Suppose we have a stream of characters, or we can consider a string and we have to find the first non-repeating character in the string. So, if the string is like “people”, the first letter whose occurrence is one is ‘o’. So, the index will be returned, that is 2 here. If there is no such character, then return -1.
To solve this, we will follow these steps −
create one frequency map
for each character c in string, do
if c is not in frequency, then insert it into frequency, and put value 1
otherwise increase the count in frequency
Scan the frequency map, if the value of specific key is 1, then return that key, otherwise return -1
Example
Let us see the following implementation to get better understanding −
class Solution(object):
def firstUniqChar(self, s):
"""
:type s: str
:rtype: int
"""
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
ob1 = Solution()
print(ob1.firstUniqChar("people"))
print(ob1.firstUniqChar("abaabba"))Input
"people" "abaabba"
Output
2 -1