Lexicographically largest string formed from the characters in range L and R

Given a string S and a range L and R, the task is to print the lexicographically largest string that can be formed from the characters in range L and R. 

Examples: 

Input: str = "thgyfh", L = 2, R = 6  
Output: yhhgf

Input: str = "striver", L = 3, R = 5
Output: vri

Approach: 

• Iterate from min(L, R) to max(L, R) and increase the frequencies of characters in a freq[] array.
• Iterate from 25 to 0 and print the number of times every character occurs to get the lexicographically largest string.

The common point of mistake that everyone does is they iterate from L to R instead of min(L, R) to max(L, R). 

Below is the implementation of the above approach: 

C++

// C++ program to print the
// lexicographically largest string that
// can be formed from the characters
// in range L and R

#include <bits/stdc++.h>
using namespace std;

// Function to return the lexicographically largest string
string printLargestString(string s, int l, int r)
{
    // hash array
    int freq[26] = { 0 };

    // make 0-based indexing
    l--;
    r--;

    // iterate and count frequencies of character
    for (int i = min(l, r); i <= max(l, r); i++) {
        freq[s[i] - 'a']++;
    }

    // ans string
    string ans = "";

    // iterate in frequency array
    for (int i = 25; i >= 0; i--) {

        // add till all characters
        // are added
        while (freq[i]) {
            ans += char('a' + i);
            freq[i]--;
        }
    }

    return ans;
}

// Driver Code
int main()
{
    string s = "striver";
    int l = 3, r = 5;
    cout << printLargestString(s, l, r);
    return 0;
}

Java

// Java program to print the
// lexicographically largest String that
// can be formed from the characters
// in range L and R

class GFG {

// Function to return the lexicographically largest String
    static String printLargestString(String s, int l, int r) {
        // hash array
        int freq[] = new int[26];

        // make 0-based indexing
        l--;
        r--;

        // iterate and count frequencies of character
        for (int i = Math.min(l, r); i <= Math.max(l, r); i++) {
            freq[s.charAt(i) - 'a']++;
        }

        // ans String
        String ans = "";

        // iterate in frequency array
        for (int i = 25; i >= 0; i--) {

            // add till all characters
            // are added
            while (freq[i] > 0) {
                ans += (char) ('a' + i);
                freq[i]--;
            }
        }

        return ans;
    }

// Driver Code
    public static void main(String[] args) {

        String s = "striver";
        int l = 3, r = 5;
        System.out.println(printLargestString(s, l, r));

    }
}
/* This JAVA code is contributed by 29AjayKumar*/

Python 3

# Python 3 program to print the
# lexicographically largest string that
# can be formed from the characters
# in range L and R

# Function to return the lexicographically
# largest string
def printLargestString(s, l, r):

    # hash array
    freq = [0] * 26

    # make 0-based indexing
    l -= 1
    r -= 1

    # iterate and count frequencies of character
    for i in range(min(l, r), max(l, r) + 1) :
        freq[ord(s[i]) - ord('a')] += 1

    # ans string
    ans = ""

    # iterate in frequency array
    for i in range(25, -1, -1):

        # add till all characters are added
        while (freq[i]):
            ans += chr(ord('a') + i)
            freq[i] -= 1

    return ans

# Driver Code
if __name__ == "__main__":

    s = "striver"
    l = 3
    r = 5
    print(printLargestString(s, l, r))

# This code is contributed by ita_c

C#

// C# program to print the lexicographically
// largest String that can be formed from the
// characters in range L and R
using System;

class GFG
{

// Function to return the lexicographically
// largest String
static String printLargestString(String s,
                                 int l, int r)
{
    // hash array
    int []freq = new int[26];

    // make 0-based indexing
    l--;
    r--;

    // iterate and count frequencies
    // of character
    for (int i = Math.Min(l, r);
             i <= Math.Max(l, r); i++)
    {
        freq[s[i] - 'a']++;
    }

    // ans String
    String ans = "";

    // iterate in frequency array
    for (int i = 25; i >= 0; i--)
    {

        // add till all characters
        // are added
        while (freq[i] > 0)
        {
            ans += (char) ('a' + i);
            freq[i]--;
        }
    }

    return ans;
}

// Driver Code
public static void Main()
{
    String s = "striver";
    int l = 3, r = 5;
    Console.Write(printLargestString(s, l, r));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// javascript program to print the
// lexicographically largest String that
// can be formed from the characters
// in range L and R
    // Function to return the lexicographically largest String
     function printLargestString( s , l , r) {
        // hash array
        var freq = Array(26).fill(0);

        // make 0-based indexing
        l--;
        r--;

        // iterate and count frequencies of character
        for (i = Math.min(l, r); i <= Math.max(l, r); i++) {
            freq[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
        }

        // ans String
        var ans = "";

        // iterate in frequency array
        for (var i = 25; i >= 0; i--) {

            // add till all characters
            // are added
            while (freq[i] > 0) {
                ans += String.fromCharCode('a'.charCodeAt(0) + i);
                freq[i]--;
            }
        }

        return ans;
    }

    // Driver Code


        var s = "striver";
        var l = 3, r = 5;
        document.write(printLargestString(s, l, r));


// This code is contributed by todaysgaurav
</script>

Output

vri

Complexity Analysis:

• Time Complexity: O(N), as we are using a loop to traverse N times for counting the frequencies.
  Each element gets added to the frequency table only once which takes O(1) and is appended to string which also takes O(1).
• Auxiliary Space: O(N), as we are using extra space for storing resultant string.