2. return
3. for i = 0 to n-2
4. if (arr[i] != 0) && (arr[i] == arr[i+1])
5. arr[i] = 2 * arr[i]
6. arr[i+1] = 0
7. i++
After modifying the array, Move all zeroes to the end of the array.
C++
// C++ implementation to rearrange the array elements after
// modification
#include <bits/stdc++.h>
using namespace std;
// function which pushes all zeros to end of an array.
void pushZerosToEnd(int arr[], int n)
{
// Count of non-zero elements
int count = 0;
// Traverse the array. If element encountered is
// non-zero, then replace the element at index 'count'
// with this element
for (int i = 0; i < n; i++)
if (arr[i] != 0)
// here count is incremented
arr[count++] = arr[i];
// Now all non-zero elements have been shifted to front
// and 'count' is set as index of first 0. Make all
// elements 0 from count to end.
while (count < n)
arr[count++] = 0;
}
// function to rearrange the array elements after
// modification
void modifyAndRearrangeArr(int arr[], int n)
{
// if 'arr[]' contains a single element only
if (n == 1)
return;
// traverse the array
for (int i = 0; i < n - 1; i++) {
// if true, perform the required modification
if ((arr[i] != 0) && (arr[i] == arr[i + 1])) {
// double current index value
arr[i] = 2 * arr[i];
// put 0 in the next index
arr[i + 1] = 0;
// increment by 1 so as to move two indexes
// ahead during loop iteration
i++;
}
}
// push all the zeros at the end of 'arr[]'
pushZerosToEnd(arr, n);
}
// function to print the array elements
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Driver program to test above
int main()
{
int arr[] = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Original array: ";
printArray(arr, n);
modifyAndRearrangeArr(arr, n);
cout << "\nModified array: ";
printArray(arr, n);
return 0;
}
// C implementation to rearrange the array elements after
// modification
#include <stdio.h>
// function which pushes all zeros to end of an array.
void pushZerosToEnd(int arr[], int n)
{
// Count of non-zero elements
int count = 0;
// Traverse the array. If element encountered is
// non-zero, then replace the element at index 'count'
// with this element
for (int i = 0; i < n; i++)
if (arr[i] != 0)
// here count is incremented
arr[count++] = arr[i];
// Now all non-zero elements have been shifted to front
// and 'count' is set as index of first 0. Make all
// elements 0 from count to end.
while (count < n)
arr[count++] = 0;
}
// function to rearrange the array elements after
// modification
void modifyAndRearrangeArr(int arr[], int n)
{
// if 'arr[]' contains a single element only
if (n == 1)
return;
// traverse the array
for (int i = 0; i < n - 1; i++) {
// if true, perform the required modification
if ((arr[i] != 0) && (arr[i] == arr[i + 1])) {
// double current index value
arr[i] = 2 * arr[i];
// put 0 in the next index
arr[i + 1] = 0;
// increment by 1 so as to move two indexes
// ahead during loop iteration
i++;
}
}
// push all the zeros at the end of 'arr[]'
pushZerosToEnd(arr, n);
}
// function to print the array elements
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
}
// Driver program to test above
int main()
{
int arr[] = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Original array: ");
printArray(arr, n);
modifyAndRearrangeArr(arr, n);
printf("\nModified array: ");
printArray(arr, n);
return 0;
}
// This code is contributed by Sania Kumari Gupta
// Java implementation to rearrange the
// array elements after modification
import java.io.*;
class GFG {
// function which pushes all
// zeros to end of an array.
static void pushZerosToEnd(int arr[], int n)
{
// Count of non-zero elements
int count = 0;
// Traverse the array. If element
// encountered is non-zero, then
// replace the element at index
// 'count' with this element
for (int i = 0; i < n; i++)
if (arr[i] != 0)
// here count is incremented
arr[count++] = arr[i];
// Now all non-zero elements
// have been shifted to front and
// 'count' is set as index of first 0.
// Make all elements 0 from count to end.
