Count the number of pairs in an array that sum to an even number.

• Iterate through the array and check each pair of elements.

• If the sum of the pair is divisible by 2, increment the count.

• Return the coun...read more

((cntEven * (cntEven – 1)) / 2) + ((cntOdd * (cntOdd – 1)) / 2)

There can be a total of n(n-1) 2 numbers of total pairs from the given arrays number

It's can be done by the location and arithmetic operator with the help of if and else condition with a loop for the location of columns numbers in array and outer loop for changing the rows .

In order to get the pair sum as even, all the even elements will be paired with only even elements and all the odd elements will be paired with only odd elements and the count will be ((cntEven * (cnt...read more

If both number are of same type(even or odd) then their sum always give even no. And everytime increase count by 1. Otherwise no increment.

The sum of two odd no always give a sum to even. Example:- 3+3=6, 3+5=8

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