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Find the Difference

Easy

Apple

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Topics:

StringsBit Manipulation

You are given two strings s and t.

String t is generated by random shuffling string s and then adding one more letter at a random position.

Return the letter that was added to t.

For example:

If s = "abcd" and t = "abcde", the output should be "e" because 'e' is the letter that was added.
If s = "" and t = "y", the output should be "y".

Write an efficient algorithm to solve this problem.

Find the Difference

Easy

Apple

2 views

Topics:

StringsBit Manipulation

You are given two strings s and t.

String t is generated by random shuffling string s and then adding one more letter at a random position.

Return the letter that was added to t.

For example:

If s = "abcd" and t = "abcde", the output should be "e" because 'e' is the letter that was added.
If s = "" and t = "y", the output should be "y".

Write an efficient algorithm to solve this problem.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the strings `s` and `t` contain characters other than lowercase English letters?
Is it guaranteed that string `t` is exactly one character longer than string `s`?
Could either `s` or `t` be null or empty strings? What should I return in those cases?
If there are multiple occurrences of the added character in `t`, should I return any one of them, or does the problem require me to handle the first such character only?
Is the added character guaranteed to be a single character, or could there be more than one extra character in `t` relative to `s`?

Brute Force Solution

Approach

The brute force approach to finding the difference involves checking every possible association between characters in the two strings. It's like a meticulous search to pinpoint the single character that doesn't match between the texts. It's not efficient, but it guarantees finding the answer.

Here's how the algorithm would work step-by-step:

Take each character from the longer string.
For each of these characters, check if that character exists in the shorter string.
If a character from the longer string is not found in the shorter string, then that is the extra character.
If you've checked all of the characters from the longer string, and still haven't found one that is unique, then there might be a bug in the approach.

Code Implementation

def find_the_difference_brute_force(longer_string, shorter_string):
    # Iterate through the longer string to find the extra character
    for char_from_longer_string in longer_string:

        # Check if the character exists in the shorter string
        if char_from_longer_string not in shorter_string:
            return char_from_longer_string

        # Remove the first occurence to handle duplicates
        shorter_string = shorter_string.replace(char_from_longer_string, '', 1)

    # If no extra character is found, something is wrong
    return None

Big(O) Analysis

Time Complexity

O(n*m) – The provided brute force approach iterates through each character of the longer string, which we can denote as length 'n'. For each of these characters, it checks if that character exists within the shorter string, which we can denote as length 'm'. In the worst-case scenario, for every character in the longer string, we might have to scan the entire shorter string to confirm its absence or presence. This results in a nested loop-like behavior where for each of the 'n' characters, 'm' comparisons might be needed. Therefore, the time complexity is proportional to n * m, or O(n*m).

Space Complexity

O(1) – The brute force approach outlined operates directly on the input strings. It does not create any auxiliary data structures like arrays, hash maps, or sets to store intermediate results or character counts. Therefore, the space complexity is constant because it only uses a few variables for iteration, regardless of the input strings' lengths. Since the space used does not depend on the input size, N, the space complexity remains O(1).

Optimal Solution

Approach

The most efficient way to find the single different character is to recognize that letters are basically numbers, and we can easily add or subtract these number representations. By summing the characters in both strings and comparing the sums, we can quickly isolate the extra character.

Here's how the algorithm would work step-by-step:

Imagine each character in both strings is a number representing its position in the alphabet.
Add up all the numbers representing characters in the first string.
Add up all the numbers representing characters in the second string.
Subtract the sum of the first string from the sum of the second string. The result will be the numeric value of the extra character.
Convert this numeric value back into a character. This character is the difference between the two strings.

Code Implementation

def find_the_difference(string1, string2):
    string1_character_sum = 0
    string2_character_sum = 0

    for char in string1:
        string1_character_sum += ord(char)

    for char in string2:
        string2_character_sum += ord(char)

    # Get the difference between the sums.
    difference_between_strings = string2_character_sum - string1_character_sum

    # Convert the ASCII difference to a character.
    different_character = chr(difference_between_strings)

    return different_character

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the first string of length n, summing the character values, which takes O(n) time. It then iterates through the second string, which has length n+1, summing its character values, also taking O(n) time. The subtraction and conversion steps are constant time operations, O(1). Therefore, the overall time complexity is dominated by the two linear traversals, resulting in O(n) + O(n) + O(1) which simplifies to O(n).

