Adding Simple Solution for Problem - 1295 - Even Numbers of Digits

Author: Vanditg

Even_Numbers_of_Digits/SimpleSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 1295
+## Problem Name: Find Numbers with Even Number of Digits
+##===================================
+#
+#Given an array nums of integers, return how many of them contain an even number of digits.
+#
+#Example 1:
+#
+#Input: nums = [12,345,2,6,7896]
+#Output: 2
+#Explanation: 
+#12 contains 2 digits (even number of digits). 
+#345 contains 3 digits (odd number of digits). 
+#2 contains 1 digit (odd number of digits). 
+#6 contains 1 digit (odd number of digits). 
+#7896 contains 4 digits (even number of digits). 
+#Therefore only 12 and 7896 contain an even number of digits.
+#
+#Example 2:
+#
+#Input: nums = [555,901,482,1771]
+#Output: 1 
+#Explanation: 
+#Only 1771 contains an even number of digits.
+class Solution:
+    def findNumbers(self, array):
+        count = 0    #Intialize count 
+        for i in range(len(array)):    #Loop through array
+            #print(array[i])
+            splitArrayDigit = [int(j) for j in str(array[i])]    #Split the array's number into each digit.
+            #print(splitArrayDigit)
+            if len(splitArrayDigit) % 2 == 0:     #Condition-check: If array's digits are even we'll enter the if condition
+                count += 1     #Update the counter by 1
+        return count    #We'll return the counter at the end of loop. 
+
+#Example:
+#array = [555, 901, 482, 1771]
+#Count = 0
+#i = 0
+#splitArrayDigit = [5, 5, 5]
+#If condition-check is false for i = 0. 
+#Count = 0
+#i = 1
+#splitArrayDigit = [9, 0, 1]
+#If condition-check is false for i = 1.
+#Count = 0
+#i = 2
+#splitArrayDigit = [4, 8, 2]
+#If condition-check is false for i = 2.
+#Count = 0
+#i = 3
+#splitArrayDigit = [1, 7, 7, 1]
+#If condition-check is true for i = 3.
+#Count = 1
+#So for this array we'll get count = 1. So even number of digits are for this example is 1.