Longest substring that starts with X and ends with Y

Last Updated : 14 Mar, 2023

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++

// 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

// 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

Python3

# 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

// 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

Output

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++

// 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

// 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

# 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

JavaScript

<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>

Output

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

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