Space Complexity

O(1) – The algorithm calculates the sum of character values using accumulator variables. The plain English explanation doesn't suggest creating auxiliary data structures like lists or hash maps to store intermediate character sums. Therefore, the space required is constant, as only a fixed number of integer variables are used regardless of the input string lengths.

Edge Cases

Case	How to Handle
One string is empty or null	If s is empty, the extra character is the first character of t, or vice versa.
Strings are identical (empty difference)	The algorithm should return the single character from the longer string if the shorter string is empty after removing the matching characters.
Very large strings which might cause integer overflow in character counts	Using character counts requires using larger integer types or a different data structure like a hash map that can handle very high counts without overflow
Unicode characters in the strings	The solution should work correctly with Unicode characters which may require special handling depending on the language and string representation.
Strings are very long, causing performance issues.	Bit manipulation can be faster and uses a constant amount of memory.
The difference character appears multiple times in the longer string.	Character counts using HashMap ensures the correct occurrence amount is found
String 't' contains all characters of string 's', plus one new character. The new character is at the beginning or end.	The algorithm accounts for leading or trailing characters efficiently as it iterates through all chars and their counts.
Null input for both strings	Throw IllegalArgumentException or return null if both strings are null, depending on the API contract

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the strings `s` and `t` contain characters other than lowercase English letters?
Is it guaranteed that string `t` is exactly one character longer than string `s`?
Could either `s` or `t` be null or empty strings? What should I return in those cases?
If there are multiple occurrences of the added character in `t`, should I return any one of them, or does the problem require me to handle the first such character only?
Is the added character guaranteed to be a single character, or could there be more than one extra character in `t` relative to `s`?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Take each character from the longer string.
For each of these characters, check if that character exists in the shorter string.
If a character from the longer string is not found in the shorter string, then that is the extra character.
If you've checked all of the characters from the longer string, and still haven't found one that is unique, then there might be a bug in the approach.

Code Implementation

def find_the_difference_brute_force(longer_string, shorter_string):
    # Iterate through the longer string to find the extra character
    for char_from_longer_string in longer_string:

        # Check if the character exists in the shorter string
        if char_from_longer_string not in shorter_string:
            return char_from_longer_string

        # Remove the first occurence to handle duplicates
        shorter_string = shorter_string.replace(char_from_longer_string, '', 1)

    # If no extra character is found, something is wrong
    return None

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine each character in both strings is a number representing its position in the alphabet.
Add up all the numbers representing characters in the first string.
Add up all the numbers representing characters in the second string.
Subtract the sum of the first string from the sum of the second string. The result will be the numeric value of the extra character.
Convert this numeric value back into a character. This character is the difference between the two strings.

Code Implementation

def find_the_difference(string1, string2):
    string1_character_sum = 0
    string2_character_sum = 0

    for char in string1:
        string1_character_sum += ord(char)

    for char in string2:
        string2_character_sum += ord(char)

    # Get the difference between the sums.
    difference_between_strings = string2_character_sum - string1_character_sum

    # Convert the ASCII difference to a character.
    different_character = chr(difference_between_strings)

    return different_character

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
One string is empty or null	If s is empty, the extra character is the first character of t, or vice versa.
Strings are identical (empty difference)	The algorithm should return the single character from the longer string if the shorter string is empty after removing the matching characters.
Very large strings which might cause integer overflow in character counts	Using character counts requires using larger integer types or a different data structure like a hash map that can handle very high counts without overflow
Unicode characters in the strings	The solution should work correctly with Unicode characters which may require special handling depending on the language and string representation.
Strings are very long, causing performance issues.	Bit manipulation can be faster and uses a constant amount of memory.
The difference character appears multiple times in the longer string.	Character counts using HashMap ensures the correct occurrence amount is found
String 't' contains all characters of string 's', plus one new character. The new character is at the beginning or end.	The algorithm accounts for leading or trailing characters efficiently as it iterates through all chars and their counts.
Null input for both strings	Throw IllegalArgumentException or return null if both strings are null, depending on the API contract