reverse an array without changing position of zero - c++

Question

I have just tried some of the data structures questions along with my friends. I encountered this problem from one of my friends who is also not able to solve it.
Question: Reverse an array without changing position of zeroes. example : if array has has 0 5 7 8 0 9 then the result should be 0 9 8 7 0 5.
I tried it but it does not do it correctly in all cases, I am sorry if code looks ugly I am a novice right now.
#include<iostream>
using namespace std;
int main()
{
int arr[100], tot, i, j, temp;
cout<<"Enter the Size for Array: ";
cin>>tot;
cout<<"Enter "<<tot<<" Array Elements: ";
for(i=0; i<tot; i++)
cin>>arr[i];
cout<<"\nThe Original Array is:\n";
for(i=0; i<tot; i++)
cout<<arr[i]<<" ";
j = tot-1;
for(i=0; i<j; i++, j--)
{
if(arr[i] == 0) {
i++;
continue;
}else if(arr[j] == 0) {
j--;
continue;
}
else {
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
cout<<"\n\nThe Reverse of Given Array is:\n";
for(i=0; i<tot; i++)
cout<<arr[i]<<" ";
cout<<endl;
return 0;
}
I have tried the above code but it does not give the correct results.

Answer 1

The issue here is the fact that you're modifying the loop variables i and j in every iteration of the loop; you need to update it only in the case of elements being swapped:
for(i=0; i<j;)
{
if(arr[i] == 0) {
i++;
}else if(arr[j] == 0) {
j--;
}
else {
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
++i;
--j;
}
}
Demo on godbolt.org

Answer 2

Seems you need the next logic inside the loop:
i = 0;
j = tot-1;
while(i<j) {
while (arr[i]==0 && i<j)
i++;
while (arr[j]==0 && j>i)
j--;
if (i < j) {
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
i++;
j--;
}

Answer 3

You can use two pointers to switch, simply skipping any 0's.
#include<iostream>
using namespace std;
int main()
{
int arr[100], tot, i, j, temp;
cout<<"Enter the Size for Array: ";
cin>>tot;
cout<<"Enter "<<tot<<" Array Elements: ";
for(i=0; i<tot; i++)
cin>>arr[i];
cout<<"\nThe Original Array is:\n";
for(i=0; i<tot; i++)
cout<<arr[i]<<" ";
i = 0, j = tot-1;
while (i < j) {
if (arr[i]!=0 && arr[j]!=0) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
} else if (arr[i] == 0) {
i++;
} else {
j--;
}
}
cout << "\n Reverse all elements of the array: " << endl;
// use for loop to print the reverse array
for ( i = 0; i < tot; i++)
{
cout << arr[i] << " ";
}
cout<<endl;
return 0;
}
You can see this in action here: https://godbolt.org/z/84q3McccW

Answer 4

In your code in the loop:
if(arr[i] == 0) {
i++;
continue;
}else if(arr[j] == 0) {
j--;
continue;
You increment i when arr[i] is a 0. But then you continue and for the next iteration i is incremented again as part of i++, j--. This makes your code skip elements.
I suggest to use std::vector and iterators:
#include <iostream>
#include <vector>
int main()
{
std::vector<int> x{1,2,0,3,4,0,0,0};
auto begin = x.begin();
auto end = std::prev(x.end());
while (begin != end) {
while(begin != end && *begin == 0) ++begin;
while(begin != end && *end == 0) --end;
if (begin != end) {
std::swap(*begin,*end);
++begin;
if (begin != end) --end;
}
}
for (const auto& e : x) std::cout << e << " ";
}
output:
4 3 0 2 1 0 0 0
The code starts with iterators to first and last element. In the loop they are advanced while they refer to a 0 and the two didn't cross each other yet. Then the elements are swapped.
If additional memory is acceptable, a much simpler solution is to copy all non zero elements to a second std::vector, std::reverse that vector, then copy back. Alternatively use the ranges library with a filter.

Answer 5

If there were no zeroes to be skipped, the code could be
for (i= 0, j= n-1; i < j; i++, j--)
{
Swap(a[i], a[j]);
}
Now it "suffices" to skip the zeroes. This gives the modified code
for (i= 0, j= n-1; i < j; i++, j--)
{
while (i < j && a[i] == 0) i++;
while (i < j && a[j] == 0) j--;
Swap(a[i], a[j]);
}
Notice that the Swap can be performed with i==j, which is useless but harmless, as 0 ≤ i ≤ j < n.
Update: as done by #fabian, you can compress the three loops in one.
for (i= 0, j= n-1; i < j; )
{
if (a[i] == 0)
i++;
else if (a[j] == 0)
j--;
else
{
Swap(a[i], a[j]);
i++; j--;
}
}

Answer 6

you need to change the logic of reverse.
#include<iostream>
using namespace std;
int main()
{
int arr[100], tot, i, j, temp;
cout<<"Enter the Size for Array: ";
cin>>tot;
cout<<"Enter "<<tot<<" Array Elements: ";
for(i=0; i<tot; i++)
cin>>arr[i];
cout<<"\nThe Original Array is:\n";
for(i=0; i<tot; i++)
cout<<arr[i]<<" ";
j = tot-1;
for ( i = 0, j = tot - 1; i < tot/2; i++, j--)
{
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
cout << "\n Reverse all elements of the array: " << endl;
// use for loop to print the reverse array
for ( i = 0; i < tot; i++)
{
cout << arr[i] << " ";
}
cout<<endl;
return 0;
}
I hope this will help