面试-在数组中找到偶数和对

Question

Given an array , how would you return the number of pairs which sum to an even number?

For example:

a[] = { 2 , -6 , 1, 3, 5 }

In this array , the nos which pair to an even sum are (2,-6), (1,3) , (1,5), (3,5)

Function should return 4 as there are 4 pairs or -1 if none.

Expected time complexity - O(N) worst case Expected space complexity - O(N) worst case

Approach 1: Brute force

Start with the first number
  Start with second number
      assign the sum to a temp variable
      check if the temp is even
          If it is increment evenPair count
      else
          increment the second index

Here the time complexity is O(N2)

Answer 1

int odd = 0, even = 0;
for (int i = 0; i < n; i++) {
    if (a[i] % 2 == 0) {
        even++;
    } else {
        odd++;
    }
}
int answer = (odd * (odd - 1) + even * (even - 1)) / 2;

Answer 2

If to use standard algorithms then the code can look the following way

#include <iostream>
#include <utility>
#include <numeric>
#include <iterator>

int main() 
{
    int a[] = { 2 , -6 , 1, 3, 5 };

    typedef size_t Odd, Even;
    auto p = std::accumulate( std::begin( a ), std::end( a ), 
                              std::pair<Odd, Even>( 0, 0 ),
                              []( std::pair<Odd, Even> &acc, int x )
                              {
                                return x & 1 ? ++acc.first : ++acc.second, acc;
                              } );

    std::cout << "There are " 
              << ( p.first * ( p.first - 1 ) + p.second * ( p.second - 1 ) ) / 2
              << " even sums" << std::endl;

    return 0;
}

The output is

There are 4 even sums

Take into account that n! / ( 2! * ( n - 2 )! ) n! / ( 2! * ( n - 2 )! ) is equivalent to ( n - 1 ) * n / 2

The advantage of using the standard algorithm is that you can use any subrange of a sequence. Also you can use a standard stream input because std::accumulate uses input iterator.

Also it would be much better if in the description of the assignment there would be written that the function should return 0 instead of -1 if there is no even sums among elements of an array.

It is not shameful to show this code in an interview though I advice never do any assignments in an interview. Interview is not an exam.

Answer 3

If a and b are even, a + b is even.
If they're odd, a + b is also even.
If one is odd and one is even, a + b is odd.

This means that we don't need to perform any additions, we only need to know how many numbers there are of each kind.
Finding this out takes linear time.

If you have k numbers, there are k - 1 pairs that include the first number, k - 2 involving the second, and so on.
Using the familiar summation formula, sum(1 .. n) = (n * (n + 1)) / 2 ,

let

ne = number of even numbers
no = number of odd numbers

then,

number of pairs = (ne-1) * ne / 2 + (no-1) * no / 2 
                = ((ne-1)*ne + (no-1)*no) / 2

That computation runs in constant time, so the time complexity is still linear.

And we only need a constant amount of extra space, which is better than the requirements.

---

Possible followup interview questions possibly worth thinking about:

What happens to the complexities (both in time and space) if:

• duplicates are only counted once, ie {1,1,1,2} only has two such pairs?
• we disregard order, ie (1,2) and (2,1) are the "same" pair?

Answer 4

A different approach would be compute the number of sums by even and odds, and then sum then together

int sumO = 0 , sumE = 0 , numO = 0 , numE = 0;
for (int i=0; i < 5; i++)
{
    if (a[i] % 2 == 0)
    {
        sumE += numE;
        numE ++;
    }
    else
    {
        sumO += numO;
        numO ++;
    }
}

printf ("Total: %d\n", sumO + sumE);

Answer 5

I got this question from an interview. No need to total even and odd both separate.

public static int evenTotalPairs(int[] A) {
int size = A.length;
int evens = 0;
for(int i = 0; i < size; i++ ){
    if(A[i] % 2 == 0){
        evens++;
    }
}
return evens*(evens-1)/2 + (size-evens)*(size-evens-1)/2;

}