CareerCup

XOR all the elements and at the same time in that pass calculate the sum and also the product. After XOR the element which occurs 3 times will be left. Suppose that element is x and the element which occurs 2 times is y.

so 3x + 2y = sum this way we can get y
if the values obtained satisfy x^3 * y^2 = product then the array satisfies the property.

I think this works in a single pass and no extra space. If anyone can get a counter example do tell me

if((a[0]^a[1]^a[2]==0)&&(a[3]^a[4] == 0))
return true;
else
return false

the numbers can be in any positions, the array is not sorted.

1. if you xor all the numbers then the no which occurs 3 times can be obtained, now an efficient way needs to be found out for finding the 2nd no.


2. when you get the odd no of times occurring element then again XOR each element with this element , there should be 3 zeros now and two similar nos.

0,0,0, 001, 001 ( 4^5 = 001)

3. Now again XOR all the elements with each other if the result == 0 (001^001 == 0) then return true

But this approach requires 3 passes so not an efficient one.

If anyone can suggest a better solution

if((a[0]^a[1]^a[2]^a[3]^a[4]==a[0])||(a[0]^a[1]^a[2]^a[3]^a[4]==a[4]))
return 1
else
return 0

Declare 2 variables min/max and two variable count_min = count_max =0
Set min/max in one traversal
Traverse array and increment count_min and count_max on min/max occurances
if((count_min == 2 && count_max == 3) || (count_min == 3 && count_max == 2))
return 1;

return 0;

boolean isConditionTrueForArray(int[] inputArr) {
   
    int firstCount = 0;
    int secondCount = 0;
    int firstLetter = inputArr[0];
    int secondLetter;
    for ( int i = 1; i < 5; i++ ) { 
          if ( firstLetter != array[i])  {
                secondLetter = array[i];
                break;
          }
    }

    for ( i = 0; i < 5; i++) {
         if ( firstLetter == inputArr[i] ) {
               firstLetterCount++;
         } else if ( secondLetter == inputArr[i] ) {
               secondLetterCount++;
         } else {
               return false;
         }
   }  
   
    if ( (firstLetterCount == 3 && secondLetterCount == 2) 
         || (firstLetterCount == 2 && secondLetterCount == 3) ) {
         return true;
    } else {
        return false;
    }

}

import java.util.*;
class Counter {
public Counter(int a[]) {
Arrays.sort(a);
int tester[]=a;
int i=0;
if (tester[i]==tester[i+1]){
if (tester[i+1]==tester[i+2]) {
if (tester [i+3]== tester [i+4]) {
System.out.println("Three of a kind");
}
else
System.out.println("Format incorrect");
}
else if(tester[i+2]==tester[i+3]){
if (tester[i+3]==tester[i+4]) {
System.out.println("Three of a kind");
}
else
System.out.println("Format incorrect");
}
}
else
System.out.println("Format incorrect");

}
}
public class TwoThree {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] a={2,3,2,3,4};
Counter obj= new Counter(a);
}
}

bool isSame(int arr[5], const int size){
    int M[10] = {0};
    bool validThree = false, validTwo = false;
    
    for(int i = 0; i < size; ++i){
    	M[arr[i]]++;
    }
    
    for(int i = 0; i < 10; ++i){
    	if(M[i] == 3){
    		validThree = true;
    	}
    	if(M[i] == 2){
    		validTwo = true;
    	}
    }
    return validThree && validTwo;
}

A little too many if elsem but it does it in O(N).
Logic:
-Start with storing first new number, increment counter for first new number.
-Store the second new number and increment the counter for second number.
-If there is a third new number return False, if the next number encountered is same as the first or second numbers, increment respective counters.
-At the end check for valid counter numbers or else return false.

def formatCorrect(inArray):
    first = -1
    firstCount = 0
    second = -1
    secondCount = 0
    for num in inArray: 
        if first != -1:
            if num == first:
                firstCount += 1
            elif second != -1:
                if num == second:
                    secondCount +=1
                else:
                    #print firstCount, secondCount
                    return False
            else:
                second = num
                secondCount += 1
        else:
            first = num
            firstCount += 1 
        #print "First Number is "+str(first)+" with count "+str(firstcount)+"."
        #print "Second Number is "+str(second)+" with count "+str(secondcount)+"."
    if (firstCount == 3 and secondCount == 2) or (firstCount == 2 and secondCount == 3):
        #print firstCount, secondCount
        return True
    else:
        #print firstCount, secondCount
        return False

a mixted strategy solution based on the comments before

bool isCount2And3(int A[]){
        int *num1 = NULL, *num2 = NULL;
        int sum = A[0];
        for(int i=1; i<5;++i){
                sum+=A[i];
                if(!num1) num1=A+i;
                if(*num1 == A[i]) continue;
                if(!num2) num2=A+i;
                if(num2 && *num2!=A[i]) return false;
        }
        return sum == (*num1)*3+(*num2)*2 || sum == (*num1)*2 + (*num2)*3;
}

Dim a As Array = {4, 5, 4, 5, 4}
        Dim i As Integer
        Dim count As Integer = 0
        Dim secCount As Integer = 0
       
        For i = 0 To a.Length - 1
            If (a(0).Equals(a(i))) Then
                count = count + 1
            End If
        Next
        For i = 0 To a.Length - 1
            If Not (a(0).Equals(a(i))) Then
                For k = 0 To a.Length - 1
                    If (a(i).Equals(a(k))) Then
                        secCount = secCount + 1
                    End If
                Next
                Exit For
            End If
        Next
        If ((count = 3 And secCount = 2) Or (count = 2 And secCount = 3)) Then
            Return True
        Else
            Return False
        End If

Dim a As Array = {4, 5, 4, 5, 4}
        Dim i As Integer
        Dim count As Integer = 0
        Dim secCount As Integer = 0
                For i = 0 To a.Length - 1
            If (a(0).Equals(a(i))) Then
                count = count + 1
            End If
        Next
        For i = 0 To a.Length - 1
            If Not (a(0).Equals(a(i))) Then
                For k = 0 To a.Length - 1
                    If (a(i).Equals(a(k))) Then
                        secCount = secCount + 1
                    End If
                Next
                Exit For
            End If
        Next
        If ((count = 3 And secCount = 2) Or (count = 2 And secCount = 3)) Then
            Return True
        Else
            Return False
        End If

def first(list):
count = 0
dict = {}
for item in list:
if item not in dict.keys():
dict[item] = 1
else:
val = dict[item]
dict[item] = val + 1
if len(dict.keys()) <= 2:
for item in dict:
if dict[item] == 2 or dict[item] == 3:
return True
else:
return False
else:
return False
value = first([4,4,4,5,4])
if value:
print("successful")
else:
print("unsuccessful")

def first(list):
        count = 0
        dict = {}
        for item in list:
                if item not in dict.keys():
                        dict[item] = 1
                else:
                        val = dict[item]
                        dict[item] = val + 1
        if len(dict.keys()) <= 2:
                for item in dict:
                        if dict[item] == 2 or dict[item] == 3:
                                return True
                        else:
                                return  False
        else:
                return False
value = first([4,4,4,5,4])
if value:
        print("successful")
else:
        print("unsuccessful")

if(a[0] == a[1] && a[3] == a[4])
    return true;
else
    return false;
}