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.
// 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 :
"; for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
} |
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)
OTHER APPROACH :
Assuming the start element as the pivot and while iterating through the array if a non-zero element is encountered then swap the element with the pivot and increment the index. This is continued till all the non-zero elements are placed towards the left and all the zeros are towards the right.
#include <iostream> using namespace std;
void swap( int A[], int i, int j)
{ int temp = A[i];
A[i] = A[j];
A[j] = temp;
} // Function to move all zeros present in an array to the end void fun( int A[], int n)
{ int j = 0;
// when we encounter a non-zero, `j` is incremented, and
// the element is placed before the pivot
for ( int i = 0; i < n; i++) {
if (A[i] != 0) // pivot is 0
{
swap(A, i, j);
j++;
}
}
} int main()
{ int A[] = { 1, 0, 2, 0, 3, 0, 4, 0, 5, 0 };
int n = sizeof (A) / sizeof (A[0]);
fun(A, n);
for ( int i = 0; i < n; i++) {
printf ( "%d " , A[i]);
}
return 0;
} |
1 2 3 4 5 0 0 0 0 0
Time Complexity: O(n) where n is number of elements in input array.
Auxiliary Space: O(1)
Please refer complete article on Move all zeroes to end of array for more details!