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# Count number of ways to get Odd Sum

• Difficulty Level : Medium
• Last Updated : 05 May, 2021

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:
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:

Approach:

• We will use dynamic programming here, where dp[i] will store number of possible ways to get even sum upto i’th pair and dp[i] will store number of possible ways to get odd sum upto i’th pair.
• cnt[i] will store count of even numbers in i’th pair and cnt[i] 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] * cnt[i] + dp[i – 1] * cnt[i].
• Ways to get odd sum upto i’th pair will be dp[i – 1] * cnt[i] + dp[i – 1] * cnt[i].

Below is the implementation of above Approach:

## C++

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

## Java

 `// 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] = 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] = 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] = 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`

## Python3

 `# 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] = 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] = 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] = 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`

## C#

 `// 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`

## Javascript

 ``
Output:
`2`

Time Complexity: O(N)

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