From 0052ec63ca70eaa6d9ffb1926c294dbfd85f8c05 Mon Sep 17 00:00:00 2001
From: Mohammad Sajid Anwar <mohammad.anwar@yahoo.com>
Date: Wed, 18 Sep 2024 20:18:52 +0100
Subject: - Added solutions by Paulo Custodio. - Added solutions by Peter
 Campbell Smith. - Added solutions by Robert Ransbottom. - Added solutions by
 Feng Chang. - Added solutions by PokGoPun. - Added solutions by Peter
 Pentchev. - Added solutions by Santiago Leyva. - Added solutions by Robbie
 Hatley. - Added solutions by Ulrich Rieke. - Added solutions by Laurent
 Rosenfeld.

---
 challenge-287/ulrich-rieke/cpp/ch-1.cpp | 71 +++++++++++++++++++++++++++++++++
 1 file changed, 71 insertions(+)
 create mode 100755 challenge-287/ulrich-rieke/cpp/ch-1.cpp

(limited to 'challenge-287/ulrich-rieke/cpp/ch-1.cpp')

diff --git a/challenge-287/ulrich-rieke/cpp/ch-1.cpp b/challenge-287/ulrich-rieke/cpp/ch-1.cpp
new file mode 100755
index 0000000000..6048d7476f
--- /dev/null
+++ b/challenge-287/ulrich-rieke/cpp/ch-1.cpp
@@ -0,0 +1,71 @@
+#include <iostream>
+#include <string>
+#include <regex>
+#include <vector>
+#include <numeric>
+
+int main( ) {
+   std::cout << "Enter a password!\n" ;
+   std::string word ;
+   std::cin >> word ;
+   int totalsteps = 0 ;
+   const std::regex rgx1("[a-z]") ;
+   const std::regex rgx2("[A-Z]") ;
+   const std::regex rgx3("\\d") ;
+   int len = word.length( ) ;
+   if ( len < 6 ) {
+      totalsteps += 6 - len ;
+   }
+   //beware : it took me quite some time to find out that 
+   //std::regex_match does not return the right result!!
+   //if you want to check whether a given regular expression
+   //such as \d matches I had to use std::regex_search!!!
+   if ( ! std::regex_search( word , rgx1 )) {
+      if ( len >= 6 ) {
+	 totalsteps++ ;
+      }
+   }
+   if ( ! std::regex_search( word , rgx2 )) {
+      if ( len >= 6 ) {
+	 totalsteps++ ;
+      }
+   }
+   if ( ! std::regex_search( word , rgx3 )) {
+      if ( len >= 6 ) {
+	 totalsteps++ ;
+      }
+   }
+   //std::regex appears not to allow backreferences...?
+   //so, to prevent too many identical neighbouring letters
+   //I group the letters , divide their lengths by 3 and sum the
+   //quotients up ( inspired by Haskell where this operation is so easy
+   //to perform!!!
+   std::vector<std::string> neighbouring_letters ;
+   std::string currentWord ;
+   //imitate Haskell's group function!
+   for ( auto it = word.begin( ) ; it != word.end( ) ; ++it ) {
+      if ( currentWord.empty( ) ) 
+	 currentWord.push_back( *it ) ;
+      else {
+	 char last_char = currentWord.back( ) ;
+	 if ( last_char == *it ) {
+	    currentWord.push_back( *it ) ;
+	 }
+	 else {
+	    neighbouring_letters.push_back( currentWord ) ;
+	    currentWord.clear( ) ;
+	    currentWord.push_back( *it ) ;
+	 }
+      }
+   }
+   neighbouring_letters.push_back( currentWord ) ;
+   std::vector<int> grouplengths ;
+   for ( auto v : neighbouring_letters ) {
+      grouplengths.push_back( v.length( ) / 3 ) ;
+   }
+   int replacements = std::accumulate( grouplengths.begin( ) , 
+	 grouplengths.end( ) , 0 ) ;
+   totalsteps += replacements ;
+   std::cout << totalsteps << '\n' ;
+   return 0 ;
+}
-- 
cgit