Given an array of random numbers, Push all the zero’s of a given array to the end of the array. For example, if the given arrays is {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0}, it should be changed to {1, 9, 8, 4, 2, 7, 6, 0, 0, 0, 0}. The order of all other elements should be same. Expected time complexity is O(n) and extra space is O(1).
Example:
Input : arr[] = {1, 2, 0, 4, 3, 0, 5, 0};
Output : arr[] = {1, 2, 4, 3, 5, 0, 0};
Input : arr[] = {1, 2, 0, 0, 0, 3, 6};
Output : arr[] = {1, 2, 3, 6, 0, 0, 0};
There can be many ways to solve this problem. Following is a simple and interesting way to solve this problem.
Traverse the given array ‘arr’ from left to right. While traversing, maintain count of non-zero elements in array. Let the count be ‘count’. For every non-zero element arr[i], put the element at ‘arr[count]’ and increment ‘count’. After complete traversal, all non-zero elements have already been shifted to front end and ‘count’ is set as index of first 0. Now all we need to do is that run a loop which makes all elements zero from ‘count’ till end of the array.
Below is the implementation of the above approach.
C++
// A C++ program to move all zeroes at the end of array #include <iostream> using namespace std; // Function which pushes all zeros to end of an array. void pushZerosToEnd(int arr[], int n) { int count = 0; // Count of non-zero elements // 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) arr[count++] = arr[i]; // here count is // incremented // 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; } // Driver program to test above function int main() { int arr[] = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}; int n = sizeof(arr) / sizeof(arr[0]); pushZerosToEnd(arr, n); cout << "Array after pushing all zeros to end of array :\n"; for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0; } |
Java
/* Java program to push zeroes to back of array */import java.io.*; class PushZero { // Function which pushes all zeros to end of an array. static void pushZerosToEnd(int arr[], int n) { int count = 0; // Count of non-zero elements // 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) arr[count++] = arr[i]; // here count is // incremented // 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; } /*Driver function to check for above functions*/ public static void main (String[] args) { int arr[] = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}; int n = arr.length; pushZerosToEnd(arr, n); System.out.println("Array after pushing zeros to the back: "); for (int i=0; i<n; i++) System.out.print(arr[i]+" "); } } /* This code is contributed by Devesh Agrawal */ |
Python3
# Python3 code to move all zeroes # at the end of array # Function which pushes all # zeros to end of an array. def pushZerosToEnd(arr, n): count = 0 # Count of non-zero elements # Traverse the array. If element # encountered is non-zero, then # replace the element at index # 'count' with this element for i in range(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 # Driver code arr = [1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9] n = len(arr) pushZerosToEnd(arr, n) print("Array after pushing all zeros to end of array:") print(arr) # This code is contributed by "Abhishek Sharma 44" |
C#
/* C# program to push zeroes to back of array */using System; class PushZero { // 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 â..countâ.. 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 â..countâ.. is set as index of first 0. // Make all elements 0 from count to end. while (count < n) arr[count++] = 0; } // Driver function public static void Main () { int []arr = {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}; int n = arr.Length; pushZerosToEnd(arr, n); Console.WriteLine("Array after pushing all zeros to the back: "); for (int i = 0; i < n; i++) Console.Write(arr[i] +" "); } } /* This code is contributed by Anant Agrawal */ |
PHP
<?php // A PHP program to move all // zeroes at the end of array // Function which pushes all // zeros to end of an array. function 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 = 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; } // Driver Code $arr = array(1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9); $n = sizeof($arr); pushZerosToEnd($arr, $n); echo "Array after pushing all " . "zeros to end of array :\n"; for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; // This code is contributed // by ChitraNayal ?> |
Output:
Array after pushing all zeros to end of array : 1 9 8 4 2 7 6 9 0 0 0 0
Time Complexity: O(n) where n is number of elements in input array.
Auxiliary Space: O(1)
This article is contributed by Chandra Prakash. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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