letter = input()  phrase = input() count = 0 for i in phrase:     if i == letter:         count = count + 1; if count != 1:     print(str(count) + " " + letter + "'s") else:     print(str(count) + " " + letter )

I need help with this python computer science DEEP BREADTH SEARCH question. I have the layout of the code but I need help filling in the parts that say "pass" and I want you to take a picture of your output and make sure it matches the sample run below.   Find a path (not the best path, but just any path) from A to Z.  You see that the green path is not the shortest but it does let us navigate from start to finish.  (The picture below) First, we need to know A and Z, our starting and ending points.  We'll pass these into our function.   I'm going to use a dictionary to represent this graph.  Each node (vertex, circle) will have a name, in this case "A" and "Z" were the names of the nodes, but in the generated maps I'm going to use "Node 1", "Node 2", "Node 3", etc.     Here is an example web_map web_map = { 'Node 1': ['Node 3', 'Node 2'],  'Node 2': ['Node 1', 'Node 4'],  'Node 3': ['Node 1'],  'Node 4': ['Node 2'] } Node 1 is connected to 2 and 3 for instance, and then also note that…

Can this python code be done without using the isdigit() function. Please try to change the following code in the picture so it doesn't use isdigit().

Write a program that takes in four positive integers and outputs the number of odd numbers. (Hint: use the modulo operator to determine if a number is odd) Ex: If the input is: 1 2 3 4 the output is: 2 Below is my code so far and I'm getting back unexpected EOF while parsing. I can't figure out what I'm doing wrong. odd_number=[]number_1=int(input())number_2=int(input())number_3=int(input())number_4=int(input())if number_1%2!=0:    odd_number.append(number_1)else:    passif number_2%2!=0:    odd_number.append(number_2)else:    passif number_3%2!=0:    odd_number.append(number_3)else:    passif number_4%2!=0:    odd_number.append(number_4)else:    passprint(len(odd_number)