# Number of Subsequences with Even and Odd Sum

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

**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 becountEVEN[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

*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]

**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

**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 <bits/stdc++.h> ` `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[0] and countODD[0] 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; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `{ countEVEN[n], countODD[n] }; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 2, 2, 3 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `// Calling the function ` ` ` ` ` `pair<` `int` `, ` `int` `> ans = countSum(arr, n); ` ` ` ` ` `cout << ` `"EvenSum = "` `<< ans.first; ` ` ` `cout << ` `" OddSum = "` `<< ans.second; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## 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[0] and countODD[0] 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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[0] and countODD[0] 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); ` ` ` `Console.WriteLine(` `"EvenSum = "` `+ ans[0]); ` ` ` `Console.WriteLine(` `"OddSum = "` `+ ans[1]); ` ` ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

*chevron_right*

*filter_none*

**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 <bits/stdc++.h> ` `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[0]); ` ` ` ` ` `// Calling the function ` ` ` ` ` `pair<` `int` `, ` `int` `> ans = countSum(arr, n); ` ` ` ` ` `cout << ` `"EvenSum = "` `<< ans.first; ` ` ` `cout << ` `" OddSum = "` `<< ans.second; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

**Output:**

EvenSum = 7 OddSum = 8

**Time Complexity: ** **O(N)**.

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

## Recommended Posts:

- Given a number as a string, find the number of contiguous subsequences which recursively add up to 9
- Given a number as a string, find the number of contiguous subsequences which recursively add up to 9 | Set 2
- Sum of all subsequences of a number
- Number of subsequences with zero sum
- Number of Subsequences with Even and Odd Sum | Set 2
- Number of subsequences of the form a^i b^j c^k
- Number of K length subsequences with minimum sum
- Number of palindromic subsequences of length k where k <= 3
- Minimum number of increasing subsequences
- Number of subsequences in a string divisible by n
- Number of subsequences as "ab" in a string repeated K times
- Number of GP (Geometric Progression) subsequences of size 3
- Count the number of subsequences of length k having equal LCM and HCF
- Count number of increasing subsequences of size k
- Number of ways to partition a string into two balanced subsequences

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.