﻿ Python: Find the first repeated character of a given string where the index of first occurrence is smallest - w3resource

# Python: Find the first repeated character of a given string where the index of first occurrence is smallest

## Python String: Exercise-54 with Solution

Write a Python program to find the first repeated character of a given string where the index of first occurrence is smallest.

Sample Solution:-

Python Code:

``````def first_repeated_char_smallest_distance(str1):
temp = {}
for ch in str1:
if ch in temp:
return ch, str1.index(ch);
else:
temp[ch] = 0
return 'None'
print(first_repeated_char_smallest_distance("abcabc"))
print(first_repeated_char_smallest_distance("abcb"))
print(first_repeated_char_smallest_distance("abcc"))
print(first_repeated_char_smallest_distance("abcxxy"))
print(first_repeated_char_smallest_distance("abc"))))
```
```

Sample Output:

```('a', 0)
('b', 1)
('c', 2)
('x', 3)
None
```

Pictorial Presentation:

Flowchart:

## Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Python Code Editor:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.

﻿

## Python: Tips of the Day

Check if a given key already exists in a dictionary:

In is the intended way to test for the existence of a key in a dict.

```d = {"key1": 10, "key2": 23}

if "key1" in d:
print("this will execute")

if "nonexistent key" in d:
print("this will not")
```

If you wanted a default, you can always use dict.get():

```d = dict()

for i in range(100):
key = i % 10
d[key] = d.get(key, 0) + 1
```

and if you wanted to always ensure a default value for any key you can either use dict.setdefault() repeatedly or defaultdict from the collections module, like so:

```from collections import defaultdict

d = defaultdict(int)

for i in range(100):
d[i % 10] += 1
```

but in general, the in keyword is the best way to do it.

Ref: https://bit.ly/2XPMRyz