Length of the longest substring with consecutive characters
Given a string str of lowercase alphabets, the task is to find the length of the longest sub-string of characters in alphabetical order i.e. string “dfabck” will return 3. Note that the alphabetical order here is considered circular i.e. a, b, c, d, e, …, x, y, z, a, b, c, ….
Examples:
Input: str = “abcabcdefabc”
Output: 6
All valid sub-strings are “abc”, “abcdef” and “abc”
And, the length of the longest of these is 6Input: str = “zabcd”
Output: 5
Approach:
- Initialize i = 0 and len = 0 and starting from i find the ending index of the longest valid sub-string and store it in end.
- Now, update len = max(end – i + 1, len) and i = end + 1 (to get the next valid sub-string starting from the index end + 1) and repeat step 1 until i < len(str).
- Print the value of len in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std; // Function to return the ending index for the // largest valid sub-string starting from index i int getEndingIndex(string str, int n, int i) { i++; while (i < n) { char curr = str[i]; char prev = str[i-1]; // If the current character appears after // the previous character according to // the given circular alphabetical order if ((curr == 'a' && prev == 'z') || (curr - prev == 1)) i++; else break; } return i - 1; } // Function to return the length of the longest // sub-string of consecutive characters from str int largestSubStr(string str, int n) { int len = 0; int i = 0; while (i < n) { // Valid sub-string exists from index // i to end int end = getEndingIndex(str, n, i); // Update the length len = max(end - i + 1, len); i = end + 1; } return len; } // Driver code int main() { string str = "abcabcdefabc"; int n = str.length(); cout << (largestSubStr(str, n)); } // This code is contributed by // Surendra_Gangwar |
Java
// Java implementation of the approach class GFG { // Function to return the ending index for the // largest valid sub-string starting from index i static int getEndingIndex(String str, int n, int i) { i++; while (i < n) { char curr = str.charAt(i); char prev = str.charAt(i - 1); // If the current character appears after // the previous character according to // the given circular alphabetical order if ((curr == 'a' && prev == 'z') || (curr - prev == 1)) i++; else break; } return i - 1; } // Function to return the length of the longest // sub-string of consecutive characters from str static int largestSubStr(String str, int n) { int len = 0; int i = 0; while (i < n) { // Valid sub-string exists from index // i to end int end = getEndingIndex(str, n, i); // Update the length len = Math.max(end - i + 1, len); i = end + 1; } return len; } // Driver code public static void main(String args[]) { String str = "abcabcdefabc"; int n = str.length(); System.out.print(largestSubStr(str, n)); } } |
Python3
# Python3 implementation of the approach # Function to return the ending index for the # largest valid sub-str1ing starting from index i def getEndingIndex(str1, n, i): i += 1 while (i < n): curr = str1[i] prev = str1[i - 1] # If the current character appears after # the previous character according to # the given circular alphabetical order if ((curr == 'a' and prev == 'z') or (ord(curr) - ord(prev) == 1)): i += 1 else: break return i - 1 # Function to return the length of the # longest sub-str1ing of consecutive # characters from str1 def largestSubstr1(str1, n): Len = 0 i = 0 while (i < n): # Valid sub-str1ing exists from # index i to end end = getEndingIndex(str1, n, i) # Update the Length Len = max(end - i + 1, Len) i = end + 1 return Len # Driver code str1 = "abcabcdefabc"n = len(str1) print(largestSubstr1(str1, n)) # This code is contributed by # Mohit Kumar 29 |
C#
// C# implementation of the approach using System; class GFG { // Function to return the ending index for the // largest valid sub-string starting from index i static int getEndingIndex(string str, int n, int i) { i++; while (i < n) { char curr = str[i]; char prev = str[i - 1]; // If the current character appears after // the previous character according to // the given circular alphabetical order if ((curr == 'a' && prev == 'z') || (curr - prev == 1)) i++; else break; } return i - 1; } // Function to return the length of the longest // sub-string of consecutive characters from str static int largestSubStr(string str, int n) { int len = 0; int i = 0; while (i < n) { // Valid sub-string exists from index // i to end int end = getEndingIndex(str, n, i); // Update the length len = Math.Max(end - i + 1, len); i = end + 1; } return len; } // Driver code public static void Main() { string str = "abcabcdefabc"; int n = str.Length; Console.Write(largestSubStr(str, n)); } } // This code is contributed // by Akanksha Rai |
PHP
<?php // PHP implementation of the approach // Function to return the ending index // for the largest valid sub-string /// starting from index i function getEndingIndex($str, $n, $i) { $i++; while ($i < $n) { $curr = $str[$i]; $prev = $str[$i - 1]; // If the current character appears after // the previous character according to // the given circular alphabetical order if (($curr == 'a' && $prev == 'z') || (ord($curr) - ord($prev) == 1)) $i++; else break; } return $i - 1; } // Function to return the length of the longest // sub-string of consecutive characters from str function largestSubStr($str, $n) { $len = 0; $i = 0; while ($i < $n) { // Valid sub-string exists from index // i to end $end = getEndingIndex($str, $n, $i); // Update the length $len = max($end - $i + 1, $len); $i = $end + 1; } return $len; } // Driver code $str = "abcabcdefabc"; $n = strlen($str); echo largestSubStr($str, $n); // This code is contributed by Ryuga ?> |
Output:
6
Time Complexity : O(n) where n is length of the input string. Note that we increase index by end.
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