Kth smallest even number in range L to R

Given two variables L and R, indicating a range of integers from L to R inclusive, and a number K, the task is to find Kth smallest even number. If K is greater than a number of even numbers in the range L to R then return -1. LLONG_MIN <= L <= R <= LLONG_MAX.

Examples:

Input: L = 3, R = 9, K = 3
Output: 8
Explanation:  The even numbers in the range are 4, 6, 8 and the 3rd smallest even number is 8

Input: L = -3, R = 3, K = 2
Output: 0

Naive Approach: The basic idea is to traverse the numbers from L to R, and then print the Kth even number.

Time Complexity: O(R-L)
Auxiliary Space: O(1)

Approach: The given problem can be solved using basic maths and using ceil and floor from cmath library. The idea is to check if L is odd or even and calculate Kth smallest even number accordingly. Below steps can be used to solve the problem:

• If K<=0 then return -1
• Initialize count to calculate the number of even numbers within the range
• If L is odd
  • count = floor((float)(R-L+1)/2)
  • If K > count return -1
  • Else return (L + 2*K – 1)
• If L is even
  • count = ceil((float)(R-L+1)/2)
  • If K > count return -1
  • Else return (L + 2*K – 2)

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <cmath>
#include <iostream>
#define ll long long
using namespace std;

// Function to return Kth smallest
// even number if it exists
ll findEven(ll L, ll R, ll K)
{

    // Base Case
    if (K <= 0)
        return -1;

    if (L % 2) {

        // Calculate count of even numbers
        // within the range
        ll Count = floor((float)(R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 1);
    }

    else {
        // Calculate count of even numbers
        // within the range

        ll Count = ceil((float)(R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 2);
    }
}

// Driver Code
int main()
{
    ll L = 3, R = 9, K = 3;

    cout << findEven(L, R, K);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to return Kth smallest
// even number if it exists
static long findEven(long L, long R, long K)
{

    // Base Case
    if (K <= 0)
        return -1;

    if (L % 2 == 1) {

        // Calculate count of even numbers
        // within the range
        long Count = (int)Math.floor((float)(R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 1);
    }

    else {
        // Calculate count of even numbers
        // within the range

        long Count = (int)Math.ceil((float)(R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 2);
    }
}

// Driver Code
public static void main(String args[])
{
    long L = 3, R = 9, K = 3;

    System.out.println(findEven(L, R, K));

}
}
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach

# Function to return Kth smallest
# even number if it exists
def findEven(L, R, K):

    # Base Case
    if (K <= 0):
        return -1

    if (L % 2):

        # Calculate count of even numbers
        # within the range
        Count = (R - L + 1) // 2

        # if k > range then kth smallest
        # even number is not in this range
        # then return -1
        return -1 if (K > Count) else (L + 2 * K - 1)

    else:
        # Calculate count of even numbers
        # within the range

        Count = (R - L + 1) // 2

        # if k > range then kth smallest
        # even number is not in this range
        # then return -1
        return -1 if (K > Count) else (L + 2 * K - 2)

# Driver Code
L = 3
R = 9
K = 3

print(findEven(L, R, K))

# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{
// Function to return Kth smallest
// even number if it exists
static long findEven(long L, long R, long K)
{

    // Base Case
    if (K <= 0)
        return -1;

    if (L % 2 == 1) {

        // Calculate count of even numbers
        // within the range
        long Count = (int)Math.Floor((float)(R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 1);
    }

    else {
        // Calculate count of even numbers
        // within the range

        long Count = (int)Math.Ceiling((float)(R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 2);
    }
}

// Driver Code
public static void Main()
{
    long L = 3, R = 9, K = 3;

    Console.Write(findEven(L, R, K));

}
}
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript program for the above approach

// Function to return Kth smallest
// even number if it exists
function findEven(L, R, K)
{

    // Base Case
    if (K <= 0)
        return -1;

    if (L % 2) {

        // Calculate count of even numbers
        // within the range
        let Count = Math.floor((R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 1);
    }

    else {
        // Calculate count of even numbers
        // within the range

        let Count = Math.ceil((float)(R - L + 1) / 2);

        // if k > range then kth smallest
        // even number is not in this range
        // then return -1
        return (K > Count) ? -1 : (L + 2 * K - 2);
    }
}

// Driver Code
let L = 3, R = 9, K = 3;

document.write(findEven(L, R, K));

// This code is contributed by Samim Hossain Mondal.
</script>

Output

8

Time Complexity: O(1)
Auxiliary Space: O(1)