Make Number of Distinct Characters Equal
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Strings
You are given two 0-indexed strings word1 and word2.
A move consists of choosing two indices i and j such that 0 <= i < word1.length and 0 <= j < word2.length and swapping word1[i] with word2[j].
Return true if it is possible to get the number of distinct characters in word1 and word2 to be equal with exactly one move. Return false otherwise.
Example 1:
Input: word1 = "ac", word2 = "b" Output: false Explanation: Any pair of swaps would yield two distinct characters in the first string, and one in the second string.
Example 2:
Input: word1 = "abcc", word2 = "aab" Output: true Explanation: We swap index 2 of the first string with index 0 of the second string. The resulting strings are word1 = "abac" and word2 = "cab", which both have 3 distinct characters.
Example 3:
Input: word1 = "abcde", word2 = "fghij" Output: true Explanation: Both resulting strings will have 5 distinct characters, regardless of which indices we swap.
Constraints:
1 <= word1.length, word2.length <= 105word1andword2consist of only lowercase English letters.
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 you clarify the data type and range of values within the input strings? Are we dealing with only lowercase English letters, or are there other characters?
- What should I return if it's impossible to make the number of distinct characters equal by performing the allowed operations?
- Are the input strings guaranteed to be non-empty, or should I handle the case where either or both strings are empty?
- By 'make number of distinct characters equal', do we want to minimize changes? If multiple solutions exist, is one considered better than another?
- Could you define what 'distinct characters' exactly refers to? Are we considering case sensitivity, for instance?
Brute Force Solution
Approach
The brute force approach is all about trying absolutely everything. We'll explore every possible character swap between the two input strings, counting the distinct characters in each string after each swap to see if we've achieved our goal.
Here's how the algorithm would work step-by-step:
- Take the first character in the first string.
- Swap it with the first character in the second string.
- Count how many different characters are now in the first string and how many different characters are now in the second string.
- If the number of different characters in both strings is now equal, we have found a valid solution and we can stop.
- If not, put the original characters back where they were to undo the swap.
- Now, swap the first character in the first string with the second character in the second string, and repeat the counting and checking process.
- Keep doing this, swapping the first character in the first string with every single character in the second string, one at a time.
- Once we've tried all possible swaps with the first character, move on to the second character in the first string and repeat the entire process, swapping it with every character in the second string.
- Continue until we have tried swapping every character in the first string with every character in the second string.
- If after trying all possible swaps, we never found a case where the number of different characters in each string was equal, then we know there is no solution.
Code Implementation
def make_number_of_distinct_characters_equal(first_string, second_string):
first_string_list = list(first_string)
second_string_list = list(second_string)
for first_string_index in range(len(first_string_list)):
for second_string_index in range(len(second_string_list)):
# Try swapping characters between the two strings.
original_first_char = first_string_list[first_string_index]
original_second_char = second_string_list[second_string_index]
first_string_list[first_string_index] = original_second_char
second_string_list[second_string_index] = original_first_char
first_string_after_swap = "".join(first_string_list)
second_string_after_swap = "".join(second_string_list)
first_distinct_count = len(set(first_string_after_swap))
second_distinct_count = len(set(second_string_after_swap))
# Check if the swap resulted in equal distinct characters.
if first_distinct_count == second_distinct_count:
return True
# Revert the swap to restore the original strings
first_string_list[first_string_index] = original_first_char
second_string_list[second_string_index] = original_second_char
# If no swap resulted in equal distinct characters return false
return FalseBig(O) Analysis
Time Complexity
O(n*m) – The algorithm iterates through each character of the first string (let's say it has size n) and attempts to swap it with every character of the second string (let's say it has size m). For each swap, the number of distinct characters in both strings is calculated, which takes O(n+m) time given the string sizes. Therefore, the total time complexity is O(n * m * (n+m)). However the problem description does not discuss how to count distinct elements, only how swaps occur. Therefore it is simply O(n*m).
Space Complexity
O(N) – The algorithm's space complexity stems from calculating the number of distinct characters in each string after every swap. To count distinct characters, a set or a similar data structure capable of storing unique elements of the strings is implicitly used. The size of this set can grow up to N, where N is the length of the string (since, in the worst case, all characters are distinct). Therefore, the auxiliary space required is O(N).
Optimal Solution
Approach
The goal is to determine if we can make the number of unique characters in two strings equal by swapping one character between them. We'll count the frequencies of characters in each string and then efficiently check if a single swap can equalize the number of distinct characters.
Here's how the algorithm would work step-by-step:
- First, count how many times each character appears in the first string and the second string separately.
- Next, find out how many different characters are in each string.
- Now, consider every possible character swap: try swapping each character from the first string with each character from the second string.
- For each potential swap, temporarily update the counts in both strings to simulate the swap.
- After each simulated swap, recalculate the number of distinct characters in each string based on the updated counts.
- If, after a swap, the two strings have the same number of distinct characters, then we know it is possible to make the number of distinct characters equal and we can stop checking further.
- If we have tried all possible swaps and haven't found a swap that makes the number of distinct characters equal, then it's not possible.
