Added Leetcode easy problems

Author: das-jishu

Leetcode/easy/excel-sheet-column-title.py

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+""" 
+# EXCEL SHEET COLUMN TITLE
+
+Given a positive integer, return its corresponding column title as appear in an Excel sheet.
+
+For example:
+
+    1 -> A
+    2 -> B
+    3 -> C
+    ...
+    26 -> Z
+    27 -> AA
+    28 -> AB 
+    ...
+Example 1:
+
+Input: 1
+Output: "A"
+
+Example 2:
+
+Input: 28
+Output: "AB"
+
+Example 3:
+
+Input: 701
+Output: "ZY" 
+"""
+
+class Solution:
+    def convertToTitle(self, n: int) -> str:
+        if n == 0:
+            return ""
+        
+        res = ""
+        while n > 0:
+            rem = n % 26
+            if rem == 0:
+                rem = 26
+            res = chr(64+rem) + res
+            n = n // 26
+            if rem == 26:
+                n -= 1
+            
+        return res

Leetcode/easy/intersection-of-two-linked-lists.py

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+""" 
+# INTERSECTION OF TWO LINKED LISTS
+
+Write a program to find the node at which the intersection of two singly linked lists begins.
+
+For example, the following two linked lists:
+
+begin to intersect at node c1.
+
+Example 1:
+
+Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
+Output: Reference of the node with value = 8
+Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
+ 
+Example 2:
+
+Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
+Output: Reference of the node with value = 2
+Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
+ 
+Example 3:
+
+Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
+Output: null
+Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
+Explanation: The two lists do not intersect, so return null.
+ 
+Notes:
+
+If the two linked lists have no intersection at all, return null.
+The linked lists must retain their original structure after the function returns.
+You may assume there are no cycles anywhere in the entire linked structure.
+Each value on each linked list is in the range [1, 10^9].
+Your code should preferably run in O(n) time and use only O(1) memory.
+"""
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        cur1=headA
+        cur2=headB
+        while cur1!=cur2:
+            if cur1==None:
+                cur1=headB
+            else:
+                cur1=cur1.next
+            if cur2==None:
+                cur2=headA
+            else:
+                cur2=cur2.next
+        return cur1

Leetcode/easy/majority-element.py

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+""" 
+# MAJORITY ELEMENT
+
+Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
+
+You may assume that the array is non-empty and the majority element always exist in the array.
+
+Example 1:
+
+Input: [3,2,3]
+Output: 3
+
+Example 2:
+
+Input: [2,2,1,1,1,2,2]
+Output: 2 
+"""
+
+class Solution:
+    def majorityElement(self, nums) -> int:
+        
+        candidate = nums[0]
+        count = 0
+        
+        for x in nums:
+            if candidate == x:
+                count += 1
+                
+            else:
+                if count != 0:
+                    count -= 1
+                else:
+                    candidate = x
+                    count = 0
+                    
+        return candidate

crobot.py

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+
+class CRobot(object):
+    def __init__(self):
+        self.typeRobot = ""
+        self.SN = 0
+        self.orientation = 1
+        self.status = False
+
+    def __init__(self, typeRobot, sn):
+        self.typeRobot = typeRobot
+        self.SN = sn
+        self.orientation = 1
+        self.status = False
+
+    def getType(self):
+        return self.typeRobot
+
+    def getSN(self):
+        return self.SN
+
+    def getOrientation(self):
+        return self.orientation
+
+    def getState(self):
+        return self.status
+
+    def setOrientation(self, x):
+        self.orientation = x
+
+    def setState(self, x):
+        self.status = x
+
+    def turn(self):
+        if self.orientation == 1:
+            self.orientation = 4
+        else:
+            self.orientation -= 1
+
+    def display(self):
+        print("Serial number :",self.SN)
+        print("State :",self.status)
+        print("Orientation :",self.orientation)
+        print("Type :",self.typeRobot)
+
+
+list_robot = []
+for i in range(1,5):
+    robot = CRobot("X-Man Robot", i)
+    list_robot.append(robot)
+
+for i in range(0, 4):
+    list_robot[i].display()
+    list_robot[i].setState(True)
+    list_robot[i].setOrientation(i+1)
+print("\n---\nUpdate of parameters\n---\n")
+for i in range(0,4):
+    list_robot[i].display()
+
+