You are given two strings of the same length s1
and s2
and a string baseStr
.
We say s1[i]
and s2[i]
are equivalent characters.
s1 = "abc"
and s2 = "cde"
, then we have 'a' == 'c'
, 'b' == 'd'
, and 'c' == 'e'
.Equivalent characters follow the usual rules of any equivalence relation:
'a' == 'a'
.'a' == 'b'
implies 'b' == 'a'
.'a' == 'b'
and 'b' == 'c'
implies 'a' == 'c'
.For example, given the equivalency information from s1 = "abc"
and s2 = "cde"
, "acd"
and "aab"
are equivalent strings of baseStr = "eed"
, and "aab"
is the lexicographically smallest equivalent string of baseStr
.
Return the lexicographically smallest equivalent string of baseStr
by using the equivalency information from s1
and s2
.
Example 1:
Input: s1 = "parker", s2 = "morris", baseStr = "parser"
Output: "makkek"
Example 2:
Input: s1 = "hello", s2 = "world", baseStr = "hold"
Output: "hdld"
Example 3:
Input: s1 = "leetcode", s2 = "programs", baseStr = "sourcecode"
Output: "aauaaaaada"
Constraints:
1 <= s1.length, s2.length, baseStr <= 1000
s1.length == s2.length
s1
, s2
, and baseStr
consist of lowercase English letters.How would you implement a function to find the lexicographically smallest equivalent string of baseStr
given s1
and s2
?
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:
The brute force approach to finding the lexicographically smallest equivalent string involves systematically trying every possible mapping between characters. We check each mapping to see if it produces an equivalent string and then determine the lexicographically smallest one from those equivalent strings.
Here's how the algorithm would work step-by-step:
def lexicographically_smallest_equivalent_string_brute_force(source_string_one, source_string_two, base_string): all_equivalent_strings = []
# Explore all possible character mappings, a combinatorial problem
import itertools
alphabet = 'abcdefghijklmnopqrstuvwxyz'
for character_mapping in itertools.permutations(alphabet):
character_map = dict(zip(alphabet, character_mapping))
# Create a string based on this character mapping
equivalent_string_one = ''.join(character_map[character] for character in source_string_one)
equivalent_string_two = ''.join(character_map[character] for character in source_string_two)
# Check if the strings are equivalent based on current mapping
if sorted(equivalent_string_one) == sorted(equivalent_string_two):
mapped_base_string = ''.join(character_map[character] for character in base_string)
all_equivalent_strings.append(mapped_base_string)
# If no equivalent strings are found, return original base string
if not all_equivalent_strings:
return base_string
# Find the lexicographically smallest equivalent string
smallest_equivalent_string = min(all_equivalent_strings)
return smallest_equivalent_string
The goal is to find the smallest possible string by making characters equivalent. We treat the characters as belonging to groups and merge groups to find the smallest representation for each character. This involves creating connections and finding the ultimate representative for each group, ensuring the smallest character represents the group.
Here's how the algorithm would work step-by-step:
def lexicographically_smallest_equivalent_string(string_one, string_two, base_string):
parent = {char: char for char in 'abcdefghijklmnopqrstuvwxyz'}
def find(character):
# Find the ultimate parent representative of the given character.
if parent[character] != character:
parent[character] = find(parent[character])
return parent[character]
def union(character_one, character_two):
parent_one = find(character_one)
parent_two = find(character_two)
# Ensure the smallest character becomes the parent.
if parent_one != parent_two:
if parent_one < parent_two:
parent[parent_two] = parent_one
else:
parent[parent_one] = parent_two
for index in range(len(string_one)):
union(string_one[index], string_two[index])
result = ''
for character in base_string:
# Find the equivalent representative for each character.
result += find(character)
return result
Case | How to Handle |
---|---|
Empty strings for s1, s2, or baseStr | Return baseStr immediately if s1 or s2 are empty, or if baseStr is empty, return an empty string. |
s1 and s2 have different lengths | Throw an IllegalArgumentException since equivalent characters are only defined for strings of the same length. |
s1 and s2 contain non-lowercase English letters | Treat these characters as distinct and independent, or throw an IllegalArgumentException depending on requirements. |
Chains of equivalencies create cycles (e.g., a~b, b~c, c~a) | The union-find algorithm correctly handles cycles by ensuring all nodes in a connected component point to the same root. |
baseStr contains characters not present in s1 or s2 | Characters not linked to any other character are left unchanged in the baseStr. |
Maximum string length for s1, s2 and baseStr stressing memory usage | Ensure the union-find data structure has enough memory allocated to handle the maximum number of characters (26 for lowercase letters); the solution scales linearly with input length if path compression is used in union-find. |
All characters in s1 and s2 are the same, creating maximum equivalence | The union-find algorithm correctly collapses all characters to a single equivalence class. |
Integer overflow during calculation of array indices based on character values. | Characters can be directly used as array indices if the character set is small and controlled, thus avoiding the risk of overflow when doing calculations. |