Given a string str, two characters X and Y. The task is to find the length of the longest substring that starts with X and ends with Y. It is given that there always exists a substring that starts with X and ends with Y.
Examples:
Input: str = “QWERTYASDFZXCV”, X = ‘A’, Y = ‘Z’
Output: 5
Explanation:
The largest substring which start with ‘A’ and end with ‘Z’ = “ASDFZ”.
Size of the substring = 5.
Input: str = “ZABCZ”, X = ‘Z’, Y = ‘Z’
Output: 3
Explanation:
The largest substring which start with ‘Z’ and end with ‘Z’ = “ZABCZ”.
Size of the substring = 5.
Naive Approach: The naive approach is to find all the substrings of the given string out of these find the largest substring which starts with X and ends with Y.
// C++ program for the naive approach #include <bits/stdc++.h> using namespace std;
// Function returns length of longest substring starting // with X and ending with Y int longestSubstring(string str, char X, char Y)
{ int n = str.size();
int ans = 0;
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j < n; j++) {
// is str[i] == X and str[j] == Y then the
// substring str[i...j] maybe longest substring
// that we required
if (str[i] == X && str[j] == Y) {
ans = max(ans, j - i + 1);
}
}
}
return ans;
} // Driver Code int main()
{ // Given string str
string str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters
char X = 'A' , Y = 'Z' ;
// Function Call
cout << longestSubstring(str, X, Y) << "\n" ;
return 0;
} // This code is contributed by ajaymakvana |
// JAVA program for the naive approach import java.util.*;
class GFG {
// Function returns length of longest substring starting
// with X and ending with Y
public static int longestSubstring(String str, char X,
char Y)
{
int n = str.length();
int ans = 0 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = i + 1 ; j < n; j++) {
// is str[i] == X and str[j] == Y then the
// substring str[i...j] maybe longest
// substring that we required
if (str.charAt(i) == X
&& str.charAt(j) == Y) {
ans = Math.max(ans, j - i + 1 );
}
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Given string str
String str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters
char X = 'A' , Y = 'Z' ;
// Function Call
System.out.println(longestSubstring(str, X, Y));
}
} // This code is contributed by Taranpreet |
# Function returns length of longest substring starting # with X and ending with Y def longest_substring( str , X, Y):
n = len ( str )
ans = 0
for i in range (n):
for j in range (i + 1 , n):
# is str[i] == X and str[j] == Y then the
# substring str[i...j] maybe longest
# substring that we required
if str [i] = = X and str [j] = = Y:
ans = max (ans, j - i + 1 )
return ans
# given string str = "HASFJGHOGAKZZFEGA"
# Starting and Ending characters X = 'A'
Y = 'Z'
# Function call print (longest_substring( str , X, Y))
|
// C# program for the naive approach using System;
public class GFG {
// Function returns length of longest substring starting
// with X and ending with Y
public static int LongestSubstring( string str, char X,
char Y)
{
int n = str.Length;
int ans = 0;
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j < n; j++) {
// is str[i] == X and str[j] == Y then the
// substring str[i...j] maybe longest
// substring that we required
if (str[i] == X && str[j] == Y) {
ans = Math.Max(ans, j - i + 1);
}
}
}
return ans;
}
// Driver Code
public static void Main()
{
// Given string str
string str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters
char X = 'A' , Y = 'Z' ;
// Function Call
Console.WriteLine(LongestSubstring(str, X, Y));
}
} //contributed by adityasha4x71 |
// Javascript program for the above approach // Function returns length of longest substring starting // with X and ending with Y function longest_substring(str, X, Y) {
let n = str.length;
let ans = 0;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
// is str[i] == X and str[j] == Y then the
// substring str[i...j] maybe longest
// substring that we required
if (str[i] == X && str[j] == Y) {
ans = Math.max(ans, j - i + 1);
}
}
}
return ans;
} // given string let str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters let X = 'A' ;
let Y = 'Z' ;
// Function call console.log(longest_substring(str, X, Y)); // This code is contributed by princekumaras |
12
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimized the above approach, the count of characters between X and Y should be the largest. So, iterate over the string using pointers start and end to find the first occurrence of X from the starting index and the last occurrence of Y from the end. Below are the steps:
- Initialize start = 0 and end = length of string – 1.
- Traverse the string from the beginning and find the first occurrence of character X. Let it be at index xPos.
- Traverse the string from the beginning and find the last occurrence of character Y. Let it be at index yPos.
- The length of the longest substring is given by (yPos – xPos + 1).
