# Check if the number is even or odd whose digits and base (radix) is given

Given an array arr[] of size N which represents the digits of a number and an integer r which is the base (radix) of the given number i.e. n = arr[n – 1] * r0 + arr[n – 2] * r1 + … + a[0] * rN – 1. The task is to find whether the given number is odd or even.
Examples:

Input: arr[] = {1, 0}, r = 2
Output: Even
(10)2 = (2)10
Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}, r = 10
Output: Odd

Naive Approach: The simplest approach is to calculate the number n by multiplying the digits with the corresponding power of base. But as the number of digits can be of the order of 105, this approach won’t work for a large n.
Efficient Approach: There are two cases possible.

1. If r is even then the final answer depends on the last digit i.e. arr[n – 1].
2. If r is odd then we have to count the number of odd digits. If the number of odd digits is even, then the sum is even. Else the sum is odd.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function that returns true if the number` `// represented by arr[] is even in base r` `bool` `isEven(``int` `arr[], ``int` `n, ``int` `r)` `{`   `    ``// If the base is even, then` `    ``// the last digit is checked` `    ``if` `(r % 2 == 0) {` `        ``if` `(arr[n - 1] % 2 == 0)` `            ``return` `true``;` `    ``}`   `    ``// If base is odd, then the` `    ``// number of odd digits are checked` `    ``else` `{`   `        ``// To store the count of odd digits` `        ``int` `oddCount = 0;` `        ``for` `(``int` `i = 0; i < n; ++i) {` `            ``if` `(arr[i] % 2 != 0)` `                ``oddCount++;` `        ``}` `        ``if` `(oddCount % 2 == 0)` `            ``return` `true``;` `    ``}`   `    ``// Number is odd` `    ``return` `false``;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 1, 0 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``int` `r = 2;`   `    ``if` `(isEven(arr, n, r))` `        ``cout << ``"Even"``;` `    ``else` `        ``cout << ``"Odd"``;`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.io.*;`   `class` `GFG` `{` `    `  `// Function that returns true if the number` `// represented by arr[] is even in base r` `static` `boolean` `isEven(``int` `arr[], ``int` `n, ``int` `r)` `{`   `    ``// If the base is even, then` `    ``// the last digit is checked` `    ``if` `(r % ``2` `== ``0``)` `    ``{` `        ``if` `(arr[n - ``1``] % ``2` `== ``0``)` `            ``return` `true``;` `    ``}`   `    ``// If base is odd, then the` `    ``// number of odd digits are checked` `    ``else` `{`   `        ``// To store the count of odd digits` `        ``int` `oddCount = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; ++i) {` `            ``if` `(arr[i] % ``2` `!= ``0``)` `                ``oddCount++;` `        ``}` `        ``if` `(oddCount % ``2` `== ``0``)` `            ``return` `true``;` `    ``}`   `    ``// Number is odd` `    ``return` `false``;` `}`   `// Driver code` `public` `static` `void` `main (String[] args) ` `{` `    `  `    ``int` `arr[] = { ``1``, ``0` `};` `    ``int` `n = arr.length;` `    ``int` `r = ``2``;`   `    ``if` `(isEven(arr, n, r))`   `        ``System.out.println (``"Even"``);` `    ``else` `    `  `        ``System.out.println(``"Odd"``);` `}` `}`   `// This code is contributed by jit_t.`

## Python3

 `# Python 3 implementation of the approach`   `# Function that returns true if the number` `# represented by arr[] is even in base r` `def` `isEven(arr, n, r):` `    `  `    ``# If the base is even, then` `    ``# the last digit is checked` `    ``if` `(r ``%` `2` `=``=` `0``):` `        ``if` `(arr[n ``-` `1``] ``%` `2` `=``=` `0``):` `            ``return` `True`   `    ``# If base is odd, then the` `    ``# number of odd digits are checked` `    ``else``:` `        ``# To store the count of odd digits` `        ``oddCount ``=` `0` `        ``for` `i ``in` `range``(n):` `            ``if` `(arr[i] ``%` `2` `!``=` `0``):` `                ``oddCount ``+``=` `1` `        ``if` `(oddCount ``%` `2` `=``=` `0``):` `            ``return` `True`   `    ``# Number is odd` `    ``return` `False`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``1``, ``0``]` `    ``n ``=` `len``(arr)` `    ``r ``=` `2`   `    ``if` `(isEven(arr, n, r)):` `        ``print``(``"Even"``)` `    ``else``:` `        ``print``(``"Odd"``)`   `# This code is contributed by` `# Surendra_Gangwar`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG` `{` `    `  `// Function that returns true if the number` `// represented by arr[] is even in base r` `static` `bool` `isEven(``int` `[]arr, ``int` `n, ``int` `r)` `{`   `    ``// If the base is even, then` `    ``// the last digit is checked` `    ``if` `(r % 2 == 0)` `    ``{` `        ``if` `(arr[n - 1] % 2 == 0)` `            ``return` `true``;` `    ``}`   `    ``// If base is odd, then the` `    ``// number of odd digits are checked` `    ``else` `    ``{`   `        ``// To store the count of odd digits` `        ``int` `oddCount = 0;` `        ``for` `(``int` `i = 0; i < n; ++i)` `        ``{` `            ``if` `(arr[i] % 2 != 0)` `                ``oddCount++;` `        ``}` `        ``if` `(oddCount % 2 == 0)` `            ``return` `true``;` `    ``}`   `    ``// Number is odd` `    ``return` `false``;` `}`   `// Driver code` `public` `static` `void` `Main () ` `{` `    `  `    ``int` `[]arr = { 1, 0 };` `    ``int` `n = arr.Length;` `    ``int` `r = 2;`   `    ``if` `(isEven(arr, n, r))`   `        ``Console.WriteLine (``"Even"``);` `    ``else` `    `  `        ``Console.WriteLine(``"Odd"``);` `}` `}`   `// This code is contributed by anuj_67...`

## PHP

 ``

## Javascript

 ``

Output:

`Even`

Time Complexity: O(n)

Auxiliary Space: O(1)

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