Given an array arr[] and an integer K, the task is to find the Kth odd element from the given array.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: 3
3 is the 2nd odd element from the given arrayInput: arr[] = {2, 4, 6, 18}, K = 5
Output: -1
There are no odd elements in the given array.
Approach: Traverse the array element by element and for every odd element encountered, decrement the value k by 1. If the value of k becomes equal to 0 then print the current element. Else after the traversal of the complete array, if the value of k is > 0 then print -1 as the total number of odd elements in the array is < k.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the kth odd element// from the arrayint kthOdd(int arr[], int n, int k){ // Traverse the array for (int i = 0; i <= n; i++) { // If current element is odd if ((arr[i] % 2) == 1) k--; // If kth odd element is found if (k == 0) return arr[i]; } // Total odd elements in the array are < k return -1;}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; cout << (kthOdd(arr, n, k)); return 0;}// This code is contributed by jit_t |
Java
// Java implementation of the approachpublic class GFG { // Function to return the kth odd element // from the array static int kthOdd(int arr[], int n, int k) { // Traverse the array for (int i = 0; i < n; i++) { // If current element is odd if (arr[i] % 2 == 1) k--; // If kth odd element is found if (k == 0) return arr[i]; } // Total odd elements in the array are < k return -1; } // Driver code public static void main(String args[]) { int arr[] = { 1, 2, 3, 4, 5 }; int n = arr.length; int k = 2; System.out.print(kthOdd(arr, n, k)); }} |
Python3
# Python3 implementation of the approach# Function to return the kth odd # element from the arraydef kthOdd (arr, n, k): # Traverse the array for i in range(n): # If current element is odd if (arr[i] % 2 == 1): k -= 1; # If kth odd element is found if (k == 0): return arr[i]; # Total odd elements in the # array are < k return -1;# Driver codearr = [ 1, 2, 3, 4, 5 ];n = len(arr); k = 2;print(kthOdd(arr, n, k));# This code is contributed by mits |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the kth odd element // from the array static int kthOdd(int []arr, int n, int k) { // Traverse the array for (int i = 0; i < n; i++) { // If current element is odd if (arr[i] % 2 == 1) k--; // If kth odd element is found if (k == 0) return arr[i]; } // Total odd elements in the array are < k return -1; } // Driver code public static void Main() { int []arr = { 1, 2, 3, 4, 5 }; int n = arr.Length; int k = 2; Console.WriteLine(kthOdd(arr, n, k)); }}// This code is contributed by SoM15242 |
Javascript
<script>// JavaScript implementation of the approach // Function to return the kth odd element // from the array function kthOdd(arr , n , k) { // Traverse the array for (i = 0; i < n; i++) { // If current element is odd if (arr[i] % 2 == 1) k--; // If kth odd element is found if (k == 0) return arr[i]; } // Total odd elements in the array are < k return -1; } // Driver code var arr = [ 1, 2, 3, 4, 5 ]; var n = arr.length; var k = 2; document.write(kthOdd(arr, n, k));// This code contributed by Rajput-Ji </script> |
PHP
<?php// PHP implementation of the approach// Function to return the kth odd // element from the arrayfunction kthOdd ($arr, $n, $k){ // Traverse the array for ($i = 0; $i < $n; $i++) { // If current element is odd if ($arr[$i] % 2 == 1) $k--; // If kth odd element is found if ($k == 0) return $arr[$i]; } // Total odd elements in the // array are < k return -1;}// Driver code$arr = array( 1, 2, 3, 4, 5 );$n = sizeof($arr); $k = 2;echo (kthOdd($arr, $n, $k));// This code is contributed by ajit..?> |
Output
3
Time Complexity: O(n)
Auxiliary Space: O(1)
Approach: Iterating the array and counting odd elements
Below is the implementation of the above approach:
C++
#include <iostream>#include <vector>int findKthOddElement(std::vector<int>& array, int K) { int count = 0; for (int element : array) { // Checking if the element is odd if (element % 2 != 0) { //increment the count count++; if (count == K) { return element; } } } return -1;}int main() { std::vector<int> array = {1, 2, 3, 4, 5}; int K = 2; int result = findKthOddElement(array, K); std::cout << result << std::endl; return 0;} |
Java
import java.util.ArrayList;import java.util.List;public class Main { public static int findKthOddElement(List<Integer> list, int K) { int count = 0; for (int element : list) { // Checking if the element is odd if (element % 2 != 0) { // Increment the count count++; //checking if count is equal to k or not if (count == K) { return element; } } } return -1; } public static void main(String[] args) { List<Integer> list = new ArrayList<>(); list.add(1); list.add(2); list.add(3); list.add(4); list.add(5); //input int K = 2; int result = findKthOddElement(list, K); System.out.println(result); }} |
Python3
# Python3 implementation of the above approachdef find_kth_odd_element(array, K): count = 0 for element in array: # Checking if the element is odd if element % 2 != 0: count += 1 if count == K: return element return -1 # Driver Codearray = [1, 2, 3, 4, 5]K = 2result = find_kth_odd_element(array, K)print(result) |
C#
using System;using System.Collections.Generic;class Program{ // Function to find the Kth odd element in a list static int FindKthOddElement(List<int> list, int K) { int count = 0; foreach (int element in list) { // Checking if the element is odd if (element % 2 != 0) { count++; // If we've found the Kth odd element, return it if (count == K) { return element; } } } // If there are fewer than K odd elements, return -1 return -1; } static void Main() { List<int> list = new List<int> { 1, 2, 3, 4, 5 }; int K = 2; int result = FindKthOddElement(list, K); if (result != -1) { Console.WriteLine("The " + K + "th odd element is: " + result); } else { Console.WriteLine("There are fewer than " + K + " odd elements in the list."); } }} |
Javascript
// JavaScript Program for the above approachfunction find_kth_odd_element(array, K) { let count = 0; for (let element of array) { // Checking if the element is odd if (element % 2 != 0) { count += 1; if (count == K) { return element; } } } return -1;}// Driver Codelet array = [1, 2, 3, 4, 5];let K = 2;let result = find_kth_odd_element(array, K);console.log(result);// This code is contributed by Kanchan Agarwal |
Output
3
Time Complexity: O(N), where N is the length of the input array.
Auxiliary Space: O(1)
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