从 L 到 R 范围内的第 K 个最小的偶数

给定两个变量L和R ，表示从L到R的整数范围，以及一个数字K ，任务是找到第 K个最小的偶数。如果K大于L到R范围内的多个偶数，则返回 -1。 LLONG_MIN <= L <= R <= LLONG_MAX 。

例子：

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：基本思想是从L到R遍历数字，然后打印第K个偶数。

时间复杂度：O(RL)
辅助空间：O(1)

方法：给定的问题可以使用基础数学和使用 cmath 库中的 ceil 和 floor 来解决。这个想法是检查L是奇数还是偶数，并相应地计算第 K 个最小的偶数。以下步骤可用于解决问题：

• 如果K<=0则返回-1
• 初始化count计算范围内偶数个数
• 如果L是奇数
  • 计数 = 楼层（（浮点数）（R-L+1）/2）
  • 如果K > 计数返回-1
  • 否则返回(L + 2*K – 1)
• 如果L是偶数
  • 计数 = ceil((浮点数)(R-L+1)/2)
  • 如果K > 计数返回-1
  • 否则返回(L + 2*K – 2)

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
#include 
#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

// 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.

输出

8

时间复杂度：O(1)
辅助空间：O(1)