leetcode

commit 27248c4b7f20d7884c788fd214243127ce0769df
parent 838ae094d71ebac895404c4c0dfe1d016b475555
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Mon, 22 May 2023 09:19:23 -0500

Completed recursive solution to count number of texts

Diffstat:
A
count-number-of-texts/count-number-of-texts.dart
 | 
44
++++++++++++++++++++++++++++++++++++++++++++

1 file changed, 44 insertions(+), 0 deletions(-)
diff --git a/count-number-of-texts/count-number-of-texts.dart b/count-number-of-texts/count-number-of-texts.dart
@@ -0,0 +1,44 @@
+//Given an input count the number of possible texts sent by pressing the keys of a phone keyboard in this order.
+//The problem is looking for all permutations of strings that can result from pressing the number keys in a certain order.
+//My code does not work for very large inputs because it recurses too deep so I will convert this to iterative and resubmit.
+
+class Solution {
+    Map<int , List<String>> options = {};
+    Map<String , int> memo = {};
+    Solution(){
+    options = {
+        2 : ["a" , "b" , "c"],
+        3 : ["d" , "e" , "f"],
+        4 : ["g" , "h" , "i"],
+        5 : ["j" , "k" , "l"],
+        6 : ["m" , "n" , "o"],
+        7 : ["p" , "q" , "r" , "s"],
+        8 : ["t" , "u" , "v"],
+        9 : ["w" , "x" , "y" , "z"]
+    };
+  }
+  int countTexts(String pressedKeys) {
+      if(pressedKeys.length == 0){
+          return 1;
+      }
+      String current_str = pressedKeys;
+      if(memo[current_str] != null){
+          return (memo[current_str] ?? 0);
+      }
+      int ret_val = 0;
+      String new_pressed = pressedKeys.substring(1);
+      ret_val += countTexts(new_pressed);
+      ret_val = (ret_val % (pow(10, 9) + 7)).toInt();
+      String current = pressedKeys.substring(0,1);
+      for(int i = 1 ; i < (options[int.parse(current)]?.length ?? 0) && i < pressedKeys.length; ++i){
+          if(pressedKeys[i] != pressedKeys[i - 1]){
+              break;
+          }
+          new_pressed = new_pressed.substring(1);
+          ret_val += countTexts(new_pressed);
+          ret_val = (ret_val % (pow(10, 9) + 7)).toInt();
+      }
+      memo[current_str] = ret_val;
+      return (ret_val % (pow(10, 9) + 7)).toInt();
+  }
+}