Monday, June 20, 2016

Find number of subarrays with even sum - GeeksforGeeks


Find number of subarrays with even sum - GeeksforGeeks
Given an array, find the number of subarrays whose sum is even.
O(n) Time and O(1) Space Method [Efficient]
If we do compute the cumulative sum array in temp[] of our input array, then we can see that the sub-array starting from i and ending at j, has an even sum if temp[] if (temp[j] – temp[i]) % 2 = 0. So, instead of building a cumulative sum array we build a cumulative sum modulo 2 array, and find how many times 0 and 1 appears in temp[] array using handshake formula. [n * (n-1) /2]
int countEvenSum(int arr[], int n)
{
    // A temporary array of size 2. temp[0] is
    // going to store count of even subarrays
    // and temp[1] count of odd.
    // temp[0] is initialized as 1 because there
    // a single even element is also counted as
    // a subarray
    int temp[2] = {1, 0};
 
    // Initialize count.  sum is sum of elements
    // under modulo 2 and ending with arr[i].
    int result = 0, sum = 0;
 
    // i'th iteration computes sum of arr[0..i]
    // under modulo 2 and increments even/odd count
    // according to sum's value
    for (int i=0; i<=n-1; i++)
    {
        // 2 is added to handle negative numbers
        sum = ( (sum + arr[i]) % 2 + 2) % 2;
 
        // Increment even/odd count
        temp[sum]++;
    }
 
    // Use handshake lemma to count even subarrays
    // (Note that an even cam be formed by two even
    // or two odd)
    result = result + (temp[0]*(temp[0]-1)/2);
    result = result + (temp[1]*(temp[1]-1)/2);
 
    return (result);
}



O(n2) time and O(1) space method [Brute Force]
We can simply generate all the possible sub-arrays and find whether the sum of all the elements in them is an even or not. If it is even then we will count that sub-array otherwise neglect it.
int countEvenSum(int arr[], int n)
{
    int result = 0;
 
    // Find sum of all subarrays and increment
    // result if sum is even
    for (int i=0; i<=n-1; i++)
    {
        int sum = 0;
        for (int j=i; j<=n-1; j++)
        {
            sum = sum + arr[j];
            if (sum % 2 == 0)
                    result++;
        }
    }
 
    return (result);
}
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