# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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].
• So we can store the dp array in this way,

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

Below is the implementation of above Approach:

 `// 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 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 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# 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 `

Output:
```2
```

Time Complexity: O(N)

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Improved By : mohit kumar 29, ihritik

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