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function returns length of longest // substring starting with X and // ending with Y int longestSubstring(string str,
char X, char Y)
{ // Length of string
int N = str.length();
int start = 0;
int end = N - 1;
int xPos = 0;
int yPos = 0;
// Find the length of the string
// starting with X from the beginning
while ( true ) {
if (str[start] == X) {
xPos = start;
break ;
}
start++;
}
// Find the length of the string
// ending with Y from the end
while ( true ) {
if (str[end] == Y) {
yPos = end;
break ;
}
end--;
}
// Longest substring
int length = (yPos - xPos) + 1;
// Print the length
cout << length;
} // Driver Code int main()
{ // Given string str
string str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters
char X = 'A' , Y = 'Z' ;
// Function Call
longestSubstring(str, X, Y);
return 0;
} |
// Java program for the above approach class GFG{
// Function returns length of longest // substring starting with X and // ending with Y public static void longestSubstring(String str,
char X, char Y)
{ // Length of string
int N = str.length();
int start = 0 ;
int end = N - 1 ;
int xPos = 0 ;
int yPos = 0 ;
// Find the length of the string
// starting with X from the beginning
while ( true )
{
if (str.charAt(start) == X)
{
xPos = start;
break ;
}
start++;
}
// Find the length of the string
// ending with Y from the end
while ( true )
{
if (str.charAt(end) == Y)
{
yPos = end;
break ;
}
end--;
}
// Longest substring
int length = (yPos - xPos) + 1 ;
// Print the length
System.out.print(length);
} // Driver code public static void main(String[] args)
{ // Given string str
String str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters
char X = 'A' , Y = 'Z' ;
// Function Call
longestSubstring(str, X, Y);
} } // This code is contributed by divyeshrabadiya07 |
# Python3 program for the above approach # Function returns length of longest # substring starting with X and # ending with Y def longestSubstring( str , X, Y):
# Length of string
N = len ( str )
start = 0
end = N - 1
xPos = 0
yPos = 0
# Find the length of the string
# starting with X from the beginning
while ( True ):
if ( str [start] = = X):
xPos = start
break
start + = 1
# Find the length of the string
# ending with Y from the end
while ( True ):
if ( str [end] = = Y):
yPos = end
break
end - = 1
# Longest substring
length = (yPos - xPos) + 1
# Print the length
print (length)
# Driver Code if __name__ = = "__main__" :
# Given string str
str = "HASFJGHOGAKZZFEGA"
# Starting and Ending characters
X = 'A'
Y = 'Z'
# Function Call
longestSubstring( str , X, Y)
# This code is contributed by sanjoy_62 |
// C# program for the above approach using System;
class GFG{
// Function returns length of longest // substring starting with X and // ending with Y static void longestSubstring( string str,
char X, char Y)
{ // Length of string
int N = str.Length;
int start = 0;
int end = N - 1;
int xPos = 0;
int yPos = 0;
// Find the length of the string
// starting with X from the beginning
while ( true )
{
if (str[start] == X)
{
xPos = start;
break ;
}
start++;
}
// Find the length of the string
// ending with Y from the end
while ( true )
{
if (str[end] == Y)
{
yPos = end;
break ;
}
end--;
}
// Longest substring
int length = (yPos - xPos) + 1;
// Print the length
Console.Write(length);
} // Driver code public static void Main()
{ // Given string str
string str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters
char X = 'A' , Y = 'Z' ;
// Function call
longestSubstring(str, X, Y);
} } // This code is contributed by sanjoy_62 |
<script> // JavaScript program for the above approach
// Function returns length of longest
// substring starting with X and
// ending with Y
function longestSubstring(str, X, Y) {
// Length of string
var N = str.length;
var start = 0;
var end = N - 1;
var xPos = 0;
var yPos = 0;
// Find the length of the string
// starting with X from the beginning
while ( true ) {
if (str[start] === X) {
xPos = start;
break ;
}
start++;
}
// Find the length of the string
// ending with Y from the end
while ( true ) {
if (str[end] === Y) {
yPos = end;
break ;
}
end--;
}
// Longest substring
var length = yPos - xPos + 1;
// Print the length
document.write(length);
}
// Driver code
// Given string str
var str = "HASFJGHOGAKZZFEGA" ;
// Starting and Ending characters
var X = "A" ,
Y = "Z" ;
// Function call
longestSubstring(str, X, Y);
</script>
|
12
Time Complexity: O(N)
Auxiliary Space: O(1)