Moving positions of strings in an array

Question

int moveToEnd(string a[], int n, int pos);

Eliminate the item at position pos by copying all elements after it one place to the left. Put the item that was thus eliminated into the last position of the array. Return the original position of the item that was moved to the end. Here's an example:

string actors[5] = { "peter", "lois", "meg", "chris", "stewie" };
int j = moveToEnd(actors, 5, 1);  // returns 1
// actors now contains:  "peter"  "meg"  "chris"  "stewie"  "lois"

This is what I have so far:

int moveToEnd(string a[], int n, int pos)
{int i = 0;
int initial;
int initial2;
int initial3;
for (i=0; i<n; i++)
    {if (i<pos)
        initial=i-1;
    if (i>pos)
        initial2=i-1;
    else 
        pos + (n-pos);
}}

it is totally wrong but I am stuck trying to figure out how to move the position to the end and than shift all the other elements to the left.

Petar Minchev · Accepted Answer · 2011-05-01 16:31:18Z

1

string save = arr[pos];

for (int i = pos; i < n - 1; i++)
   arr[i] = arr[i + 1];

arr[n - 1] = save;

edited May 1, 2011 at 16:31

answered May 1, 2011 at 16:14

Petar Minchev

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6 Comments

Brandyn Over a year ago

Thank You! What do I return at the end because it is giving me a message saying no value returned. I was thinking that the new position called save needs to be but that is a string instead of an int

Petar Minchev Over a year ago

@Brandyn - arr is the input array. In your case it is called a.

Brandyn Over a year ago

Slowly getting somewhere. It now compiles but once it does so it gives an error saying the project is not working. this is what i formatted it to be.

Brandyn Over a year ago

{ int i; string save = a[pos]; for (i = pos; i < n - 1; i++) a[i] = a[i + 1]; a[n] = save; return pos; }

Petar Minchev Over a year ago

@Brandyn - It should be arr[n - 1] = save.

|

Brian Roach · Accepted Answer · 2011-05-01 16:21:06Z

1

You don't need your second argument, as you can get the length of the array with a.length

The basic logic is this:

store the element at a[pos] in a temp String variable.
iterate from a[pos] to a[a.length - 2] storing a[pos + 1] to a[pos]
store the temp String to a[a.length - 1]

edited May 1, 2011 at 16:21

answered May 1, 2011 at 16:15

Brian Roach

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