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: t 
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.
• Keep 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++ Program to implement
// the above approach
#include <bits/stdc++.h>
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 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 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# 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

<script>

// Javascript program for the above approach

// Function to return the
// nearest smaller letacter
function bs(ar, n, ele)
{
    let start = 0;
    let end = n - 1;

    // Stores the nearest smaller
    // letacter
    let ch = '@';

    // Iterate till starts cross end
    while (start <= end) 
    {

        // Find the mid element
        let mid = start + Math.floor((end - start) / 2);

        // Check if K is found
        if (ar[mid] == ele)
            end = mid - 1;

        // Check if current letacter
        // 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 letacter
    return ch;
}

// Driver Code
    let ar = [ 'e', 'g', 't', 'y' ];
    let n = ar.length;

    let K = 'u';

    let ch = bs(ar, n, K);

    if (ch == '@')
        document.write("-1");
    else
        document.write(ch);

// This code is contributed by sanjoy_62.
</script>

t

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