solves student attendancerecord I

Author: anishLearnsToCode

Diff for: README.md

@@ -144,7 +144,7 @@
 | 538 | [Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree) | Easy | [![Java](assets/java.png)](src/ConvertBSTToGreaterTree.java) [![Python](assets/python.png)](python/convert_bst_to_greater_tree.py) |
 | 541 | [Reverse String II](https://leetcode.com/problems/reverse-string-ii) | Easy | [![Java](assets/java.png)](src/ReverseStringII.java/) [![Python](assets/python.png)](python/reverse_string_ii.py) |
 | 543 | [Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree) | Easy | [![Java](assets/java.png)](src/DiameterOfBinaryTree.java) [![Python](assets/python.png)](python/diameter_of_binary_tree.py) |
-| 551 | [Student Attendance Record I](https://leetcode.com/problems/student-attendance-record-i) | Easy | |
+| 551 | [Student Attendance Record I](https://leetcode.com/problems/student-attendance-record-i) | Easy | [![Java](assets/java.png)](src/StudentAttendanceRecordI.java) [![Python](assets/python.png)](python/student_attendance_record_I.py) |
 | 557 | [Reverse Words in a String III](https://leetcode.com/problems/reverse-words-in-a-string-iii) | Easy | |
 | 559 | [Maximum Depth of N-Ary Tree](https://leetcode.com/problems/maximum-depth-of-n-ary-tree) | Easy | |
 | 561 | [Array Partition I](https://leetcode.com/problems/array-partition-i) | Easy | |

Diff for: python/student_attendance_record_I.py

@@ -0,0 +1,11 @@
+class Solution:
+    def checkRecord(self, s: str) -> bool:
+        consecutiveLate, absent = 0, 0
+        for character in s:
+            if character == 'L':
+                consecutiveLate += 1
+            else:
+                if character == 'A': absent += 1
+                consecutiveLate = 0
+            if consecutiveLate >= 3 or absent >= 2: return False
+        return True

Diff for: src/StudentAttendanceRecordI.java

@@ -0,0 +1,17 @@
+public class StudentAttendanceRecordI {
+    public boolean checkRecord(String s) {
+        int consecutiveLate = 0, absentees = 0;
+        for (int index = 0 ; index < s.length() ; index++) {
+            if (s.charAt(index) == 'L') {
+                consecutiveLate++;
+            } else {
+                if (s.charAt(index) == 'A') absentees++;
+                consecutiveLate = 0;
+            }
+            if (absentees >= 2 || consecutiveLate >= 3) {
+                return false;
+            }
+        }
+        return true;
+    }
+}