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Longest substring with no pair of adjacent characters are adjacent English alphabets

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Given a string S consisting of lowercase English alphabets, the task is to find the longest substring from the given string such that no two adjacent characters are neighbouring English alphabets.

Examples:

Input: S = “aabdml”
Output: “bdm”
Explanation: Substring “bdm” is the longest substring which satisfies the given condition.

Input: S = “abc”
Output: “a”
Explanation: Substrings “a”, “b”, “c” satisfies the given condition. Print any one of them.

Approach: Follow the steps below to solve the problem:

  1. Initialize an empty string, say T, to store all the temporary substrings during iteration.
  2. Initialize another string, say ans, to store the longest substring as per the given conditions and a variable, say len, to store the length of the longest substring.
  3. Append the first character of the string to ans.
  4. Traverse the string in the range of indices [1, S.length()] and perform the following operations:
    • If the absolute difference between ASCII values of the adjacent characters is 1, then update ans and the length of the longest string and set T equal to the current character for the next iteration.
    • Otherwise, append the current character to string T.
  5. Check again for the longest substring and update the values accordingly.
  6. Print the longest substring obtained in variable ans.

Below is the implementation of the above approach:

C++




// C++ program for
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the longest substring
// satisfying the given condition
void findSubstring(string S)
{
    // Stores all temporary substrings
    string T = "";
 
    // Stores the longest substring
    string ans = "";
 
    // Stores the length
    // of the substring T
    int l = 0;
 
    // Stores the first
    // character of string S
    T += S[0];
 
    // Traverse the string
    for (int i = 1; i < S.length(); i++) {
 
        // If the absolute difference is 1
        if (abs(S[i] - S[i - 1]) == 1) {
 
            // Update the length of
            // substring T
            l = T.length();
 
            // Update the longest substring
            if (l > ans.length()) {
                ans = T;
            }
 
            T = "";
            T += S[i];
        }
 
        // Otherwise, stores the current
        // character
        else {
            T += S[i];
        }
    }
 
    // Again checking for
    // longest substring
    // and update accordingly
    l = (int)T.length();
 
    if (l > (int)ans.length()) {
        ans = T;
    }
 
    // Print the longest substring
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    // Given string
    string S = "aabdml";
 
    // Function call to find
    // the longest substring
    // satisfying given condition
    findSubstring(S);
 
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
class GFG
{
 
  // Function to find the longest substring
  // satisfying the given condition
  static void findSubstring(String S)
  {
 
    // Stores all temporary substrings
    String T = "";
 
    // Stores the longest substring
    String ans = "";
 
    // Stores the length
    // of the substring T
    int l = 0;
 
    // Stores the first
    // character of string S
    T += S.charAt(0);
 
    // Traverse the string
    for (int i = 1; i < S.length(); i++) {
 
      // If the absolute difference is 1
      if (Math.abs(S.charAt(i) - S.charAt(i - 1))
          == 1) {
 
        // Update the length of
        // substring T
        l = T.length();
 
        // Update the longest substring
        if (l > ans.length()) {
          ans = T;
        }
 
        T = "";
        T += S.charAt(i);
      }
 
      // Otherwise, stores the current
      // character
      else {
        T += S.charAt(i);
      }
    }
 
    // Again checking for
    // longest substring
    // and update accordingly
    l = T.length();
    if (l > ans.length()) {
      ans = T;
    }
 
    // Print the longest substring
    System.out.println(ans);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given string
    String S = "aabdml";
 
    // Function call to find
    // the longest substring
    // satisfying given condition
    findSubstring(S);
  }
}
 
// This code is contributed by Dharanendra L V.


Python3




# Python3 program for
# the above approach
 
# Function to find the longest substring
# satisfying the given condition
def findSubstring(S):
   
    # Stores all temporary substrings
    T = ""
 
    # Stores the longest substring
    ans = ""
 
    # Stores the length
    # of the subT
    l = 0
 
    # Stores the first
    # character of S
    T += S[0]
 
    # Traverse the string
    for i in range(1,len(S)):
 
        # If the absolute difference is 1
        if (abs(ord(S[i]) - ord(S[i - 1])) == 1):
 
            # Update the length of
            # subT
            l = len(T)
 
            # Update the longest substring
            if (l > len(ans)):
                ans = T
            T = ""
            T += S[i]
 
        # Otherwise, stores the current
        # character
        else:
            T += S[i]
 
    # Again checking for
    # longest substring
    # and update accordingly
    l = len(T)
    if (l > len(ans)):
        ans = T
 
    # Print the longest substring
    print (ans)
 
# Driver Code
if __name__ == '__main__':
   
    # Given string
    S = "aabdml"
 
    # Function call to find
    # the longest substring
    # satisfying given condition
    findSubstring(S)
 
    # This code is contributed by mohit kumar 29.


C#




// C# program for the above approach
using System;
public class GFG
{
   
// Function to find the longest substring
// satisfying the given condition
static void findSubstring(string S)
{
   
    // Stores all temporary substrings
    string T = "";
 
    // Stores the longest substring
    string ans = "";
 
    // Stores the length
    // of the substring T
    int l = 0;
 
    // Stores the first
    // character of string S
    T += S[0];
 
    // Traverse the string
    for (int i = 1; i < S.Length; i++)
    {
 
        // If the absolute difference is 1
        if (Math.Abs(S[i] - S[i - 1]) == 1)
        {
 
            // Update the length of
            // substring T
            l = T.Length;
 
            // Update the longest substring
            if (l > ans.Length)
            {
                ans = T;
            }
 
            T = "";
            T += S[i];
        }
 
        // Otherwise, stores the current
        // character
        else
        {
            T += S[i];
        }
    }
 
    // Again checking for
    // longest substring
    // and update accordingly
    l = T.Length;
    if (l > ans.Length)
    {
        ans = T;
    }
 
    // Print the longest substring
    Console.Write(ans);
}
 
 
// Driver Code
public static void Main(String[] args)
{
   
    // Given string
    string S = "aabdml";
 
    // Function call to find
    // the longest substring
    // satisfying given condition
    findSubstring(S);
}
}
 
// This code is contributed by code_hunt.


Javascript




<script>
 
      // JavaScript program for the above approach
       
      // Function to find the longest substring
      // satisfying the given condition
      function findSubstring(S) {
        // Stores all temporary substrings
        var T = "";
 
        // Stores the longest substring
        var ans = "";
 
        // Stores the length
        // of the substring T
        var l = 0;
 
        // Stores the first
        // character of string S
        T += S[0];
 
        // Traverse the string
        for (var i = 1; i < S.length; i++) {
          // If the absolute difference is 1
          if (Math.abs(S[i].charCodeAt(0) -
          S[i - 1].charCodeAt(0)) === 1)
          {
            // Update the length of
            // substring T
            l = T.length;
 
            // Update the longest substring
            if (l > ans.length) {
              ans = T;
            }
 
            T = "";
            T += S[i];
          }
 
          // Otherwise, stores the current
          // character
          else {
            T += S[i];
          }
        }
 
        // Again checking for
        // longest substring
        // and update accordingly
        l = T.length;
        if (l > ans.length) {
          ans = T;
        }
 
        // Print the longest substring
        document.write(ans);
      }
 
      // Driver Code
      // Given string
      var S = "aabdml";
 
      // Function call to find
      // the longest substring
      // satisfying given condition
      findSubstring(S);
       
</script>


Output: 

bdm

 

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



Last Updated : 18 May, 2021
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