Related Articles
Number of Subsequences with Even and Odd Sum
• Difficulty Level : Medium
• Last Updated : 02 Jan, 2020

Given an array, find the number of subsequences whose sum is even and the number of subsequences whose sum is odd.

Example :

Input: arr[] = {1, 2, 2, 3}
Output: EvenSum = 7, OddSum = 8
There are possible subsequences.
The subsequences with even sum is
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.

Naive Approach:
One simple approach is to generate all possible subsequences recursively and count the number of subsequences with even sum and then subtract from total subsequences and the number will be of odd subsequence. The time complexity of this approach will be .

Better Approach:
A better approach will be using Dynamic programming.

• We would be calculating the count of even subsequences as we iterate through the array. we create 2 arrays countODD[N] and countEVEN[N], where countODD[i] denotes the number of subsequences with odd sum in range and countEVEN[i] denotes the number of subsequences with even sum in range • If we are at position i, and the number is ODD then the total number of subsequences with even sum would be
``` countEVEN[i] = countEVEN[i-1] + countODD[i-1]
countODD[i] = countODD[i-1] + countEVEN[i-1] + 1
```
• For countEVEN[i], the i-th number is not paired with any other subseuence (i.e. even subsequences till position) + ith number is paired with all other odd subsequences till position (odd+odd=even)
• For countODD[i], the i-th number is not paired with any other subseuence(i.e. odd subsequences till position) + ith number is paired with all other even subsequences till position (odd+even=odd) + one subsequence with only 1 element i.e the ith number itself
• If we are at position i, and the number is EVEN then the total number of subsequences with even sum would be
``` countEVEN[i] = countEVEN[i-1] + countEVEN[i-1] + 1
countODD[i] = countODD[i-1] + countODD[i-1]
```
• For countEVEN[i], the i-th number is not paired with any other subseuence (i.e. even subsequences till position) + i-th number is paired with all other even subsequences till position (even+even=even) + one subsequence with only 1 element i.e the i-th number itself
• For countODD[i], the i-th number is not paired with any other subseuence (i.e. odd subsequences till position) + i-th number is paired with all other odd subsequences till position (even+odd=odd)

Below is the implementation of above approach:

