Rectangular chocolate bar Create at least one piece which consists of exactly nTiles tiles

Rectangular chocolate bar Create at least one piece which consists of exactly nTiles tiles.

You have a rectangular chocolate bar that consists of width x height square tiles. You can split it into two rectangular pieces by creating a single vertical or horizontal break along tile edges. For example, a 2x2 chocolate bar can be divided into two 2x1 pieces, but it cannot be divided into two pieces, where one of them is 1x1. You can repeat the split operation as many times as you want, each time splitting a single rectangular piece into two rectangular pieces.
Your goal is to create at least one piece which consists of exactly nTiles tiles. Return the minimal number of split operations necessary to reach this goal. If it is impossible, return -1.

Complete the function getMinSplit, which takes in 3 integers as parameters. The first parameter is width of the chocolate, the second is height of the chocolate and third is nTiles, the number of tiles required.

Constraints
- width will be between 1 and 109, inclusive.
- hight will be between 1 and 109, inclusive.
- nTiles will be between 1 and 109, inclusive.

Example 1 Example 1 Example 1 Example 1 Example 1
5  12 2 17 226800000
4  10 2 19 10000000
12 120 1 111 938071715
Returns: 1 Returns: -1 Returns: 2 Returns: -1 Returns: 1

ANSWER

Logic : 

First Try to understand that
Answers can be only 
-1 ( not possible )
1 (in only one cut)
2 (in two cut - One Horizontally and One Vertically )
Why max 2  ? Think Its very easy. (as we have only 2D)
for 3D cube Max can be 3. :P Now you got ?

Now 
With Following Algorithm code is very easy every one can write .. 

W = No Width
H = No Height
nT = nTiles given..

Thanks to Dhaval for suggesting this approach and code

#include <iostream>
#include <stdio.h>
 
inline int minimum (int a, int b) { return a < b ? a : b; }
int fact[100]={0};
void getFactors(int n){
int i,j=0;
for (i=2;i<n;i++){
         if(n%i==0) {
         fact[j]=i;
         j++;
         }
}//for
printf("factors");
i=0;
while(fact[i]>0){
 printf(" %d ",fact[i]);
 i++;
}
}//getFactor
int getMinTiles( int width , int height, int nTiles)
{
int w=width;
int h=height;
int n=nTiles;
int i,j,l,fact1,fact2;
int min=10; //min any number greater thn 2 
 
 
if( w*h <= n ) return -1; 
// w*h total number of tiles are more or equal to required, return -1
 
getFactors(n);
//store all factors of a nTiles except 1 and nTiles.
i=l=0;
while(fact[i]>0){
 i++;
 
}//while
l=i;
printf("\nLength - %d ",l);
//l = length(fact);
 
//for each pair of factors
for (i=0,j=l ; i <=j ; i++,j--)
{
fact1=fact[i];
fact2=fact[j];
 
if ( fact1 == w && fact2 < h ) {min = 1;}
else if ( fact2 == h && fact1 < w )  {min = 1;}
 
// eg : fact1=4,fact2=3 ; h=4 ,w=5 , n=12;
// When row or col number are matching with width or column , with only one cut we can produce desired result.
// if another fact is smaller than column or width (opposite)
 
else if ( (fact1 < w && fact1 < h) ||  (fact2 < w && fact2 < h) )  
{ min = minimum (min, 2); }
// both factors are very small then height and width then 2 cut are required one horizontally and one vertically) 
 
 
else { min = 0;}
 
}//for 
 
return min;
 
}//Function
 
int main() {
 int cut;
 cut = getMinTiles( 5, 4, 18);
 printf("\nNumber of cut required %d ",cut);
 return 0;
}

You can check it at http://ideone.com/tvR7yx
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