Given a name and a typed-name of a person. Sometimes, when typing a vowel [aeiou], the key might get long pressed, and the character will be typed 1 or more times. The task is to examine the typed-name and tell if it is possible that typed name was of person’s name, with some characters (possibly none) being long pressed. Return ‘True‘ if it is, else ‘False‘.
Note: name and typed-name are separated by space with no space in between individuals names. Each character of the name is unique.
Examples:
Input: str = “neveropen”, typed = “geeeeks”
Output: True
The vowel ‘e’ repeats more times in typed and all other characters match.Input: str = “alice”, typed = “aallicce”
Output: False
Here ‘l’ and ‘c’ are repeated which not a vowel.
Hence name and typed-name represents different names.Input: str = “alex”, typed = “aaalaeex”
Output: False
A vowel ‘a’ is extra in typed.
Approach: The idea is based on Run Length Encoding. We consider only vowels and count their consecutive occurrences in str and typed. The count of occurrences in str must be less.
Below is the implementation of above approach.
C++
// CPP program to implement run length encoding#include <bits/stdc++.h>using namespace std;// Check if the character is vowel or notbool isVowel(char c){ string vowel = "aeiou"; for (int i = 0; i < vowel.length(); ++i) if (vowel[i] == c) return true; return false;}// Returns true if 'typed' is a typed name// given strbool printRLE(string str, string typed){ int n = str.length(), m = typed.length(); // Traverse through all characters of str. int j = 0; for (int i = 0; i < n; i++) { // If current characters do not match if (str[i] != typed[j]) return false; // If not vowel, simply move ahead in both if (isVowel(str[i]) == false) { j++; continue; } // Count occurrences of current vowel in str int count1 = 1; while (i < n - 1 && str[i] == str[i + 1]) { count1++; i++; } // Count occurrences of current vowel in // typed. int count2 = 1; while (j < m - 1 && typed[j] == str[i]) { count2++; j++; } if (count1 > count2) return false; } return true;}int main(){ string name = "alex", typed = "aaalaeex"; if (printRLE(name, typed)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to implement run length encodingpublic class Improve { // Check if the character is vowel or not static boolean isVowel(char c) { String vowel = "aeiou"; for (int i = 0; i < vowel.length(); ++i) if (vowel.charAt(i) == c) return true; return false; } // Returns true if 'typed' is a typed name // given str static boolean printRLE(String str, String typed) { int n = str.length(), m = typed.length(); // Traverse through all characters of str. int j = 0; for (int i = 0; i < n; i++) { // If current characters do not match if (str.charAt(i) != typed.charAt(j)) return false; // If not vowel, simply move ahead in both if (isVowel(str.charAt(i)) == false) { j++; continue; } // Count occurrences of current vowel in str int count1 = 1; while (i < n - 1 && str.charAt(i) == str.charAt(i+1)) { count1++; i++; } // Count occurrences of current vowel in // typed. int count2 = 1; while (j < m - 1 && typed.charAt(j) == str.charAt(i)) { count2++; j++; } if (count1 > count2) return false; } return true; } public static void main(String args[]) { String name = "alex", typed = "aaalaeex"; if (printRLE(name, typed)) System.out.println("Yes"); else System.out.println("No"); } // This code is contributed by ANKITRAI1} |
Python3
# Python3 program to implement run # length encoding# Check if the character is # vowel or notdef isVowel(c): vowel = "aeiou" for i in range(len(vowel)): if(vowel[i] == c): return True return False # Returns true if 'typed' is a # typed name given strdef printRLE(str, typed): n = len(str) m = len(typed) # Traverse through all # characters of str j = 0 for i in range(n): # If current characters do # not match if str[i] != typed[j]: return False # If not vowel, simply move # ahead in both if isVowel(str[i]) == False: j = j + 1 continue # Count occurrences of current # vowel in str count1 = 1 while (i < n - 1 and (str[i] == str[i + 1])): count1 = count1 + 1 i = i + 1 # Count occurrence of current # vowel in typed count2 = 1 while(j < m - 1 and typed[j] == str[i]): count2 = count2 + 1 j = j + 1 if count1 > count2: return False return True# Driver codename = "alex"typed = "aaalaeex"if (printRLE(name, typed)): print("Yes")else: print("No") # This code is contributed# by Shashank_Sharma |
C#