## C++

 `// C++ implementation``#include ``using` `namespace` `std;`` ` `// returns the count of odd and``// even subsequences``pair<``int``, ``int``> countSum(``int` `arr[], ``int` `n)``{``    ``int` `result = 0;`` ` `    ``// Arrays to store the count of even``    ``// subsequences and odd subsequences``    ``int` `countODD[n + 1], countEVEN[n + 1];`` ` `    ``// Initialising countEVEN and countODD to 0``    ``// since as there is no subsequence before the``    ``// iteration with even or odd count.``    ``countODD = 0;``    ``countEVEN = 0;`` ` `    ``// Find sum of all subsequences with even count``    ``// and odd count storing them as we iterate.`` ` `    ``// Here countEVEN[i] denotes count of``    ``// even subsequences till i`` ` `    ``// Here countODD[i] denotes count of``    ``// odd subsequences till i`` ` `    ``for` `(``int` `i = 1; i <= n; i++) {`` ` `        ``// if the number is even``        ``if` `(arr[i - 1] % 2 == 0) {``            ``countEVEN[i] = countEVEN[i - 1]``                           ``+ countEVEN[i - 1] + 1;`` ` `            ``countODD[i] = countODD[i - 1]``                          ``+ countODD[i - 1];``        ``}``        ``// if the number is odd``        ``else` `{``            ``countEVEN[i] = countEVEN[i - 1]``                           ``+ countODD[i - 1];`` ` `            ``countODD[i] = countODD[i - 1]``                          ``+ countEVEN[i - 1] + 1;``        ``}``    ``}`` ` `    ``return` `{ countEVEN[n], countODD[n] };``}`` ` `// 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 implementation to find the number ``// of Subsequences with Even and Odd Sum ``import` `java.util.*; ``import` `java.lang.*;`` ` `class` `GFG ``{``    ``public` `static` `int``[] countSum(``int` `arr[], ``int` `n)``    ``{``        ``int` `result = ``0``; `` ` `        ``// Arrays to store the count of even ``        ``// subsequences and odd subsequences ``        ``int``[] countODD = ``new` `int``[n + ``1``];``        ``int``[] countEVEN = ``new` `int``[n + ``1``];``         ` `        ``// Initialising countEVEN and countODD to 0 ``        ``// since as there is no subsequence before the ``        ``// iteration with even or odd count. ``        ``countODD[``0``] = ``0``; ``        ``countEVEN[``0``] = ``0``; ``         ` `        ``// Find sum of all subsequences with even count ``        ``// and odd count storing them as we iterate. ``     ` `        ``// Here countEVEN[i] denotes count of ``        ``// even subsequences till i ``     ` `        ``// Here countODD[i] denotes count of ``        ``// odd subsequences till i ``        ``for` `(``int` `i = ``1``; i <= n; i++) ``        ``{ `` ` `            ``// if the number is even ``            ``if` `(arr[i - ``1``] % ``2` `== ``0``)``            ``{ ``                ``countEVEN[i] = countEVEN[i - ``1``] + ``                               ``countEVEN[i - ``1``] + ``1``; ``     ` `                ``countODD[i] = countODD[i - ``1``] + ``                              ``countODD[i - ``1``]; ``            ``} ``             ` `            ``// if the number is odd ``            ``else``            ``{ ``                ``countEVEN[i] = countEVEN[i - ``1``] + ``                               ``countODD[i - ``1``]; ``     ` `                ``countODD[i] = countODD[i - ``1``] + ``                              ``countEVEN[i - ``1``] + ``1``; ``            ``} ``        ``}``         ` `        ``int``[] ans = ``new` `int``[``2``];``        ``ans[``0``] = countEVEN[n];``        ``ans[``1``] = countODD[n];``        ``return` `ans;``    ``} `` ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args) ``    ``{``        ``int``[] arr = ``new` `int``[]{ ``1``, ``2``, ``2``, ``3` `}; ``        ``int` `n = ``4``;``        ``int``[] ans = countSum(arr, n);``        ``System.out.println(``"EvenSum = "` `+ ans[``0``]);``        ``System.out.println(``"OddSum = "` `+ ans[``1``]);``    ``}``}`` ` `// This code is contributed by Shivam Sharma `

## Python3

 `# Python3 implementation of above approach`` ` `# Returns the count of odd and``# even subsequences``def` `countSum(arr, n):``     ` `    ``result ``=` `0`` ` `    ``# Variables to store the count of even``    ``# subsequences and odd subsequences`` ` `    ``# Initialising count_even and count_odd to 0``    ``# since as there is no subsequence before the``    ``# iteration with even or odd count.``    ``count_odd ``=` `0``    ``count_even ``=` `0`` ` `    ``# Find sum of all subsequences with even count``    ``# and odd count and storing them as we iterate.`` ` `    ``for` `i ``in` `range``(n):`` ` `        ``# if the number is even``        ``if` `arr[i ``-` `1``] ``%` `2` `=``=` `0``:``            ``count_even ``=` `count_even ``+` `count_even ``+` `1``            ``count_odd ``=` `count_odd ``+` `count_odd`` ` `        ``# if the number is odd``        ``else``:``            ``temp ``=` `count_even``            ``count_even ``=` `count_even ``+` `count_odd``            ``count_odd ``=` `count_odd ``+` `temp ``+` `1``         ` `    ``return` `[count_even, count_odd]`` ` `# Driver code``arr ``=` `[ ``1``, ``2``, ``2``, ``3` `]``n ``=` `len``(arr)`` ` `# Calling the function``ans ``=` `countSum(arr, n)`` ` `print``(``'EvenSum ='``, ans[``0``], ``      ``'OddSum ='``, ans[``1``])`` ` `# This code is contributed ``# by Saurabh_shukla`

