Lexicographically largest string formed from the characters in range L and R
Last Updated :
31 Aug, 2023
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
Naive Approach:
- Extract the substring from index L to R from the given string.
- Sort the substring in descending order using any sorting algorithm.
- Return the sorted substring as the lexicographically largest string.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
string printLargestString(string s, int L, int R) {
string sub = s.substr(L - 1, R - L + 1);
sort(sub.rbegin(), sub.rend());
return sub;
}
int main() {
string s = "striver";
int L = 3, R = 5;
cout << printLargestString(s, L, R) << endl;
return 0;
}
import java.util.Arrays;
public class GFG {
static String printLargestString(String s, int L, int R) {
String sub = s.substring(L - 1, R);
char[] charArr = sub.toCharArray();
Arrays.sort(charArr);
return new StringBuilder(new String(charArr)).reverse().toString();
}
public static void main(String[] args) {
String s = "striver";
int L = 3, R = 5;
System.out.println(printLargestString(s, L, R));
}
}
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
# function to print the largest string
def printLargestString(s, L, R):
sub = s[L - 1:R]
sub = ''.join(sorted(sub, reverse=True))
return sub
# Driver Program
s = "striver"
L = 3
R = 5
print(printLargestString(s, L, R))
# THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
using System;
class GFG {
static string printLargestString(string s, int L, int R) {
// Extract the substring from index L-1 to R
string sub = s.Substring(L - 1, R - L + 1);
// Convert the substring to a char array and sort it in descending order
char[] chars = sub.ToCharArray();
Array.Sort(chars);
Array.Reverse(chars);
// Convert the sorted char array back to a string
string result = new string(chars);
return result;
}
static void Main(string[] args) {
string s = "striver";
int L = 3, R = 5;
// Call the PrintLargestString function and print the result
Console.WriteLine(printLargestString(s, L, R));
}
}
// function to print the largest string
function printLargestString(s, L, R) {
let sub = s.substring(L - 1, R);
let subArray = sub.split('');
subArray.sort((a, b) => b.localeCompare(a));
return subArray.join('');
}
// Driver Program
let s = "striver";
let L = 3, R = 5;
console.log(printLargestString(s, L, R));
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
Time Complexity: O(N log N), where N is the length of the substring (R - L + 1).
Auxiliary Space: O(R - L + 1) since we are creating a substring of length (R - L + 1) to sort it.
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 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 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# 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
<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>
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.
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