Given an array of n numbers. The problem is to move all the 0’s to the end of the array while maintaining the order of the other elements. Only single traversal of the array is required.
Examples:
Input : arr[] = {1, 2, 0, 0, 0, 3, 6}
Output : 1 2 3 6 0 0 0
Input: arr[] = {0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}
Output: 1 9 8 4 2 7 6 9 0 0 0 0 0
Algorithm:
moveZerosToEnd(arr, n) Initialize count = 0 for i = 0 to n-1 if (arr[i] != 0) then swap(arr[count++], arr[i])
// Java implementation to move
// all zeroes at the end of array
import java.io.*;
class GFG {
// function to move all zeroes at
// the end of array
static void moveZerosToEnd( int arr[], int n) {
// Count of non-zero elements
int count = 0 ;
int temp;
// Traverse the array. If arr[i] is
// non-zero, then swap the element at
// index 'count' with the element at
// index 'i'
for ( int i = 0 ; i < n; i++) {
if ((arr[i] != 0 )) {
temp = arr[count];
arr[count] = arr[i];
arr[i] = temp;
count = count + 1 ;
}
}
}
// 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] + " " );
}
// Driver program to test above
public static void main(String args[]) {
int arr[] = { 0 , 1 , 9 , 8 , 4 , 0 , 0 , 2 ,
7 , 0 , 6 , 0 , 9 };
int n = arr.length;
System.out.print( "Original array: " );
printArray(arr, n);
moveZerosToEnd(arr, n);
System.out.print( "\nModified array: " );
printArray(arr, n);
}
}
Output:
Original array: 0 1 9 8 4 0 0 2 7 0 6 0 9 Modified array: 1 9 8 4 2 7 6 9 0 0 0 0 0
Time Complexity: O(n).
Auxiliary Space: O(1).