Given an array of random numbers, Push all the zero’s of the given array to the end of the array. For example, if the given arrays is {1, 0, 2, 6, 0, 4}, it should be changed to {1, 2, 6, 4, 0, 0}. The order of all other elements should be the same.
Examples:
Input: arr[]={8, 9, 0, 1, 2, 0, 3} Output: arr[]={8, 9, 1, 2, 3, 0, 0} Explanation: Swap {0 ,1} -> Resulting array {8, 9, 1, 0, 2, 0, 3} Swap {0 ,2} -> Resulting array {8, 9, 1, 2, 0, 0, 3} Swap {0 ,3} -> Final array {8, 9, 1, 2, 3, 0, 0} Input: arr[]={4, 5, 0, 0, 0, 0, 6, 7} Output: arr[]={4, 5, 6, 7, 0, 0, 0, 0}
Approach:
- Iterate the array from 0 to N.
- Keep two pointers, one for zero elements and other for non-zero elements.
- Swap every zero elements with the non-zero element that comes just after it.
C
// C implementation to move all zeroes at // the end of array #include<stdio.h> // Function to move all zeroes at // the end of array void moveZerosToEnd( int arr[], int n) { int j=0, temp, i; // Traverse the array. If arr[i] is // non-zero and arr[j] is zero, // then swap both the element for (i=0;i<n;i++) { if (arr[i]!=0 && arr[j]==0) { temp=arr[i]; arr[i]=arr[j]; arr[j]=temp; } if (arr[j]!=0) j+=1; } } // Function to print the array elements void printArray( int arr[], int n) { for ( int i = 0; i < n; i++) printf ( "%d " , arr[i]); } // Driver Code int main() { int arr[] = {8, 9, 0, 1, 2, 0, 3}; int n = sizeof (arr) / sizeof (arr[0]); printf ( "Original array: " ); printArray(arr, n); moveZerosToEnd(arr, n); printf ( "\nModified array: " ); printArray(arr, n); return 0; } |
C++
// C++ implementation to move all zeroes at // the end of array #include <iostream> using namespace std; // Function to move all zeroes at // the end of array void moveZerosToEnd( int arr[], int n) { int j=0, temp, i; // Traverse the array. If arr[i] is // non-zero and arr[j] is zero, // then swap both the element for (i=0;i<n;i++) { if (arr[i]!=0 && arr[j]==0) { swap(arr[i],arr[j]); } if (arr[j]!=0) j+=1; } } // Function to print the array elements void printArray( int arr[], int n) { for ( int i = 0; i < n; i++) cout << arr[i] << " " ; } // Driver Code int main() { int arr[] = {8, 9, 0, 1, 2, 0, 3}; int n = sizeof (arr) / sizeof (arr[0]); cout << "Original array: " ; printArray(arr, n); moveZerosToEnd(arr, n); cout << "\nModified array: " ; printArray(arr, n); return 0; } |
Java
// Java implementation to move all zeroes at // the end of array class GFG { // Function to move all zeroes at // the end of array static void moveZerosToEnd( int arr[], int n) { int j = 0 , i; // Traverse the array. If arr[i] is // non-zero and arr[j] is zero, // then swap both the element for (i = 0 ; i < n; i++) { if (arr[i] != 0 && arr[j] == 0 ) { arr = swap(arr, i, j); } if (arr[j] != 0 ) j += 1 ; } } static int [] swap( int [] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr; } // 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 Code public static void main(String[] args) { int arr[] = { 8 , 9 , 0 , 1 , 2 , 0 , 3 }; int n = arr.length; System.out.print( "Original array: " ); printArray(arr, n); moveZerosToEnd(arr, n); System.out.print( "\nModified array: " ); printArray(arr, n); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to move all zeroes at # the end of array # Function to move all zeroes at # the end of array def moveZerosToEnd(nums): j = 0 for i in range ( len (nums)): if nums[i]! = 0 and nums[j] = = 0 : nums[i], nums[j] = nums[j], nums[i] if nums[j]! = 0 : j + = 1 # 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 = [ 8 , 9 , 0 , 1 , 2 , 0 , 3 ] n = len (arr) print ( "Original array:" , end = " " ) printArray(arr, n) moveZerosToEnd(arr) print ( "\nModified array: " , end = " " ) printArray(arr, n) |
C#
// C# implementation to move all zeroes // at the end of array using System; class GFG{ // Function to move all zeroes // at the end of array static void moveZerosToEnd( int []arr, int n) { int j = 0, i; // Traverse the array. If arr[i] // is non-zero and arr[j] is zero, // then swap both the element for (i = 0; i < n; i++) { if (arr[i] != 0 && arr[j] == 0) { arr = swap(arr, i, j); } if (arr[j] != 0) j += 1; } } static int [] swap( int [] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr; } // Function to print the array elements static void printArray( int []arr, int n) { for ( int i = 0; i < n; i++) Console.Write(arr[i] + " " ); } // Driver Code public static void Main(String[] args) { int []arr = { 8, 9, 0, 1, 2, 0, 3 }; int n = arr.Length; Console.Write( "Original array: " ); printArray(arr, n); moveZerosToEnd(arr, n); Console.Write( "\nModified array: " ); printArray(arr, n); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation to move all zeroes at // the end of array // Function to move all zeroes at // the end of array function moveZerosToEnd(arr, n) { let j=0, temp, i; // Traverse the array. If arr[i] is // non-zero and arr[j] is zero, // then swap both the element for (i=0;i<n;i++) { if (arr[i]!=0 && arr[j]==0) { let temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } if (arr[j]!=0) j+=1; } } // Function to print the array elements function printArray(arr, n) { for (let i = 0; i < n; i++) document.write(arr[i] + " " ); } // Driver Code let arr = [8, 9, 0, 1, 2, 0, 3]; let n = arr.length; document.write( "Original array: " ); printArray(arr, n); moveZerosToEnd(arr, n); document.write( "<br>Modified array: " ); printArray(arr, n); // This code is contributed by Surbhi Tyagi. </script> |
Output:
Original array: 8 9 0 1 2 0 3 Modified array: 8 9 1 2 3 0 0
Time Complexity: O(N).
Auxiliary Space: O(1)
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