/*Problem description
Given N natural number(s), separate 0's and bring them to the left side of the containers.
Time Complexity  : O(n)
Memory Complexity: O(1)
*/

#include <iostream>
using namespace std;

int main() {
	int* storage;			//Use dynamic allocated array
	int numOfNum;			//Number of input(s)

	//Take in the input and display it
	cin >> numOfNum;
	storage = new int[numOfNum];
	for (int indx = 0; indx < numOfNum; indx++) {
		cin >> storage[indx];
	}
	//Display the initial list
	cout << "Your initial list: ";
	for (int indx = 0; indx < numOfNum; indx++) {
		cout << storage[indx] << " ";
	}
	cout << endl;

	//Go from the last index to the first one, shift all the non zero to the right most position 
	//	(next to another right most or end of the array)
	int lastZeroPos = -1;	//The index of the last 0 position
	//First, find the position of the last 0
	for (lastZeroPos = numOfNum - 1; lastZeroPos >= 0 && storage[lastZeroPos] != 0; lastZeroPos--);
	if (lastZeroPos <= 0) {
		//Nothing to shift
		cout << "Your list doesn't need to be shifted!" << endl;
	} else {
		//Then shift
		for (int indx = lastZeroPos - 1; indx >= 0; indx--) {
			//If we encounter something non-zero then...
			if (storage[indx] != 0) {
				//Swap!
				storage[lastZeroPos--] = storage[indx];
				storage[indx] = 0;
			}
		}
		//And, finally, re-display
		cout << "Your shifted list: ";
		for (int indx = 0; indx < numOfNum; indx++) {
			cout << storage[indx] << " ";
		}
		cout << endl;
	}

	delete[] storage;					//Delete after you play!
	return 0;
}