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])
CPP
// 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) { // Count of non-zero elements int count = 0; // 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) swap(arr[count++], arr[i]); } // 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, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "Original array: "; printArray(arr, n); moveZerosToEnd(arr, n); cout << "
Modified array: "; printArray(arr, n); return 0; } |
Java
// 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("
Modified array: "); printArray(arr, n); } } // This code is contributed by Nikita Tiwari. |
Python3
# Python implementation to move all zeroes at # the end of array # function to move all zeroes at # the end of array def moveZerosToEnd (arr, n): # Count of non-zero elements count = 0; # Traverse the array. If arr[i] is non-zero, then # swap the element at index 'count' with the # element at index 'i' for i in range(0, n): if (arr[i] != 0): arr[count], arr[i] = arr[i], arr[count] count+=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 = [ 0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9 ] n = len(arr) print("Original array:", end=" ") printArray(arr, n) moveZerosToEnd(arr, n) print("
Modified array: ", end=" ") printArray(arr, n) # This code is contributed by # Smitha Dinesh Semwal |
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) { // 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++) Console.Write(arr[i] + " "); } // Driver program to test above public static void Main() { int[] arr = { 0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9 }; int n = arr.Length; Console.Write("Original array: "); printArray(arr, n); moveZerosToEnd(arr, n); Console.Write("
Modified array: "); printArray(arr, n); } } // This code is contributed by Sam007 |
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).
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