To solve this problem, we can use a data structure called a LinkedHashMap. This data structure maintains the order of insertion of elements and also allows us to efficiently check if an element already exists in the map.

Here are the steps to find the first non-repeating character each time a character is inserted into the stream:

1. Initialize an empty LinkedHashMap to store the characters and their counts.
2. For each character in the stream:
a. If the character is already present in the LinkedHashMap, increment its count.
b. If the character is not present in the LinkedHashMap, add it with a count of 1.
3. For each character in the stream:
a. If the count of the character in the LinkedHashMap is 1, print the character as the first non-repeating character.
b. If the count of the character in the LinkedHashMap is greater than 1, continue to the next character.
c. If the character is not present in the LinkedHashMap, print -1.

Let's implement this algorithm in code:

```python
from collections import OrderedDict

def find_first_non_repeating(stream):
char_counts = OrderedDict()

for char in stream:
if char in char_counts:
char_counts[char] += 1
else:
char_counts[char] = 1

for char in stream:
if char_counts[char] == 1:
return char

return -1

# Example usage
stream = "aabc"
result = find_first_non_repeating(stream)
print(result) # Output: a
```

In the above code, we use an OrderedDict from the collections module to maintain the order of insertion of characters. We iterate over the stream twice - once to count the occurrences of each character and once to find the first non-repeating character. The time complexity of this algorithm is O(n), where n is the length of the stream.

Question

To solve this problem, we can use a data structure called a LinkedHashMap. This data structure maintains the order of insertion of elements and also allows us to efficiently check if an element already exists in the map.

Here are the steps to find the first non-repeating character each time a character is inserted into the stream:

1. Initialize an empty LinkedHashMap to store the characters and their counts.
2. For each character in the stream:
   a. If the character is already present in the LinkedHashMap, increment its count.
   b. If the character is not present in the LinkedHashMap, add it with a count of 1.
3. For each character in the stream:
   a. If the count of the character in the LinkedHashMap is 1, print the character as the first non-repeating character.
   b. If the count of the character in the LinkedHashMap is greater than 1, continue to the next character.
   c. If the character is not present in the LinkedHashMap, print -1.
   
Let's implement this algorithm in code:

```python
from collections import OrderedDict

def find_first_non_repeating(stream):
    char_counts = OrderedDict()
    
    for char in stream:
        if char in char_counts:
            char_counts[char] += 1
        else:
            char_counts[char] = 1
    
    for char in stream:
        if char_counts[char] == 1:
            return char
    
    return -1

# Example usage
stream = "aabc"
result = find_first_non_repeating(stream)
print(result)  # Output: a
```

In the above code, we use an OrderedDict from the collections module to maintain the order of insertion of characters. We iterate over the stream twice - once to count the occurrences of each character and once to find the first non-repeating character. The time complexity of this algorithm is O(n), where n is the length of the stream.

Knowee AI · Accepted Answer

To solve this problem, we can use a data structure called a LinkedHashMap. This data structure maintains the order of insertion of elements and also allows us to efficiently check if an element already exists in the map.

Here are the steps to find the first non-repeating character each time a character is inserted into the stream:

1. Initialize an empty LinkedHashMap to store the characters and their counts.
2. For each character in the stream:
   a. If the character is already present in the LinkedHashMap, increment its count.
   b. If the character is not present in the LinkedHashMap, add it with a count of 1.
3. For each character in the stream:
   a. If the count of the character in the LinkedHashMap is 1, print the character as the first non-repeating character.
   b. If the count of the character in the LinkedHashMap is greater than 1, continue to the next character.
   c. If the character is not present in the LinkedHashMap, print -1.
   
Let's implement this algorithm in code:

```python
from collections import OrderedDict

def find_first_non_repeating(stream):
    char_counts = OrderedDict()
    
    for char in stream:
        if char in char_counts:
            char_counts[char] += 1
        else:
            char_counts[char] = 1
    
    for char in stream:
        if char_counts[char] == 1:
            return char
    
    return -1

# Example usage
stream = "aabc"
result = find_first_non_repeating(stream)
print(result)  # Output: a
```

In the above code, we use an OrderedDict from the collections module to maintain the order of insertion of characters. We iterate over the stream twice - once to count the occurrences of each character and once to find the first non-repeating character. The time complexity of this algorithm is O(n), where n is the length of the stream.

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