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Number of Subsequences with Even and Odd Sum | Set 2
• Last Updated : 16 Sep, 2019

Given an array arr[] of size N. The task is to find the number of subsequences whose sum is even and the number of subsequences whose sum is odd.

Examples:

Input: arr[] = {1, 2, 2, 3}
Output: EvenSum = 7, OddSum = 8
There are 2N-1 possible subsequences.
The subsequences with even sum are
1) {1, 3} Sum = 4
2) {1, 2, 2, 3} Sum = 8
3) {1, 2, 3} Sum = 6 (Of index 1)
4) {1, 2, 3} Sum = 6 (Of index 2)
5) {2} Sum = 2 (Of index 1)
6) {2, 2} Sum = 4
7) {2} Sum = 2 (Of index 2)
and the rest subsequence is of odd sum.

Input: arr[] = { 2, 2, 2, 2 }
Output: EvenSum = 15, OddSum = 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This article already exists here having time complexity of where N is the size of array. Visit here before moving further.

• If we can find the number of odd subsequences then we can easily find the number of even subsequences.
• The odd subsequences can be formed in two ways:
1. By taking odd number odd times.
2. Taking even number and odd number odd time.
• Below are some variables and their definition:
• N = Total number of element in the array.
• Even = Total number of evens in the array.
• Odd = Total number of odd in the array.
• Tseq = Total number of subsequences.
• Oseq = Total number of subsequences with only odd number.
• Eseq = Total number of subsequences with even number.
• OddSumSeq = Total number of subsequences with odd sum.
• EvenSumSeq = Total number of subsequences with even sum.

Tseq = 2N – 1 = Even Subsequence + OddSubsequence

Eseq = 2Even

Oseq = OddC1 + OddC3 + OddC5 + . . . .
where OddC1 = Choosing 1 Odd
OddC3 = Choosing 3 Odd and so on

We can reduce the above equation by this identity, Hence

Oseq = 2Odd – 1

So, hence for the total odd sum subsequence, it can be calculated by multiplying these above result

OddSumSeq = 2Even * 2Odd – 1

EvenSumSeq = 2N – 1 – OddSumSeq

Below is the implementation of the above approach:

## C++

 `// CPP program to find number of``// Subsequences with Even and Odd Sum``#include ``using` `namespace` `std;`` ` `// Function to find number of``// Subsequences with Even and Odd Sum``pair<``int``, ``int``> countSum(``int` `arr[], ``int` `n)``{``    ``int` `NumberOfOdds = 0, NumberOfEvens = 0;`` ` `    ``// Counting number of odds``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] & 1)``            ``NumberOfOdds++;`` ` `    ``// Even count``    ``NumberOfEvens = n - NumberOfOdds;`` ` `    ``int` `NumberOfOddSubsequences = (1 << NumberOfEvens)``                                  ``* (1 << (NumberOfOdds - 1));`` ` `    ``// Total Subsequences is (2^n - 1)``    ``// For NumberOfEvenSubsequences subtract``    ``// NumberOfOddSubsequences from total``    ``int` `NumberOfEvenSubsequences = (1 << n) - 1``                                   ``- NumberOfOddSubsequences;`` ` `    ``return` `{ NumberOfEvenSubsequences,``             ``NumberOfOddSubsequences };``}`` ` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 2, 3 };`` ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`` ` `    ``// Calling the function``    ``pair<``int``, ``int``> ans = countSum(arr, n);`` ` `    ``cout << ``"EvenSum = "` `<< ans.first;``    ``cout << ``" OddSum = "` `<< ans.second;`` ` `    ``return` `0;``}`

