Maximum number of characters between any two same character in a string
Given a string, find the maximum number of characters between any two characters in the string. If no character repeats, print -1.
Examples:
Input : str = "abba"
Output : 2
The maximum number of characters are
between two occurrences of 'a'.
Input : str = "baaabcddc"
Output : 3
The maximum number of characters are
between two occurrences of 'b'.
Input : str = "abc"
Output : -1
Approach 1 (Simple): Use two nested loops. The outer loop picks characters from left to right, the inner loop finds the farthest occurrence and keeps track of the maximum.
C++
#include <bits/stdc++.h>
using namespace std;
int maximumChars(string& str)
{
int n = str.length();
int res = -1;
for ( int i = 0; i < n - 1; i++)
for ( int j = i + 1; j < n; j++)
if (str[i] == str[j])
res = max(res, abs (j - i - 1));
return res;
}
int main()
{
string str = "abba" ;
cout << maximumChars(str);
return 0;
}
|
Java
class GFG {
static int maximumChars(String str)
{
int n = str.length();
int res = - 1 ;
for ( int i = 0 ; i < n - 1 ; i++)
for ( int j = i + 1 ; j < n; j++)
if (str.charAt(i) == str.charAt(j))
res = Math.max(res,
Math.abs(j - i - 1 ));
return res;
}
public static void main(String[] args)
{
String str = "abba" ;
System.out.println(maximumChars(str));
}
}
|
Python3
def maximumChars( str ):
n = len ( str )
res = - 1
for i in range ( 0 , n - 1 ):
for j in range (i + 1 , n):
if ( str [i] = = str [j]):
res = max (res, abs (j - i - 1 ))
return res
if __name__ = = '__main__' :
str = "abba"
print (maximumChars( str ))
|
C#
using System;
public class GFG {
static int maximumChars( string str)
{
int n = str.Length;
int res = -1;
for ( int i = 0; i < n - 1; i++)
for ( int j = i + 1; j < n; j++)
if (str[i] == str[j])
res = Math.Max(res,
Math.Abs(j - i - 1));
return res;
}
static public void Main ()
{
string str = "abba" ;
Console.WriteLine(maximumChars(str));
}
}
|
Javascript
<script>
function maximumChars(str)
{
let n = str.length;
let res = -1;
for (let i = 0; i < n - 1; i++)
for (let j = i + 1; j < n; j++)
if (str[i] == str[j])
res = Math.max(res,
Math.abs(j - i - 1));
return res;
}
let str = "abba" ;
document.write(maximumChars(str));
</script>
|
Output:
2
Time Complexity : O(n^2)
Auxiliary Space: O(1), since no extra space has been taken.
Approach 2 (Efficient) : Initialize an array”FIRST” of length 26 in which we have to store the first occurrence of an alphabet in the string and another array “LAST” of length 26 in which we will store the last occurrence of the alphabet in the string. Here, index 0 corresponds to alphabet a, 1 for b and so on . After that, we will take the difference between the last and first arrays to find the max difference if they are not at the same position.
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 256;
int maximumChars(string& str)
{
int n = str.length();
int res = -1;
int firstInd[MAX_CHAR];
for ( int i = 0; i < MAX_CHAR; i++)
firstInd[i] = -1;
for ( int i = 0; i < n; i++) {
int first_ind = firstInd[str[i]];
if (first_ind == -1)
firstInd[str[i]] = i;
else
res = max(res, abs (i - first_ind - 1));
}
return res;
}
int main()
{
string str = "abba" ;
cout << maximumChars(str);
return 0;
}
|
Java
import java.io.*;
public class GFG {
static int MAX_CHAR = 256 ;
static int maximumChars(String str)
{
int n = str.length();
int res = - 1 ;
int []firstInd = new int [MAX_CHAR];
for ( int i = 0 ; i < MAX_CHAR; i++)
firstInd[i] = - 1 ;
for ( int i = 0 ; i < n; i++)
{
int first_ind = firstInd[str.charAt(i)];
if (first_ind == - 1 )
firstInd[str.charAt(i)] = i;
else
res = Math.max(res, Math.abs(i
- first_ind - 1 ));
}
return res;
}
static public void main (String[] args)
{
String str = "abba" ;
System.out.println(maximumChars(str));
}
}
|
Python3
MAX_CHAR = 256
def maximumChars(str1):
n = len (str1)
res = - 1
firstInd = [ - 1 for i in range (MAX_CHAR)]
for i in range (n):
first_ind = firstInd[ ord (str1[i])]
if (first_ind = = - 1 ):
firstInd[ ord (str1[i])] = i
else :
res = max (res, abs (i - first_ind - 1 ))
return res
str1 = "abba"
print (maximumChars(str1))
|
C#
using System;
public class GFG {
static int MAX_CHAR = 256;
static int maximumChars( string str)
{
int n = str.Length;
int res = -1;
int []firstInd = new int [MAX_CHAR];
for ( int i = 0; i < MAX_CHAR; i++)
firstInd[i] = -1;
for ( int i = 0; i < n; i++)
{
int first_ind = firstInd[str[i]];
if (first_ind == -1)
firstInd[str[i]] = i;
else
res = Math.Max(res, Math.Abs(i
- first_ind - 1));
}
return res;
}
static public void Main ()
{
string str = "abba" ;
Console.WriteLine(maximumChars(str));
}
}
|
Javascript
<script>
let MAX_CHAR = 256;
function maximumChars(str)
{
let n = str.length;
let res = -1;
let firstInd = new Array(MAX_CHAR);
for (let i = 0; i < MAX_CHAR; i++)
firstInd[i] = -1;
for (let i = 0; i < n; i++)
{
let first_ind = firstInd[str[i].charCodeAt(0)];
if (first_ind == -1)
firstInd[str[i].charCodeAt(0)] = i;
else
res = Math.max(res, Math.abs(i
- first_ind - 1));
}
return res;
}
let str = "abba" ;
document.write(maximumChars(str));
</script>
|
Output:
2
Time Complexity : O(n)
Auxiliary Space: O(256) since 256 extra space has been taken.
Last Updated :
19 Aug, 2022
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