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📜  排序数组中最接近字符K 的较小字符

📅  最后修改于: 2021-09-06 06:38:30             🧑  作者: Mango

给定一个排序的字符数组arr[]和一个字符K ，任务是从给定的数组中找到最接近的 ASCII 值小于K的字符。如果没有发现 ASCII 值小于K 的字符，则打印-1 。
例子：

天真的方法：
解决这个问题最简单的方法是遍历数组，找到ASCII值小于K的字符，并且它们的ASCII值之间的差异最小。
时间复杂度： O(N)
辅助空间： O(1)
有效的方法：
这个想法是使用二分搜索来找出地板元素(最大的字符只是小于键)。请按照以下步骤解决问题：

  • 对数组执行二分查找。
  • 检查当前中间元素( mid )是否等于字符K 。
  • 如果是这样，那么将start设置为mid – 1 ，因为我们只需要找到更小的元素。
  • 如果arr[mid]小于K ，则将其存储在某个变量中，例如ch 。将start设置为mid + 1以搜索更接近K的较小字符。
  • 否则，将end设置为mid-1 。
  • 不断重复以上步骤。最后，打印完成搜索后得到的字符。如果未获得字符，则打印-1 。

下面是上述方法的实现：

C++
// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to return the
// nearest smaller character
char bs(char ar[], int n, int ele)
{
    int start = 0;
    int end = n - 1;
 
    // Stores the nearest smaller
    // character
    char ch = '@';
 
    // Iterate till starts cross end
    while (start <= end) {
 
        // Find the mid element
        int mid = start + (end - start) / 2;
 
        // Check if K is found
        if (ar[mid] == ele)
            end = mid - 1;
 
        // Check if current character
        // is less than K
        else if (ar[mid] < ele) {
            ch = ar[mid];
 
            // Increment the start
            start = mid + 1;
        }
 
        // Otherwise
        else
 
            // Increment end
            end = mid - 1;
    }
    // Return the character
    return ch;
}
 
// Driver Code
int main()
{
    char ar[] = { 'e', 'g', 't', 'y' };
    int n = sizeof(ar) / sizeof(ar[0]);
 
    char K = 'u';
 
    char ch = bs(ar, n, K);
 
    if (ch == '@')
        cout << "-1";
    else
        cout << ch;
 
    return 0;
}

Java
// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to return the
// nearest smaller character
static char bs(char ar[], int n, int ele)
{
    int start = 0;
    int end = n - 1;
 
    // Stores the nearest smaller
    // character
    char ch = '@';
 
    // Iterate till starts cross end
    while (start <= end)
    {
 
        // Find the mid element
        int mid = start + (end - start) / 2;
 
        // Check if K is found
        if (ar[mid] == ele)
            end = mid - 1;
 
        // Check if current character
        // is less than K
        else if (ar[mid] < ele)
        {
            ch = ar[mid];
 
            // Increment the start
            start = mid + 1;
        }
 
        // Otherwise
        else
 
            // Increment end
            end = mid - 1;
    }
     
    // Return the character
    return ch;
}
 
// Driver Code
public static void main(String[] args)
{
    char ar[] = { 'e', 'g', 't', 'y' };
    int n = ar.length;
 
    char K = 'u';
 
    char ch = bs(ar, n, K);
 
    if (ch == '@')
        System.out.print("-1");
    else
        System.out.print(ch);
}
}
 
// This code is contributed by 29AjayKumar

Python3
# Python3 program to implement
# the above approach
 
# Function to return the
# nearest smaller character
def bs(a, n, ele):
     
    start = 0
    end = n - 1
     
    # Stores the nearest smaller
    # character
    ch = '@'
     
    # Iterate till starts cross end
    while (start <= end):
         
        # Find the mid element
        mid = start + (end - start) // 2;
 
        # Check if K is found
        if(ar[mid] == ele):
            end = mid - 1
             
        # Check if current character
        # is less than K
        elif(ar[mid] < ele):
            ch = ar[mid]
             
            # Increment the start
            start = mid + 1;
         
        # Otherwise
        else:
             
            # Increment end
            end = mid - 1;
             
    # Return the character
    return ch
 
# Driver code
if __name__=='__main__':
     
    ar = [ 'e', 'g', 't', 'y' ]
    n = len(ar)
    K = 'u';
 
    ch = bs(ar, n, K);
 
    if (ch == '@'):
        print('-1')
    else:
        print(ch)
         
# This code is contributed by rutvik_56

C#
// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to return the
// nearest smaller character
static char bs(char []ar, int n, int ele)
{
    int start = 0;
    int end = n - 1;
 
    // Stores the nearest smaller
    // character
    char ch = '@';
 
    // Iterate till starts cross end
    while (start <= end)
    {
 
        // Find the mid element
        int mid = start + (end - start) / 2;
 
        // Check if K is found
        if (ar[mid] == ele)
            end = mid - 1;
 
        // Check if current character
        // is less than K
        else if (ar[mid] < ele)
        {
            ch = ar[mid];
 
            // Increment the start
            start = mid + 1;
        }
 
        // Otherwise
        else
 
            // Increment end
            end = mid - 1;
    }
     
    // Return the character
    return ch;
}
 
// Driver Code
public static void Main(String[] args)
{
    char []ar = { 'e', 'g', 't', 'y' };
    int n = ar.Length;
 
    char K = 'u';
    char ch = bs(ar, n, K);
 
    if (ch == '@')
        Console.Write("-1");
    else
        Console.Write(ch);
}
}
 
// This code is contributed by 29AjayKumar

输出： 
t

时间复杂度： O(logN)
辅助空间： O(1)

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