Count number of ways to get Odd Sum

Given N pairs of numbers. The task is to count ways to choose exactly one number from each pair such that the sum of those numbers is odd.

Examples:

Input:
N = 2
3 4
1 2
Output: 2
Explanation:
We can choose 3 from the first pair and 2 from the second pair, and their sum is 5 which is odd.
Also, we can choose 4 from the first pair and 1 from the second pair, and their sum is 5 which is odd.
So the total possible ways will be 2.



Input:
N = 2
2 2
2 2
Output: 0

Approach:

  • We will use dynamic programming here, where dp[i][0] will store number of possible ways to get even sum upto i’th pair and dp[i][1] will store number of possible ways to get odd sum upto i’th pair.
  • cnt[i][0] will store count of even numbers in i’th pair and cnt[i][1] will store count of odd numbers in i’th pair.
  • It is known that the sum of two even or sum of two odd will always be even and the sum of one even and one odd will always be odd.
  • We apply this to store the count in the DP array.
  • Ways to get even sum upto i’th pair will be dp[i – 1][0] * cnt[i][0] + dp[i – 1][1] * cnt[i][1].
  • Ways to get odd sum upto i’th pair will be dp[i – 1][1] * cnt[i][0] + dp[i – 1][0] * cnt[i][1].
  • So we can store the dp array in this way,

    dp[i][0] = dp[i - 1][0] * cnt[i][0] + dp[i - 1][1] * cnt[i][1]
    dp[i][1] = dp[i - 1][0] * cnt[i][1] + dp[i - 1][1] * cnt[i][0]
    

Below is the implementation of above Approach:

C++

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// C++ implementation
#include <bits/stdc++.h>
using namespace std;
  
// Count the ways to sum up with odd
// by choosing one element form each
// pair
int CountOfOddSum(int a[][2], int n)
{
    int dp[n][2], cnt[n][2];
  
    // Initialize two array with 0
    memset(dp, 0, sizeof(dp));
    memset(cnt, 0, sizeof(cnt));
  
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < 2; j++) {
  
            // if element is even
            if (a[i][j] % 2 == 0) {
  
                // store count of even
                // number in i'th pair
                cnt[i][0]++;
            }
  
            // if the element is odd
            else {
  
                // store count of odd
                // number in i'th pair
                cnt[i][1]++;
            }
        }
    }
  
    // Initial state of dp array
    dp[0][0] = cnt[0][0], dp[0][1] = cnt[0][1];
  
    for (int i = 1; i < n; i++) {
  
        // dp[i][0] = total number of ways
        // to get even sum upto i'th pair
        dp[i][0] = (dp[i - 1][0] * cnt[i][0]
                    + dp[i - 1][1] * cnt[i][1]);
  
        // dp[i][1] = total number of ways
        // to odd even sum upto i'th pair
        dp[i][1] = (dp[i - 1][0] * cnt[i][1]
                    + dp[i - 1][1] * cnt[i][0]);
    }
  
    // dp[n - 1][1] = total number of ways
    // to get odd sum upto n'th pair
    return dp[n - 1][1];
}
  
// Driver code
int main()
{
  
    int a[][2] = { { 1, 2 }, { 3, 6 } };
    int n = sizeof(a) / sizeof(a[0]);
  
    int ans = CountOfOddSum(a, n);
  
    cout << ans << "\n";
  
    return 0;
}

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Java

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// Java implementation of above approach
class GFG
{
    // Count the ways to sum up with odd
    // by choosing one element form each
    // pair
    static int CountOfOddSum(int a[][], int n)
    {
        int [][]dp = new int[n][2];
        int [][]cnt = new int[n][2];
      
        for (int i = 0; i < n; i++) 
        {
            for (int j = 0; j < 2; j++) 
            {
      
                // if element is even
                if (a[i][j] % 2 == 0
                {
      
                    // store count of even
                    // number in i'th pair
                    cnt[i][0]++;
                }
      
                // if the element is odd
                else
                {
      
                    // store count of odd
                    // number in i'th pair
                    cnt[i][1]++;
                }
            }
        }
      
