Number of Subsequences with Even and Odd Sum | Set 2
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:
- By taking odd number odd times.
- 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 <bits/stdc++.h> 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[0]); // 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
<script> // JavaScript program to find number of // Subsequences with Even and Odd Sum // Function to find number of // Subsequences with Even and Odd Sum function countSum(arr, n) { let NumberOfOdds = 0, NumberOfEvens = 0; // Counting number of odds for (let i = 0; i < n; i++) if (arr[i] & 1) NumberOfOdds++; // Even count NumberOfEvens = n - NumberOfOdds; let NumberOfOddSubsequences = (1 << NumberOfEvens) * (1 << (NumberOfOdds - 1)); // Total Subsequences is (2^n - 1) // For NumberOfEvenSubsequences subtract // NumberOfOddSubsequences from total let NumberOfEvenSubsequences = (1 << n) - 1 - NumberOfOddSubsequences; return [NumberOfEvenSubsequences, NumberOfOddSubsequences]; } // Driver code let arr = [1, 2, 2, 3]; let n = arr.length; // Calling the function let ans = countSum(arr, n); document.write( "EvenSum = " + ans[0]); document.write( " OddSum = " + ans[1]); // This code is contributed by _saurabh_jaiswal </script> |
EvenSum = 7 OddSum = 8
Time Complexity: O(n)
Auxiliary Space: O(1)
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