# Number of Subsequences with Even and Odd Sum | Set 2

• Last Updated : 29 Mar, 2022

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

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`

## Javascript

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Output:

`EvenSum = 7 OddSum = 8`

Time Complexity: O(n)

Auxiliary Space: O(1)

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