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
// 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. |
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).
Please refer complete article on Move all zeroes to end of array | Set-2 (Using single traversal) for more details!
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