while (count < n)
arr[count++] = 0;
}
// function to rearrange the array
// elements after modification
static void modifyAndRearrangeArr(int arr[], int n)
{
// if 'arr[]' contains a single element
// only
if (n == 1)
return;
// traverse the array
for (int i = 0; i < n - 1; i++) {
// if true, perform the required modification
if ((arr[i] != 0) && (arr[i] == arr[i + 1]))
{
// double current index value
arr[i] = 2 * arr[i];
// put 0 in the next index
arr[i + 1] = 0;
// increment by 1 so as to move two
// indexes ahead during loop iteration
i++;
}
}
// push all the zeros at
// the end of 'arr[]'
pushZerosToEnd(arr, n);
}
// function to print the array elements
static void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
System.out.println();
}
// Driver program to test above
public static void main(String[] args)
{
int arr[] = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 };
int n = arr.length;
System.out.print("Original array: ");
printArray(arr, n);
modifyAndRearrangeArr(arr, n);
System.out.print("Modified array: ");
printArray(arr, n);
}
}
// This code is contributed
// by prerna saini
# Python3 implementation to rearrange
# the array elements after modification
# function which pushes all zeros
# to end of an array.
def pushZerosToEnd(arr, n):
# Count of non-zero elements
count = 0
# Traverse the array. If element
# encountered is non-zero, then
# replace the element at index
# 'count' with this element
for i in range(0, n):
if arr[i] != 0:
# here count is incremented
arr[count] = arr[i]
count+=1
# Now all non-zero elements have been
# shifted to front and 'count' is set
# as index of first 0. Make all
# elements 0 from count to end.
while (count < n):
arr[count] = 0
count+=1
# function to rearrange the array
# elements after modification
def modifyAndRearrangeArr(ar, n):
# if 'arr[]' contains a single
# element only
if n == 1:
return
# traverse the array
for i in range(0, n - 1):
# if true, perform the required modification
if (arr[i] != 0) and (arr[i] == arr[i + 1]):
# double current index value
arr[i] = 2 * arr[i]
# put 0 in the next index
arr[i + 1] = 0
# increment by 1 so as to move two
# indexes ahead during loop iteration
i+=1
# push all the zeros at the end of 'arr[]'
pushZerosToEnd(arr, n)
# function to print the array elements
def printArray(arr, n):
for i in range(0, n):
print(arr[i],end=" ")
# Driver program to test above
arr = [ 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 ]
n = len(arr)
print("Original array:",end=" ")
printArray(arr, n)
modifyAndRearrangeArr(arr, n)
print("\nModified array:",end=" ")
printArray(arr, n)
# This code is contributed by Smitha Dinesh Semwal
// C# implementation to rearrange the
// array elements after modification
using System;
class GFG {
// function which pushes all
// zeros to end of an array.
static void pushZerosToEnd(int[] arr, int n)
{
// Count of non-zero elements
int count = 0;
// Traverse the array. If element
// encountered is non-zero, then
// replace the element at index
// 'count' with this element
for (int i = 0; i < n; i++)
if (arr[i] != 0)
// here count is incremented
arr[count++] = arr[i];
// Now all non-zero elements
// have been shifted to front and
// 'count' is set as index of first 0.
// Make all elements 0 from count to end.
while (count < n)
arr[count++] = 0;
}
// function to rearrange the array
// elements after modification
static void modifyAndRearrangeArr(int[] arr, int n)
{
// if 'arr[]' contains a single element
// only
if (n == 1)
return;
// traverse the array
for (int i = 0; i < n - 1; i++) {
// if true, perform the required modification
if ((arr[i] != 0) && (arr[i] == arr[i + 1])) {
// double current index value
arr[i] = 2 * arr[i];
// put 0 in the next index
arr[i + 1] = 0;
// increment by 1 so as to move two
// indexes ahead during loop iteration
i++;
}
}
// push all the zeros at
// the end of 'arr[]'
pushZerosToEnd(arr, n);
}
// function to print the array elements
static void printArray(int[] arr, int n)
{
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
Console.WriteLine();
}
// Driver program to test above
public static void Main()
{
int[] arr = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 };
int n = arr.Length;
Console.Write("Original array: ");
printArray(arr, n);
modifyAndRearrangeArr(arr, n);
Console.Write("Modified array: ");
printArray(arr, n);
}
}
// This code is contributed by Sam007
<script>
// JavaScript implementation to rearrange the array
// elements after modification
// function which pushes all zeros to end of
// an array.
function pushZerosToEnd(arr, n)
{
// Count of non-zero elements
var count = 0;
// Traverse the array. If element encountered
// is non-zero, then replace the element at
// index 'count' with this element
for (var i = 0; i < n; i++)
if (arr[i] != 0)
// here count is incremented
arr[count++] = arr[i];
// Now all non-zero elements have been shifted
// to front and 'count' is set as index of
// first 0. Make all elements 0 from count
// to end.
