CareerCup

int main()
{
   int a[]= { 5, 5, 5, 2, 2 };
int num1=a[0],count1=1;
int num2,count2=0;
int i;
for(i=1;i<5;i++)
{
  if(a[i]==num1)
  count1++;
  else
   {
    if(count2==0)
    {
      count2=1;
      num2=a[i];
    }
    else if(a[i]==num2)
     count2++;
   }
}

if(count1==3&&count2==2)
printf("true");
else if(count2==3&&count1==2)
printf("true");
else
printf("false");
   return 0;
}

1. Use extra array to count each number times. For porker, int array[15]={0}
2. Go through each number i , array[i]++;
3. Go through array[], it contains only 2 or 3.

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

def pocker_func(arr);
number_count=defaultdict(int)
for val in arr:
number_count[val]+=1
inters = set(number_count.values()).intersection(set([2,3])
if inters == set([2,3]): return True
else: return False

<pre lang="" line="1" title="CodeMonkey5827" class="run-this">#include <stdio.h>
#include <conio.h>
bool func1(int a[]);
void main()
{
int a[5],i;
printf("Enter");
for(i=0;i<5;i++)
scanf("%d",&a[i]);
printf("\nReturns %d",func1(a));
getch();
}
bool func1(int s[5])
{
int i,j,temp=0,count=0,y=1,z=0;
printf("\nSorted Elements");
for(i=0;i<5;i++)
{
for(j=i;j<5;j++)
{
if(s[i]>s[j])
{
temp=s[i];
s[i]=s[j];
s[j]=temp;
}
}
printf("\n%d",s[i]);

}
printf("\nDistinct Elements ");
for(i=0;i<5;i++)
{
if(s[i]!=s[i+1])
{
printf("\n %d",s[i]);
count++;

}

}
printf("\nCount=%d",count);
for(i=0;i<5;i++)
{
if(s[i]==s[i+1])
{
y++;
z=y;
}
else
{
printf("\n%d repeated %d times",s[i],y);
y=1;
}
}
if(count==2 && z>1 && z<4){
printf("\nz=%d",z);
return true;}
else
return false;
}

</pre><pre title="CodeMonkey5827" input="yes">
</pre>

1) The return can be true only if there are only 2 distinct integers coming in the input array.
2) Use a 2x2 array to store the number and frequency of the number.
3) After the scan through the input, see if the frequency matches the required condition for a true return.

public bool PokerFun(int[] inputArray)
        {
            int[][] containerArray = new int[2][];
            for (int i = 0; i < 2; i++)
            {
                containerArray[i] = new int[2];
                for (int j = 0; j < 2; j++)
                {
                    containerArray[i][j] = 0;
                }
            }

            for (int i = 0; i < inputArray.Length; i++)
            {
                if (containerArray[0][0] != 0 && containerArray[0][0] == inputArray[i])
                {
                    containerArray[0][1]++;
                }
                else if (containerArray[0][0] == 0)
                {
                    containerArray[0][0] = inputArray[i];
                    containerArray[0][1]++;
                }
                else if (containerArray[1][0] != 0 && containerArray[1][0] == inputArray[i])
                {
                    containerArray[1][1]++;
                }
                else if (containerArray[1][0] == 0)
                {
                    containerArray[1][0] = inputArray[i];
                    containerArray[1][1]++;
                }
                else
                {
                    return false;
                }
            }

            if ((containerArray[0][1] == 2 || containerArray[0][1] == 3)
                && (containerArray[1][1] == 2 || containerArray[1][1] == 3))
            {
                return true;
            }

            return false;
        }

Test input arrays:

int[] inputArray1 = new int [] { 1, 1, 2, 2, 2 };

int[] inputArray2 = new int[] { 1, 2, 2, 2, 2 };

int[] inputArray3 = new int[] { 1, 1, 3, 2, 2 };

int[] inputArray4 = new int[] { 1, 1, 2, 5, 4 };

int[] inputArray5 = new int[] { 1,3, 2, 4, 5 };
Worked well with the tests I have.

This iz perfect

Kamal's answer seems perfect.

Kamal's Answer could have been better if he just places that else if block before else and letting else block at the last!!!

class Counter {
	public  Counter(int a[]) {
		int first=a[0],flag=0,flag2=0,second=0;
		for (int i=1;i<5;i++) {
			if (first==a[i])
				flag++;
			else{
				if (flag2==0)
					second=a[i];
				if(second==a[i])
					flag2++;
				else
					break;
			}
		}
		if ((flag==1&&flag2==3)||(flag==2&&flag2==2))
			System.out.println("Three of a kind");
		else
			System.out.println("Incorrect format");
	}
}
public class TwoThree {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] a={1,2,1,2,2};
		Counter obj= new Counter(a);
	}

}

How about using Hashmap, to store the frequency of occurrences as a value?? I can see solution with O(n) space complexity, is there a better solution to this?

Thanks in advance