Finding consecutive integers in an Array - objective-c

Question

I am looking for an algorithm to determine if there are at least three consecutive numbers in an array. I have found several and tweaked them a bit, but it does not seem to be working. Below is what I am currently doing. The array is sorted using an NSSortDescriptor prior to the loop.
For example:
in an array [5,6,101,102,103] the three consecutive numbers are [101,102,103]
and the below function should return YES.
int c = [checkArray count];
int a,b = 0;
int cnt = 1;
for (int i = 1; i < c; i++)
{
a = [[checkArray objectAtIndex:i] intValue];
b = [[checkArray objectAtIndex:i-1] intValue] - 1;
if (a == b)
{
cnt++;
if (cnt == 3)
return YES;
} else {
cnt = 1;
}
}
return NO;

Answer 1

This is an alternative approach. I haven't tested it, but you should get the idea.
int c = [checkArray count];
int a,b,c = 0;
a = [[checkArray objectAtIndex:0] intValue];
b = [[checkArray objectAtIndex:1] intValue];
for (int i = 2; i < c; i++)
{
c = [[checkArray objectAtIndex:i] intValue];
if (a+2 == b+1 && b+1 == c)
return YES;
a = b;
b = c;
}
return NO;
It has the additional advantage of being easily optimized.

Answer 2

You should be able to do something like...
for (int i = 1; i < checkArray.count-1; i++)
{
int lower = [checkArray[i-1] intValue] +1;
int mid = [checkArray[i] intValue];
int upper = [checkArray[i+1] intValue] -1;
if (lower == mid && mid == upper) {
return YES;
}
}
return NO;
You only need to check each number once. No need to count anything. Each number you check is the middle of three numbers. If the one before and the one after are one less and one greater respectively then the three numbers are consecutive.