while (count < n) arr[count++] = 0;
}
// function to rearrange the array elements
// after modification
function modifyAndRearrangeArr(arr, n)
{
// if 'arr[]' contains a single element
// only
if (n == 1) return;
// traverse the array
for (var i = 0; i < n - 1; i++)
{
// if true, perform the required modification
if (arr[i] != 0 && arr[i] == arr[i + 1]) {
// double current index value
arr[i] = 2 * arr[i];
// put 0 in the next index
arr[i + 1] = 0;
// increment by 1 so as to move two
// indexes ahead during loop iteration
i++;
}
}
// push all the zeros at the end of 'arr[]'
pushZerosToEnd(arr, n);
}
// function to print the array elements
function printArray(arr, n)
{
for (var i = 0; i < n; i++) document.write(arr[i] + " ");
}
// Driver program to test above
var arr = [0, 2, 2, 2, 0, 6, 6, 0, 0, 8];
var n = arr.length;
document.write("Original array: ");
printArray(arr, n);
modifyAndRearrangeArr(arr, n);
document.write("<br>");
document.write("Modified array: ");
printArray(arr, n);
// This code is contributed by rdtank.
</script>
OutputOriginal array: 0 2 2 2 0 6 6 0 0 8
Modified array: 4 2 12 8 0 0 0 0 0 0
Time Complexity: O(n).
Auxiliary Space: O(1)
Approach with efficient zero shiftings:
Although the above solution is efficient, we can further optimise it in shifting zero algorithms by reducing the number of operations.
In the above shifting algorithms, we scan some elements twice when we set the count index to last index element to zero.
Efficient Zero Shifting Algorithms:
int lastSeenPositiveIndex = 0;
for( index = 0; index < n; index++)
{
if(array[index] != 0)
{
swap(array[index], array[lastSeenPositiveIndex]);
lastSeenPositiveIndex++;
}
}
C++
// Utility Function For Swapping Two Element Of An Array
void swap(int& a, int& b) { a = b + a - (b = a); }
// shift all zero to left side of an array
void shiftAllZeroToLeft(int array[], int n)
{
// Maintain last index with positive value
int lastSeenNonZero = 0;
for (index = 0; index < n; index++)
{
// If Element is non-zero
if (array[index] != 0)
{
// swap current index, with lastSeen non-zero
swap(array[index], array[lastSeenNonZero]);
// next element will be last seen non-zero
lastSeenNonZero++;
}
}
}
// This snippet is contributed By: Faizanur Rahman
import java.io.*;
class GFG {
// Function For Swapping Two Element Of An Array
public static void swap(int[] A, int i, int j)
{
int temp = A[i];
A[i] = A[j];
A[j] = temp;
}
// shift all zero to left side of an array
static void shiftAllZeroToLeft(int array[], int n)
{
// Maintain last index with positive value
int lastSeenNonZero = 0;
for (int index = 0; index < n; index++) {
// If Element is non-zero
if (array[index] != 0) {
// swap current index, with lastSeen
// non-zero
swap(array, array[index],
array[lastSeenNonZero]);
// next element will be last seen non-zero
lastSeenNonZero++;
}
}
}
}
// This code is contributed By sam_2200
# Maintain last index with positive value
def shiftAllZeroToLeft(arr, n):
lastSeenNonZero = 0
for index in range(0, n):
# If Element is non-zero
if (array[index] != 0):
# swap current index, with lastSeen
# non-zero
array[index], array[lastSeenNonZero] = array[lastSeenNonZero], array[index]
# next element will be last seen non-zero
lastSeenNonZero++
# This code is contributed By sam_2200
using System;
class GFG
{
// Function For Swapping Two Element Of An Array
public static void swap(int[] A, int i, int j)
{
int temp = A[i];
A[i] = A[j];
A[j] = temp;
}
// shift all zero to left side of an array
static void shiftAllZeroToLeft(int[] array, int n)
{
// Maintain last index with positive value
int lastSeenNonZero = 0;
for (int index = 0; index < n; index++)
{
// If Element is non-zero
if (array[index] != 0)
{
// swap current index, with lastSeen
// non-zero
swap(array, array[index],
array[lastSeenNonZero]);
// next element will be last seen non-zero
lastSeenNonZero++;
}
}
}
}
// This code is contributed By Saurabh Jaiswal
<script>
// Function For Swapping Two Element Of An Array
function swap(A,i,j)
{
let temp = A[i];
A[i] = A[j];
A[j] = temp;
}
// shift all zero to left side of an array
function shiftAllZeroToLeft(array,n)
{
// Maintain last index with positive value
let lastSeenNonZero = 0;
for (let index = 0; index < n; index++) {
// If Element is non-zero
if (array[index] != 0) {
// swap current index, with lastSeen
// non-zero
swap(array, array[index],
array[lastSeenNonZero]);
// next element will be last seen non-zero
lastSeenNonZero++;
}
}
}
}
// This code is contributed by sravan kumar Gottumukkala
</script>
Time Complexity: O(n)
Auxiliary Space: O(1)
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