## C#

 `// C# implementation to find the number ``// of Subsequences with Even and Odd Sum ``using` `System;``class` `GFG ``{``    ``public` `static` `int``[] countSum(``int` `[]arr, ``int` `n)``    ``{`` ` `        ``// Arrays to store the count of even ``        ``// subsequences and odd subsequences ``        ``int``[] countODD = ``new` `int``[n + 1];``        ``int``[] countEVEN = ``new` `int``[n + 1];``         ` `        ``// Initialising countEVEN and countODD to 0 ``        ``// since as there is no subsequence before the ``        ``// iteration with even or odd count. ``        ``countODD = 0; ``        ``countEVEN = 0; ``         ` `        ``// Find sum of all subsequences with even count ``        ``// and odd count storing them as we iterate. ``     ` `        ``// Here countEVEN[i] denotes count of ``        ``// even subsequences till i ``     ` `        ``// Here countODD[i] denotes count of ``        ``// odd subsequences till i ``        ``for` `(``int` `i = 1; i <= n; i++) ``        ``{ `` ` `            ``// if the number is even ``            ``if` `(arr[i - 1] % 2 == 0)``            ``{ ``                ``countEVEN[i] = countEVEN[i - 1] + ``                               ``countEVEN[i - 1] + 1; ``     ` `                ``countODD[i] = countODD[i - 1] + ``                              ``countODD[i - 1]; ``            ``} ``             ` `            ``// if the number is odd ``            ``else``            ``{ ``                ``countEVEN[i] = countEVEN[i - 1] + ``                               ``countODD[i - 1]; ``     ` `                ``countODD[i] = countODD[i - 1] + ``                              ``countEVEN[i - 1] + 1; ``            ``} ``        ``}``         ` `        ``int``[] ans = ``new` `int``;``        ``ans = countEVEN[n];``        ``ans = countODD[n];``        ``return` `ans;``    ``} `` ` `    ``// Driver Code``    ``public` `static` `void` `Main (String[] args) ``    ``{``        ``int``[] arr = ``new` `int``[]{ 1, 2, 2, 3 }; ``        ``int` `n = 4;``        ``int``[] ans = countSum(arr, n);``        ``Console.WriteLine(``"EvenSum = "` `+ ans);``        ``Console.WriteLine(``"OddSum = "` `+ ans);``    ``}``}`` ` `// This code is contributed by Rajput-Ji`
Output:
```EvenSum = 7 OddSum = 8
```

Time Complexity: O(N).
Space Complexity: O(N) where N is the number of elements in the array.

Efficient Approach:
Instead of making countEVEN[N] and countODD[N] arrays we only need the count_even variable and count_odd variable and changing it the same way as we did earlier.
Below is the implementation of above approach:

## C++

 `// C++ implementation``#include ``using` `namespace` `std;`` ` `// Returns the count of odd and``// even subsequences``pair<``int``, ``int``> countSum(``int` `arr[], ``int` `n)``{``    ``int` `result = 0;`` ` `    ``// Variables to store the count of even``    ``// subsequences and odd subsequences``    ``int` `count_odd, count_even;`` ` `    ``// Initialising count_even and count_odd to 0``    ``// since as there is no subsequence before the``    ``// iteration with even or odd count.``    ``count_odd = 0;``    ``count_even = 0;`` ` `    ``// Find sum of all subsequences with even count``    ``// and odd count and storing them as we iterate.`` ` `    ``for` `(``int` `i = 1; i <= n; i++) {`` ` `        ``// if the number is even``        ``if` `(arr[i - 1] % 2 == 0) {``            ``count_even = count_even + count_even + 1;``            ``count_odd = count_odd + count_odd;``        ``}`` ` `        ``// if the number is odd``        ``else` `{``            ``int` `temp = count_even;``            ``count_even = count_even + count_odd;``            ``count_odd = count_odd + temp + 1;``        ``}``    ``}`` ` `    ``return` `{ count_even, count_odd };``}`` ` `// 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 get minimum cost to sort``// strings by reversal operation``class` `GFG``{`` ` `static` `class` `pair``{ ``    ``int` `first, second; ``    ``public` `pair(``int` `first, ``int` `second) ``    ``{ ``        ``this``.first = first; ``        ``this``.second = second; ``    ``} ``} `` ` `// Returns the count of odd and``// even subsequences``static` `pair countSum(``int` `arr[], ``int` `n)``{``    ``int` `result = ``0``;`` ` `    ``// Variables to store the count of even``    ``// subsequences and odd subsequences``    ``int` `count_odd, count_even;`` ` `    ``// Initialising count_even and count_odd to 0``    ``// since as there is no subsequence before the``    ``// iteration with even or odd count.``    ``count_odd = ``0``;``    ``count_even = ``0``;`` ` `    ``// Find sum of all subsequences with even count``    ``// and odd count and storing them as we iterate.``    ``for` `(``int` `i = ``1``; i <= n; i++)``    ``{`` ` `        ``// if the number is even``        ``if` `(arr[i - ``1``] % ``2` `== ``0``)``        ``{``            ``count_even = count_even + count_even + ``1``;``            ``count_odd = count_odd + count_odd;``        ``}`` ` `        ``// if the number is odd``        ``else``        ``{``            ``int` `temp = count_even;``            ``count_even = count_even + count_odd;``            ``count_odd = count_odd + temp + ``1``;``        ``}``    ``}``    ``return` `new` `pair(count_even, count_odd );``}`` ` `// 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 29AjayKumar`

