Leetcode 3138: Minimum Length of Anagram Concatenation
You are given a string
s, which is a concatenation of several anagrams of some string t. Your task is to find the minimum possible length of the string t. An anagram is formed by rearranging the letters of a string. For example, ‘abc’ and ‘cab’ are anagrams of each other. The string t is the original string that has been rearranged multiple times to form s.Problem
Approach
Steps
Complexity
Input: The input consists of a string `s`.
Example: Example 1:
Input: s = "abcabc"
Output: 3
Constraints:
• 1 <= s.length <= 10^5
• s consists only of lowercase English letters.
Output: Return the minimum length of the string `t` that can be used to form the concatenated anagrams in `s`.
Example: Example 1:
Input: s = "abcabc"
Output: 3
Constraints:
• The result must be an integer representing the minimum length of the string `t`.
Goal: To find the minimum possible length of `t`, we need to determine the greatest common divisor (GCD) of the frequency of characters in the string.
Steps:
• Count the frequency of each character in the string `s`.
• Find the GCD of the frequency counts of the characters in `s`.
• The length of `t` will be the length of `s` divided by the GCD of the frequencies.
Goal: The constraints for the input string are defined below.
Steps:
• 1 <= s.length <= 10^5
• s consists only of lowercase English letters.
Assumptions:
• The string `s` is guaranteed to be a concatenation of anagrams.
• The string `s` contains only lowercase English letters.
• Input: Example 1:
• Explanation: For the string 'abcabc', the frequency of each character is 2. The greatest common divisor (GCD) of 2 is 2, so the minimum length of `t` is 3 (which is the length of `s` divided by 2).
• Input: Example 2:
• Explanation: For the string 'abcdabcd', the frequency of each character is 2. The GCD of 2 is 2, so the minimum length of `t` is 4.
Approach: To solve this problem, we will use a frequency count of the characters in the string and apply the GCD method to determine the minimum possible length of `t`.
Observations:
• The problem is asking for the smallest possible string `t` that, when rearranged, can form the input string `s` through multiple concatenations.
• By counting the frequency of characters and applying the GCD, we can identify the smallest repeating unit in `s`.
Steps:
• Count the frequency of each character in the string `s` using a hashmap.
• Compute the GCD of all the frequency values.
• Return the length of the string `s` divided by the GCD.
Empty Inputs:
• The input string is always guaranteed to have at least one character.
Large Inputs:
• Ensure that the solution can handle input strings of length up to 10^5 efficiently.
Special Values:
• The solution must handle cases where all characters in the string are the same.
Constraints:
• Make sure to handle cases where characters have varying frequencies.
int minAnagramLength(string s) {
// 12, 4
// aaxxbb
// bbxxaa
int n = s.size();
map<char, int> mp;
for(char x: s)
mp[x]++;
int mn = mp[s[0]];
for(auto it: mp)
mn = __gcd(mn, it.second);
return n / mn;
}
1 : Function Definition
int minAnagramLength(string s) {
Defines the function `minAnagramLength`, which takes a string `s` as input and returns an integer representing the minimum length of an anagram that can be formed from `s`.
2 : Variable Initialization
int n = s.size();
Initializes the variable `n` to store the length of the string `s`.
3 : Map Initialization
map<char, int> mp;
Initializes a map `mp` to store the frequency of each character in the string `s`.
4 : For Loop
for(char x: s)
Starts a loop that iterates through each character in the string `s`.
5 : Map Update
mp[x]++;
For each character `x` in the string, it increments its count in the map `mp`.
6 : Variable Initialization
int mn = mp[s[0]];
Initializes `mn` to the frequency of the first character in the string `s` to start finding the GCD of character frequencies.
7 : For Loop
for(auto it: mp)
Starts a loop to iterate through the map `mp`, which contains the frequencies of each character in the string.
8 : GCD Calculation
mn = __gcd(mn, it.second);
For each character frequency in the map, it calculates the GCD of `mn` and the current character's frequency (`it.second`). This finds the greatest common divisor of all character frequencies.
9 : Return Statement
return n / mn;
Returns the result of dividing the string length `n` by the GCD `mn`, which gives the minimum length of the anagram that can be made from the string.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear, where `n` is the length of the input string `s`, since we are only iterating over the string once and performing constant time operations for each character.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) for storing the frequency counts of the characters in the string.
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