## Java

 `// Java program to find number of``// Subsequences with Even and Odd Sum``import` `java.util.*;`` ` `class` `GFG ``{``static` `class` `pair``{ ``    ``int` `first, second; ``    ``public` `pair(``int` `first, ``int` `second) ``    ``{ ``        ``this``.first = first; ``        ``this``.second = second; ``    ``} ``}`` ` `// Function to find number of``// Subsequences with Even and Odd Sum``static` `pair countSum(``int` `arr[], ``int` `n)``{``    ``int` `NumberOfOdds = ``0``, NumberOfEvens = ``0``;`` ` `    ``// Counting number of odds``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``if` `(arr[i] % ``2` `== ``1``)``            ``NumberOfOdds++;`` ` `    ``// Even count``    ``NumberOfEvens = n - NumberOfOdds;`` ` `    ``int` `NumberOfOddSubsequences = (``1` `<< NumberOfEvens) * ``                                  ``(``1` `<< (NumberOfOdds - ``1``));`` ` `    ``// Total Subsequences is (2^n - 1)``    ``// For NumberOfEvenSubsequences subtract``    ``// NumberOfOddSubsequences from total``    ``int` `NumberOfEvenSubsequences = (``1` `<< n) - ``1` `- ``                                    ``NumberOfOddSubsequences;`` ` `    ``return` `new` `pair(NumberOfEvenSubsequences,``                    ``NumberOfOddSubsequences);``}`` ` `// Driver code``public` `static` `void` `main(String[] args) ``{``    ``int` `arr[] = { ``1``, ``2``, ``2``, ``3` `};`` ` `    ``int` `n = arr.length;`` ` `    ``// Calling the function``    ``pair ans = countSum(arr, n);`` ` `    ``System.out.print(``"EvenSum = "` `+ ans.first);``    ``System.out.print(``" OddSum = "` `+ ans.second);``}``} `` ` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 program to find number of ``# Subsequences with Even and Odd Sum `` ` `# Function to find number of ``# Subsequences with Even and Odd Sum ``def` `countSum(arr, n) : `` ` `    ``NumberOfOdds ``=` `0``; NumberOfEvens ``=` `0``; `` ` `    ``# Counting number of odds ``    ``for` `i ``in` `range``(n) : ``        ``if` `(arr[i] & ``1``) : ``            ``NumberOfOdds ``+``=` `1``; `` ` `    ``# Even count ``    ``NumberOfEvens ``=` `n ``-` `NumberOfOdds; `` ` `    ``NumberOfOddSubsequences ``=` `(``1` `<< NumberOfEvens) ``*` `\``                              ``(``1` `<< (NumberOfOdds ``-` `1``)); `` ` `    ``# Total Subsequences is (2^n - 1) ``    ``# For NumberOfEvenSubsequences subtract ``    ``# NumberOfOddSubsequences from total ``    ``NumberOfEvenSubsequences ``=` `(``1` `<< n) ``-` `1` `-` `\``                                ``NumberOfOddSubsequences; `` ` `    ``return` `(NumberOfEvenSubsequences, ``            ``NumberOfOddSubsequences); `` ` `# Driver code ``if` `__name__ ``=``=` `"__main__"``: `` ` `    ``arr ``=` `[ ``1``, ``2``, ``2``, ``3` `]; `` ` `    ``n ``=` `len``(arr); `` ` `    ``# Calling the function ``    ``ans ``=` `countSum(arr, n); `` ` `    ``print``(``"EvenSum ="``, ans[``0``], end ``=` `" "``); ``    ``print``(``"OddSum ="``, ans[``1``]); `` ` `# This code is contributed by AnkitRai01`

## C#

 `// C# program to find number of``// Subsequences with Even and Odd Sum``using` `System;``     ` `class` `GFG ``{``public` `class` `pair``{ ``    ``public` `int` `first, second; ``    ``public` `pair(``int` `first, ``int` `second) ``    ``{ ``        ``this``.first = first; ``        ``this``.second = second; ``    ``} ``}`` ` `// Function to find number of``// Subsequences with Even and Odd Sum``static` `pair countSum(``int` `[]arr, ``int` `n)``{``    ``int` `NumberOfOdds = 0, NumberOfEvens = 0;`` ` `    ``// Counting number of odds``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(arr[i] % 2 == 1)``            ``NumberOfOdds++;`` ` `    ``// Even count``    ``NumberOfEvens = n - NumberOfOdds;`` ` `    ``int` `NumberOfOddSubsequences = (1 << NumberOfEvens) * ``                                  ``(1 << (NumberOfOdds - 1));`` ` `    ``// Total Subsequences is (2^n - 1)``    ``// For NumberOfEvenSubsequences subtract``    ``// NumberOfOddSubsequences from total``    ``int` `NumberOfEvenSubsequences = (1 << n) - 1 - ``                                    ``NumberOfOddSubsequences;`` ` `    ``return` `new` `pair(NumberOfEvenSubsequences,``                    ``NumberOfOddSubsequences);``}`` ` `// Driver code``public` `static` `void` `Main(String[] args) ``{``    ``int` `[]arr = { 1, 2, 2, 3 };`` ` `    ``int` `n = arr.Length;`` ` `    ``// Calling the function``    ``pair ans = countSum(arr, n);`` ` `    ``Console.Write(``"EvenSum = "` `+ ans.first);``    ``Console.Write(``" OddSum = "` `+ ans.second);``}``} `` ` `// This code is contributed by 29AjayKumar`
Output:
```EvenSum = 7 OddSum = 8
```

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