## Python3

 `# Python3 implementation of above approach`` ` `# Returns the count of odd and ``# even subsequences ``def` `countSum(arr, n):``    ``result ``=` `0``     ` `    ``# Variables to store the count of even ``    ``# subsequences and odd subsequences ``    ``# Initialising count_even and count_odd to 0 ``    ``# since as there is no subsequence before the ``    ``# iteration with even or odd count. ``    ``count_odd ``=` `0``    ``count_even ``=` `0`` ` `    ``# Find sum of all subsequences with even count ``    ``# and odd count and storing them as we iterate. `` ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):``         ` `        ``# if the number is even ``        ``if` `(arr[i ``-` `1``] ``%` `2` `=``=` `0``):``            ``count_even ``=` `count_even ``+` `count_even ``+` `1``            ``count_odd ``=` `count_odd ``+` `count_odd`` ` `        ``# if the number is odd ``        ``else``:``            ``temp ``=` `count_even``            ``count_even ``=` `count_even ``+` `count_odd``            ``count_odd ``=` `count_odd ``+` `temp ``+` `1`` ` `    ``return` `(count_even, count_odd) ``     ` `# Driver code ``arr ``=` `[``1``, ``2``, ``2``, ``3``]; ``n ``=` `len``(arr)`` ` `# Calling the function ``count_even, count_odd ``=` `countSum(arr, n); `` ` `print``(``"EvenSum = "``, count_even, ``      ``" OddSum = "``, count_odd)``       ` `# This code is contributed ``# by ANKITKUMAR34`

## C#

 `// C# program to get minimum cost to sort``// strings by reversal operation``using` `System;``     ` `class` `GFG``{`` ` `public` `class` `pair``{ ``    ``public` `int` `first, second; ``    ``public` `pair(``int` `first, ``int` `second) ``    ``{ ``        ``this``.first = first; ``        ``this``.second = second; ``    ``} ``} `` ` `// Returns the count of odd and``// even subsequences``static` `pair countSum(``int` `[]arr, ``int` `n)``{``    ``// Variables to store the count of even``    ``// subsequences and odd subsequences``    ``int` `count_odd, count_even;`` ` `    ``// Initialising count_even and count_odd to 0``    ``// since as there is no subsequence before the``    ``// iteration with even or odd count.``    ``count_odd = 0;``    ``count_even = 0;`` ` `    ``// Find sum of all subsequences with even count``    ``// and odd count and storing them as we iterate.``    ``for` `(``int` `i = 1; i <= n; i++)``    ``{`` ` `        ``// if the number is even``        ``if` `(arr[i - 1] % 2 == 0)``        ``{``            ``count_even = count_even + count_even + 1;``            ``count_odd = count_odd + count_odd;``        ``}`` ` `        ``// if the number is odd``        ``else``        ``{``            ``int` `temp = count_even;``            ``count_even = count_even + count_odd;``            ``count_odd = count_odd + temp + 1;``        ``}``    ``}``    ``return` `new` `pair(count_even, count_odd );``}`` ` `// 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 PrinciRaj1992`
Output:
```EvenSum = 7 OddSum = 8
```

Time Complexity: O(N).
Space Complexity: O(1), where N is the number of elements in the array.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up