Count subarrays with same even and odd elements

Given an array of N integers, count number of even-odd subarrays. An even – odd subarray is a subarray that contains the same number of even as well as odd integers.

Examples :

Input : arr[] = {2, 5, 7, 8} 
Output : 3
Explanation : There are total 3 even-odd subarrays.
               1) {2, 5}
               2) {7, 8}
               3) {2, 5, 7, 8}

Input : arr[] = {3, 4, 6, 8, 1, 10} 
Output : 3
Explanation : In this case, 3 even-odd subarrays are:
               1) {3, 4}
               2) {8, 1}
               3) {1, 10}

This problem is mainly a variation of count subarrays with equal number of 0s and 1s.

A naive approach would be to check for all possible subarrays using two loops, whether they are even-odd subarrays or not. This approach will take O(N^2) time.

An Efficient approach solves the problem in O(N) time and it is based on following ideas:

  • Even-odd subarrays will always be of even length.
  • Maintaining track of the difference between the frequency of even and odd integers.
  • Hashing of this difference of frequencies is useful in finding number of even-odd subarrays.

The basic idea is to use the difference between the frequency of odd and even numbers to obtain an optimal solution. We will maintain two integer hash arrays for the positive and negative value of the difference.
-> Example to understand in better way :
-> Consider difference = freq(odd) – freq(even)
-> To calculate this difference, increment the value of ‘difference’ when there is
an odd integer and decrement it when there is an even integer. (initially, difference = 0)
arr[] = {3, 4, 6, 8, 1, 10}

index 0 1 2 3 4 5 6

array 3 4 6 8 1 10

difference 0 1 0 -1 -2 -1 -2

-> Observe that whenever a value ‘k’ repeats in the ‘difference’ array, there exists an
even-odd subarray for each previous occurrence of that value i.e. subarray exists from
index i + 1 to j where difference[i] = k and difference[j] = k.

-> Value ‘0’ is repeated in ‘difference’ array at index 2 and hence subarray exists for
(0, 2] indexes. Similarly, for repetition of values ‘-1’ (at indexes 3 and 5) and ‘-2’ (at
indexes 4 and 6), subarray exists for (3, 5] and (4, 6] indexes.

Below is the C++ implementation of the O(N) solution described above.

[sourcecode language=”CPP” highlight=”6-57″]
/*C++ program to find total number of
even-odd subarrays present in given array*/
#include <iostream>
using namespace std;

// function that returns the count of subarrays that
// contain equal number of odd as well as even numbers
int countSubarrays(int arr[], int n)
{
// initialize difference and answer with 0
int difference = 0;
int ans = 0;

// create two auxiliary hash arrays to count frequency
// of difference, one array for non-negative difference
// and other array for negative difference. Size of these
// two auxiliary arrays is ‘n+1’ because difference can
// reach maximum value ‘n’ as well as minimum value ‘-n’
int hash_positive[n + 1], hash_negative[n + 1];

// initialize these auxiliary arrays with 0
fill_n(hash_positive, n + 1, 0);
fill_n(hash_negative, n + 1, 0);

// since the difference is initially 0, we have to
// initialize hash_positive[0] with 1
hash_positive[0] = 1;

// for loop to iterate through whole
// array (zero-based indexing is used)
for (int i = 0; i < n ; i++)
{
// incrementing or decrementing difference based on
// arr[i] being even or odd, check if arr[i] is odd
if (arr[i] & 1 == 1)
difference++;
else
difference–;

// adding hash value of ‘difference’ to our answer
// as all the previous occurrences of the same
// difference value will make even-odd subarray
// ending at index ‘i’. After that, we will increment
// hash array for that ‘difference’ value for
// its occurrence at index ‘i’. if difference is
// negative then use hash_negative
if (difference < 0)
{
ans += hash_negative[-difference];
hash_negative[-difference]++;
}

// else use hash_positive
else
{
ans += hash_positive[difference];
hash_positive[difference]++;
}
}

// return total number of even-odd subarrays
return ans;
}

// Driver code
int main()
{
int arr[] = {3, 4, 6, 8, 1, 10, 5, 7};
int n = sizeof(arr) / sizeof(arr[0]);

// Printing total number of even-odd subarrays
cout << "Total Number of Even-Odd subarrays"
" are " << countSubarrays(arr,n);

return 0;
}
[/sourcecode]

Output:

Total Number of Even-Odd subarrays are 7

Time Complexity : O(N), where N is the number of integers.
Auxiliary Space : O(2N), where N is the number of integers.



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