Leetcode 2531: Make Number of Distinct Characters Equal
You are given two strings,
word1 and word2. In each move, you swap one character from word1 with one from word2. Determine if it is possible to make the number of distinct characters in both strings equal with exactly one swap.Problem
Approach
Steps
Complexity
Input: The input consists of two strings `word1` and `word2`.
Example: "abcc", "aab"
Constraints:
• 1 <= word1.length, word2.length <= 10^5
• Both word1 and word2 consist of only lowercase English letters.
Output: The output is a boolean value indicating whether it is possible to make the number of distinct characters in both strings equal with exactly one swap.
Example: true
Constraints:
• The output will be either `true` or `false`.
Goal: The goal is to determine if it is possible to make the number of distinct characters in both strings equal with exactly one swap.
Steps:
• Count the distinct characters in both strings `word1` and `word2`.
• For each character in `word1` and `word2`, attempt to swap characters and check if the number of distinct characters in both strings becomes equal.
• Return `true` if a swap results in equal distinct character counts, otherwise return `false`.
Goal: Ensure that the solution handles strings up to 100,000 characters long efficiently.
Steps:
• 1 <= word1.length, word2.length <= 10^5
• word1 and word2 consist of only lowercase English letters.
Assumptions:
• You are guaranteed that both strings contain only lowercase English letters.
• Input: "ac", "b"
• Explanation: In this case, swapping any characters between the two strings will not make the number of distinct characters in both strings equal.
• Input: "abcc", "aab"
• Explanation: After swapping index 2 of `word1` with index 0 of `word2`, both strings will have 3 distinct characters.
• Input: "abcd", "efgh"
• Explanation: Both strings already have 4 distinct characters, so swapping any character will not affect the result.
Approach: The approach involves counting the distinct characters in both strings and then checking if swapping any characters can equalize the distinct character count.
Observations:
• We need to efficiently check the distinct characters in both strings.
• The solution needs to handle large strings, so an O(n) solution for checking distinct characters is ideal.
Steps:
• Initialize two maps to count the frequency of each character in `word1` and `word2`.
• For each character in `word1` and `word2`, attempt a swap and check if it results in equal distinct character counts.
• If any swap results in equal distinct character counts, return `true`. If no such swap is found, return `false`.
Empty Inputs:
• The strings will not be empty, as the length of both strings is at least 1.
Large Inputs:
• Ensure that the solution handles strings with lengths up to 100,000 characters.
Special Values:
• Handle cases where all characters in one or both strings are the same.
Constraints:
• The solution should be optimized for time complexity, given the constraints of up to 100,000 characters.
bool isItPossible(string w1, string w2) {
map<char, int> ma1, ma2;
for(int x: w1)
ma1[x]++;
for(int x: w2)
ma2[x]++;
for(auto it1: ma1) {
for(auto it2: ma2) {
map<char, int> d1 = ma1, d2 = ma2;
d1[it2.first]++;
d2[it1.first]++;
d1[it1.first]--;
d2[it2.first]--;
if(d1[it1.first] == 0)
d1.erase(it1.first);
if(d2[it2.first] == 0)
d2.erase(it2.first);
if(d1.size() == d2.size()) return true;
}
}
return false;
}
1 : Function Definition
bool isItPossible(string w1, string w2) {
The function 'isItPossible' is defined, accepting two strings 'w1' and 'w2'. It will return a boolean value indicating whether it's possible to make the strings anagrams by swapping characters.
2 : Variable Initialization
map<char, int> ma1, ma2;
Two maps, 'ma1' and 'ma2', are initialized to count the occurrences of characters in 'w1' and 'w2' respectively.
3 : Loop
for(int x: w1)
A for loop begins to iterate over the characters of the string 'w1'.
4 : Map Operation
ma1[x]++;
For each character in 'w1', the corresponding count is incremented in the map 'ma1'.
5 : Loop
for(int x: w2)
A similar loop starts for the string 'w2', iterating over each character.
6 : Map Operation
ma2[x]++;
For each character in 'w2', the corresponding count is incremented in the map 'ma2'.
7 : Loop
for(auto it1: ma1) {
A for loop begins to iterate over each element of 'ma1' (character-frequency pair).
8 : Loop
for(auto it2: ma2) {
A nested for loop begins to iterate over each element of 'ma2' (character-frequency pair).
9 : Variable Assignment
map<char, int> d1 = ma1, d2 = ma2;
Two temporary maps 'd1' and 'd2' are created as copies of 'ma1' and 'ma2'. These will be modified during the comparison.
10 : Map Update
d1[it2.first]++;
The count of the character from 'it2' in 'd1' is incremented.
11 : Map Update
d2[it1.first]++;
Similarly, the count of the character from 'it1' in 'd2' is incremented.
12 : Map Update
d1[it1.first]--;
The count of the character from 'it1' in 'd1' is decremented.
13 : Map Update
d2[it2.first]--;
Similarly, the count of the character from 'it2' in 'd2' is decremented.
14 : Condition Check
if(d1[it1.first] == 0)
A check is performed to see if the frequency of the character 'it1.first' in 'd1' is zero.
15 : Map Operation
d1.erase(it1.first);
If the frequency is zero, the character 'it1.first' is removed from 'd1'.
16 : Condition Check
if(d2[it2.first] == 0)
A similar check is performed for the character 'it2.first' in 'd2'.
17 : Map Operation
d2.erase(it2.first);
If the frequency is zero, the character 'it2.first' is removed from 'd2'.
18 : Condition Check
if(d1.size() == d2.size()) return true;
If the sizes of 'd1' and 'd2' are equal, it means the strings can be rearranged to form anagrams by swapping characters. The function returns true.
19 : Return Statement
return false;
If no matching swaps were found after all iterations, the function returns false.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) where n is the length of the strings since we perform operations that are linear in nature.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the use of maps for counting characters.
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