Certainly! Let's analyze and solve the problem step-by-step.
1. Initialization: We start with an empty dictionary `sol` which will be used to store the count of each character in the given sentence.
2. Iterating through Characters: We iterate through each character `char` in the sentence. The input sentence here is "all".
3. Ignoring Spaces: As per the condition, we ignore any spaces in the sentence. Since there are no spaces in "all", we continue to the next step.
4. Counting Characters:
- For the first character, 'a':
- Check if 'a' is already a key in the dictionary `sol`. Since it's not, we add 'a' to the dictionary with a count of 1 (`sol['a'] = 1`).
- For the second character, 'l':
- Check if 'l' is already a key in the dictionary `sol`. Since it's not, we add 'l' to the dictionary with a count of 1 (`sol['l'] = 1`).
- For the third character, 'l':
- Check if 'l' is already a key in the dictionary `sol`. This time it is already present, so we increase its count by 1 (`sol['l'] += 1`).
5. Final Dictionary: After processing all characters, the dictionary `sol` will contain the count of each character in the sentence.
The resulting dictionary will be:
```
{'a': 1, 'l': 2}
```
Thus, the final counts of the characters in the sentence "all" are:
- 'a' appears 1 time.
- 'l' appears 2 times.
This detailed explanation aligns with the resulting dictionary:
```
{'a': 1, 'l': 2}
```
Which confirms our step-by-step process.