        // Initial state of dp array
        dp[0][0] = cnt[0][0];
        dp[0][1] = cnt[0][1];
      
        for (int i = 1; i < n; i++) 
        {
      
            // dp[i][0] = total number of ways
            // to get even sum upto i'th pair
            dp[i][0] = (dp[i - 1][0] * cnt[i][0] + 
                        dp[i - 1][1] * cnt[i][1]);
      
            // dp[i][1] = total number of ways
            // to odd even sum upto i'th pair
            dp[i][1] = (dp[i - 1][0] * cnt[i][1] +
                        dp[i - 1][1] * cnt[i][0]);
        }
      
        // dp[n - 1][1] = total number of ways
        // to get odd sum upto n'th pair
        return dp[n - 1][1];
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int a[][] = {{ 1, 2 }, { 3, 6 }};
        int n = a.length;
      
        int ans = CountOfOddSum(a, n);
      
        System.out.println(ans);
    }
}
  
// This code is contributed by ihritik

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Python3

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# Python3 implementation of the above approach
  
# Count the ways to sum up with odd
# by choosing one element form each
# pair
def CountOfOddSum(a, n):
  
    dp = [[0 for i in range(2)] 
             for i in range(n)]
    cnt = [[0 for i in range(2)]
              for i in range(n)]
  
    # Initialize two array with 0
    for i in range(n):
        for j in range(2):
              
            # if element is even
            if (a[i][j] % 2 == 0):
  
                #store count of even
                #number in i'th pair
                cnt[i][0] += 1
  
            # if the element is odd
            else :
  
                # store count of odd
                # number in i'th pair
                cnt[i][1] += 1
  
    # Initial state of dp array
    dp[0][0] = cnt[0][0]
    dp[0][1] = cnt[0][1]
  
    for i in range(1, n):
  
        # dp[i][0] = total number of ways
        # to get even sum upto i'th pair
        dp[i][0] = (dp[i - 1][0] * cnt[i][0] + 
                    dp[i - 1][1] * cnt[i][1])
  
        # dp[i][1] = total number of ways
        # to odd even sum upto i'th pair
        dp[i][1] = (dp[i - 1][0] * cnt[i][1] + 
                    dp[i - 1][1] * cnt[i][0])
  
    # dp[n - 1][1] = total number of ways
    # to get odd sum upto n'th pair
    return dp[n - 1][1]
  
# Driver code
a = [[1, 2] , [3, 6] ]
n = len(a)
  
ans = CountOfOddSum(a, n)
  
print(ans)
      
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of above approach
using System;
  
class GFG
{
    // Count the ways to sum up with odd
    // by choosing one element form each
    // pair
    static int CountOfOddSum(int [ , ] a, int n)
    {
        int [ , ]dp = new int[n, 2];
        int [ , ]cnt = new int[n, 2];
      
        for (int i = 0; i < n; i++) 
        {
            for (int j = 0; j < 2; j++) 
            {
      
                // if element is even
                if (a[i, j] % 2 == 0)
                {
      
                    // store count of even
                    // number in i'th pair
                    cnt[i, 0]++;
                }
      
                // if the element is odd
                else 
                {
      
                    // store count of odd
                    // number in i'th pair
                    cnt[i, 1]++;
                }
            }
        }
      
        // Initial state of dp array
        dp[0, 0] = cnt[0, 0];
        dp[0, 1] = cnt[0, 1];
      
        for (int i = 1; i < n; i++) 
        {
      
            // dp[i, 0] = total number of ways
            // to get even sum upto i'th pair
            dp[i, 0] = (dp[i - 1, 0] * cnt[i, 0] + 
                        dp[i - 1, 1] * cnt[i, 1]);
      
            // dp[i, 1] = total number of ways
            // to odd even sum upto i'th pair
            dp[i, 1] = (dp[i - 1, 0] * cnt[i, 1] + 
                        dp[i - 1, 1] * cnt[i, 0]);
        }
      
        // dp[n - 1, 1] = total number of ways
        // to get odd sum upto n'th pair
        return dp[n - 1, 1];
    }
      
    // Driver code
    public static void Main () 
    {
        int [ , ] a = { { 1, 2 }, { 3, 6 } };
        int n = a.GetLength(1);
      
        int ans = CountOfOddSum(a, n);
      
        Console.WriteLine(ans);
    }
}
  
// This code is contributed by ihritik

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Output:

2

Time Complexity: O(N)



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Improved By : mohit kumar 29, ihritik