first-unique-character.py

# Given a string, find the first non-repeating character in it and return its index. 
# If it doesn't exist, return -1. # Note: all the input strings are already lowercase.

#Approach 1
def solution(s):
    frequency = {}
    for i in s:
        if i not in frequency:
            frequency[i] = 1
        else:
            frequency[i] +=1
    for i in range(len(s)):
        if frequency[s[i]] == 1:
            return i
    return -1

print(solution('alphabet'))
print(solution('barbados'))
print(solution('crunchy'))

print('###')

#Approach 2
import collections

def solution(s):
    # build hash map : character and how often it appears
    count = collections.Counter(s) # <-- gives back a dictionary with words occurrence count 
                                         #Counter({'l': 1, 'e': 3, 't': 1, 'c': 1, 'o': 1, 'd': 1})
    # find the index
    for idx, ch in enumerate(s):
        if count[ch] == 1:
            return idx     
    return -1

print(solution('alphabet'))
print(solution('barbados'))
print(solution('crunchy'))