Code Implementation
def make_number_of_distinct_characters_equal(first_string, second_string):
first_string_character_counts = {}
second_string_character_counts = {}
for char in first_string:
first_string_character_counts[char] = first_string_character_counts.get(char, 0) + 1
for char in second_string:
second_string_character_counts[char] = second_string_character_counts.get(char, 0) + 1
first_string_distinct_character_count = len(first_string_character_counts)
second_string_distinct_character_count = len(second_string_character_counts)
for first_string_char in set(first_string):
for second_string_char in set(second_string):
# Simulate swapping characters between strings
first_string_character_counts[first_string_char] -= 1
if first_string_character_counts[first_string_char] == 0:
first_string_distinct_character_count -= 1
first_string_character_counts[second_string_char] = first_string_character_counts.get(second_string_char, 0) + 1
if first_string_character_counts[second_string_char] == 1:
first_string_distinct_character_count += 1
second_string_character_counts[second_string_char] -= 1
if second_string_character_counts[second_string_char] == 0:
second_string_distinct_character_count -= 1
second_string_character_counts[first_string_char] = second_string_character_counts.get(first_string_char, 0) + 1
if second_string_character_counts[first_string_char] == 1:
second_string_distinct_character_count += 1
# If the distinct character counts are equal, return True
if first_string_distinct_character_count == second_string_distinct_character_count:
return True
# Undo the swap to prepare for the next iteration
first_string_character_counts[first_string_char] += 1
if first_string_character_counts[first_string_char] == 1:
first_string_distinct_character_count += 1
if first_string_character_counts[second_string_char] == 1:
first_string_distinct_character_count -= 1
first_string_character_counts[second_string_char] -= 1
if first_string_character_counts[second_string_char] == 0:
del first_string_character_counts[second_string_char]
second_string_character_counts[second_string_char] += 1
if second_string_character_counts[second_string_char] == 1:
second_string_distinct_character_count += 1
if second_string_character_counts[first_string_char] == 1:
second_string_distinct_character_count -= 1
second_string_character_counts[first_string_char] -= 1
if second_string_character_counts[first_string_char] == 0:
del second_string_character_counts[first_string_char]
# If no swap resulted in equal distinct character counts, return False
return FalseBig(O) Analysis
Time Complexity
O(n*m) – Let n be the length of the first string and m be the length of the second string. The solution iterates through all possible pairs of characters, one from the first string and one from the second string, simulating a swap for each pair. For each possible swap, the number of distinct characters is recalculated in both strings, which takes a constant amount of time because the size of character set is constant (26 for lowercase alphabets). Therefore, since we iterate through all pairs from the first string (n) to the second string (m), the total number of operations is proportional to n * m, which gives a time complexity of O(n*m).
Space Complexity
O(1) – The algorithm primarily uses frequency counters for each string. Assuming a fixed character set (e.g., ASCII), the size of these counters is constant, independent of the input string lengths. Specifically, we need space to store character counts, which is determined by the character set size rather than the string length, hence constant. The number of distinct characters and a few temporary variables also occupy constant space. Therefore, the auxiliary space used is O(1).
Edge Cases
| Case | How to Handle |
|---|---|
| Empty string for either input string | Return true if both strings are empty, otherwise return false as removing a character would never make equal distinct characters |
| Strings with only one character each | Return true if the characters are different, and false if they are the same. |
| Strings are identical and have only one distinct character | Return false because removing a character would result in an empty string or a single character string with one distinct character, while the other would still be a single character string with one distinct character. |
| One string has one distinct character and the other has two | Check if removing a character from the string with two distinct characters results in the same character as the string with only one distinct character. |
| Strings with very long lengths, potentially exceeding typical string limits | Ensure algorithm has linear time complexity based on string length, avoiding quadratic or exponential complexities. |
| Strings with a very large number of distinct characters | The algorithm should still work efficiently with many distinct characters; hashing or counting occurrences is recommended. |
| Strings containing unicode or special characters. | The code needs to be able to handle unicode characters correctly when counting the distinct characters. |
| No valid solution exists (no characters can be removed to make the number of distinct characters equal) | Return false after checking all possible removals. |
Empty string for either input string
How to Handle:
Return true if both strings are empty, otherwise return false as removing a character would never make equal distinct characters
Strings with only one character each
How to Handle:
Return true if the characters are different, and false if they are the same.
Strings are identical and have only one distinct character
How to Handle:
Return false because removing a character would result in an empty string or a single character string with one distinct character, while the other would still be a single character string with one distinct character.
One string has one distinct character and the other has two
How to Handle:
Check if removing a character from the string with two distinct characters results in the same character as the string with only one distinct character.
Strings with very long lengths, potentially exceeding typical string limits
How to Handle:
Ensure algorithm has linear time complexity based on string length, avoiding quadratic or exponential complexities.
Strings with a very large number of distinct characters
How to Handle:
The algorithm should still work efficiently with many distinct characters; hashing or counting occurrences is recommended.
Strings containing unicode or special characters.
How to Handle:
The code needs to be able to handle unicode characters correctly when counting the distinct characters.
No valid solution exists (no characters can be removed to make the number of distinct characters equal)
How to Handle:
Return false after checking all possible removals.
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