Nearest smaller character to a character K from a Sorted Array

Given a sorted array of characters arr[] and a character K, the task is to find the character with nearest smaller ASCII value than K from the given array. Print -1 if no character is found to be having smaller ASCII value than K.
Examples: 

Input: arr[] = {‘e’, ‘g’, ‘t’, ‘y’}, K = ‘u’ 
Output:
Explanation: 
The character with nearest smaller ASCII value as of ‘u’ is ‘t’.

Input: arr[] = {‘e’, ‘g’, ‘t’, ‘y’}, K = ‘a’ 
Output: -1 
Explanation: 
No character exists with ASCII value smaller than that of ‘a’. 
 

Naive Approach: 
The simplest approach to solve the problem is to iterate over the array and find the character having smaller ASCII value than that of K and the difference between their ASCII values is minimum. 
Time Complexity: O(N) 
Auxiliary Space: O(1) 

Efficient Approach: 
The idea is to use Binary Search to find out the floor element (largest character just smaller than key). Follow the steps below to solve the problem: 

  • Perform binary search on the array.
  • Check if the current middle element(mid) is equal to the character K or not.
  • If so, then set start to mid – 1, because we need to find just smaller element.
  • If the arr[mid] is smaller than K, then store it in some variable say ch. Set start to mid + 1 to search for a smaller character nearer to K.
  • Otherwise, set end to mid-1.
  • Kepp repeating the above steps. Finally, print the character obtained after completing the search. If no character is obtained, print -1.

Below is the implementation of the above approach:



C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the
// nearest smaller charcter
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to implement
// the above approach
import java.util.*;
  
class GFG{
  
// Function to return the
// nearest smaller charcter
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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to implement
# the above approach
  
# Function to return the
# nearest smaller charcter
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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to implement
// the above approach
using System;
  
class GFG{
  
// Function to return the
// nearest smaller charcter
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

chevron_right


Output: 

t

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.