MaxCupCake in codefights

this is good problem.

https://codefights.com/challenge/X33imTX2FSqhSLTtJ


My code is

        int c, i,t;
 int MaxCupCakes(int N, int[] P, int K) {
  for (; i < P.length; i++ ) {
   t= (P[i] -c -1);
   if(K<=t) break;
   K=K-t;
   c=P[i];
  }  
  return c+K<= N ?c+K : -1;  
 }

best code is here

 int MaxCupCakes(int n, int[] P, int k) {
     for(int i:P) 
         if(i<=k)k++;
           
  return  k>n ?-1 : k ;
 }

Max and Caroline, two girls in their mid-twenties, work at a Brooklyn restaurant as waitresses. Together, they dream of starting up their cupcake business.

One day Max comes with a box of N cupcakes numbered according to their quality from 1 to N. Caroline has a list of cupcakes P that should be removed from the box.

Your task it to find the quality of the Kth cupcake after the cupcakes from the list P are removed from the box.

If it is not possible to get the Kth cupcake, return -1 instead.

Example

For N = 4, P = [1] and K = 2, the output should be
MaxCupCakes(N, P, K) = 3.

Initially there were cupcakes of the following quality:
1, 2, 3, 4.
According to P, the cupcake with quality 1 should be removed, so only the following cupcakes are left:
2, 3, 4.
The 2nd cupcake in this list is 3, thus the output should be 3 as well.


[input] integer N

The number of cupcakes, 4 ≤ N ≤ 109.


[input] array.integer P

A sorted array of positive integers, the cupcakes to be removed.
0 ≤ P.length ≤ 500,
1 ≤ P[i] ≤ N.


[input] integer K

A positive integer, the 1-based number of the cupcake to find.


[output] integer

The quality of the Kth cupcake, or -1 if less than K cupcakes are left.