Added Leetcode easy problems

Author: das-jishu

.vscode/settings.json

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+{
+    "python.pythonPath": "C:\\Users\\Subhan\\AppData\\Local\\Microsoft\\WindowsApps\\python.exe"
+}

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

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+""" 
+# EXCEL SHEET COLUMN NUMBER
+
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+For example:
+
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28 
+    ...
+Example 1:
+
+Input: "A"
+Output: 1
+
+Example 2:
+
+Input: "AB"
+Output: 28
+
+Example 3:
+
+Input: "ZY"
+Output: 701
+ 
+Constraints:
+
+1 <= s.length <= 7
+s consists only of uppercase English letters.
+s is between "A" and "FXSHRXW". 
+"""
+
+class Solution:
+    def titleToNumber(self, s: str) -> int:
+        k = 0
+        res = 0
+        while s:
+            temp = ord(s[-1]) - 64
+            res += temp * 26 ** k
+            k += 1
+            s = s[:-1]
+            
+        return res

Leetcode/easy/factorial-trailing-zeros.py

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+""" 
+# FACTORIAL TRAILING ZEROS
+
+Given an integer n, return the number of trailing zeroes in n!.
+
+Follow up: Could you write a solution that works in logarithmic time complexity?
+
+Example 1:
+
+Input: n = 3
+Output: 0
+Explanation: 3! = 6, no trailing zero.
+
+Example 2:
+
+Input: n = 5
+Output: 1
+Explanation: 5! = 120, one trailing zero.
+
+Example 3:
+
+Input: n = 0
+Output: 0
+
+Constraints:
+
+0 <= n <= 104 
+"""
+
+class Solution:
+    def trailingZeroes(self, n: int) -> int:
+        c = 0
+        while n >= 5:
+            c += n // 5
+            n = n // 5
+            
+        return c

Leetcode/easy/reverse-bits.py

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+""" 
+# REVERSE BITS
+
+Reverse bits of a given 32 bits unsigned integer.
+
+Note:
+
+Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
+In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.
+Follow up:
+
+If this function is called many times, how would you optimize it?
+
+Example 1:
+
+Input: n = 00000010100101000001111010011100
+Output:    964176192 (00111001011110000010100101000000)
+Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
+Example 2:
+
+Input: n = 11111111111111111111111111111101
+Output:   3221225471 (10111111111111111111111111111111)
+Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.
+ 
+Constraints:
+
+The input must be a binary string of length 32 
+"""
+
+class Solution:
+    def reverseBits(self, n: int) -> int:      
+      l = 1 << 31
+      r = 1 
+      
+      # Continue until the bits from each bitmask meet each other.
+      while l > r:
+        # Check if current bits from the left and right are 1 or 0.
+        l_bit = 1 if n & l else 0
+        r_bit = 1 if n & r else 0
+        
+        # If two bits are different, we need to inverse them.
+        if l_bit != r_bit:
+          # Invert the bits in original number usin XOR.
+          n = n ^ l ^ r
+        
+        # Shiftin left bitmask to the right and the right one to the left.
+        l = l >> 1
+        r = r << 1
+        
+      return n

test-panda.py

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+
+import pandas as pd 
+  
+# Define a dictionary containing employee data 
+data = {'Name':['Jai', 'Princi', 'Gaurav', 'Anuj'], 
+        'Age':[27, 24, 22, 32], 
+        'Address':['Delhi', 'Kanpur', 'Allahabad', 'Kannauj'], 
+        'Qualification':['Msc', 'MA', 'MCA', 'Phd']} 
+  
+# Convert the dictionary into DataFrame  
+df = pd.DataFrame(data) 
+  
+# select all rows  
+# and second to fourth column 
+df[df.columns[1:4]]