// C# program to implement run // length encoding using System;class GFG{// Check if the character is// vowel or not public static bool isVowel(char c){ string vowel = "aeiou"; for (int i = 0; i < vowel.Length; ++i) { if (vowel[i] == c) { return true; } } return false;}// Returns true if 'typed' is// a typed name given str public static bool printRLE(string str, string typed){ int n = str.Length, m = typed.Length; // Traverse through all // characters of str. int j = 0; for (int i = 0; i < n; i++) { // If current characters // do not match if (str[i] != typed[j]) { return false; } // If not vowel, simply move // ahead in both if (isVowel(str[i]) == false) { j++; continue; } // Count occurrences of current // vowel in str int count1 = 1; while (i < n - 1 && str[i] == str[i + 1]) { count1++; i++; } // Count occurrences of current // vowel in typed int count2 = 1; while (j < m - 1 && typed[j] == str[i]) { count2++; j++; } if (count1 > count2) { return false; } } return true;}// Driver Codepublic static void Main(string[] args){ string name = "alex", typed = "aaalaeex"; if (printRLE(name, typed)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); }}}// This code is contributed // by Shrikant13 |
PHP
<?php// PHP program to implement // run length encoding// Check if the character is vowel or notfunction isVowel($c){ $vowel = "aeiou"; for ($i = 0; $i < strlen($vowel); ++$i) if ($vowel[$i] == $c) return true; return false;}// Returns true if 'typed' // is a typed name// given strfunction printRLE($str, $typed){ $n = strlen($str); $m = strlen($typed); // Traverse through all // characters of str. $j = 0; for ($i = 0; $i < $n; $i++) { // If current characters // do not match if ($str[$i] != $typed[$j]) return false; // If not vowel, simply // move ahead in both if (isVowel($str[$i]) == false) { $j++; continue; } // Count occurrences of // current vowel in str $count1 = 1; while ($i < $n - 1 && $str[$i] == $str[$i + 1]) { $count1++; $i++; } // Count occurrences of // current vowel in typed. $count2 = 1; while ($j < $m - 1 && $typed[$j] == $str[$i]) { $count2++; $j++; } if ($count1 > $count2) return false; } return true;}// Driver code$name = "alex";$typed = "aaalaeex";if (printRLE($name, $typed)) echo "Yes";else echo "No"; // This code is contributed// by Shivi_Aggarwal ?> |
Javascript
<script>// Javascript program to implement// run length encoding // Check if the character is vowel or not function isVowel(c){ let vowel = "aeiou"; for(let i = 0; i < vowel.length; ++i) { if (vowel[i] == c) { return true; } } return false;}// Returns true if 'typed' is// a typed name given str function printRLE(str, typed){ let n = str.length, m = typed.length; // Traverse through all // characters of str. let j = 0; for(let i = 0; i < n; i++) { // If current characters // do not match if (str[i] != typed[j]) { return false; } // If not vowel, simply move // ahead in both if (isVowel(str[i]) == false) { j++; continue; } // Count occurrences of current // vowel in str let count1 = 1; while (i < n - 1 && str[i] == str[i + 1]) { count1++; i++; } // Count occurrences of current // vowel in typed let count2 = 1; while (j < m - 1 && typed[j] == str[i]) { count2++; j++; } if (count1 > count2) { return false; } } return true;}// Driver codelet name = "alex", typed = "aaalaeex";if (printRLE(name, typed)){ document.write("Yes");}else{ document.write("No");} // This code is contributed by decode2207</script> |
Output
No
Complexity Analysis:
- Time Complexity : O(m + n)
- Auxiliary Complexity: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
