Leetcode 1297: Maximum Number of Occurrences of a Substring
Given a string
s, return the maximum number of occurrences of any substring that satisfies the following conditions: The number of unique characters in the substring must be less than or equal to maxLetters, and the substring length must be between minSize and maxSize inclusive.Problem
Approach
Steps
Complexity
Input: The input consists of a string `s` and three integers: `maxLetters`, `minSize`, and `maxSize`.
Example: Input: s = "abcabcabc", maxLetters = 2, minSize = 2, maxSize = 3
Constraints:
• 1 <= s.length <= 10^5
• 1 <= maxLetters <= 26
• 1 <= minSize <= maxSize <= min(26, s.length)
Output: Return the maximum number of occurrences of any substring that satisfies the given conditions.
Example: Output: 2
Constraints:
• Return an integer representing the maximum number of occurrences of a valid substring.
Goal: Find the maximum number of occurrences of a valid substring.
Steps:
• Generate all substrings of length between `minSize` and `maxSize`.
• For each substring, check if it has unique characters less than or equal to `maxLetters`.
• Count the occurrences of valid substrings and track the maximum count.
Goal: The algorithm must handle large input sizes efficiently and adhere to the given constraints.
Steps:
• 1 <= s.length <= 10^5
• 1 <= maxLetters <= 26
• 1 <= minSize <= maxSize <= min(26, s.length)
Assumptions:
• The string `s` consists of lowercase English letters only.
• The input is valid, and constraints are adhered to.
• Input: Input: s = "abcabcabc", maxLetters = 2, minSize = 2, maxSize = 3
• Explanation: The substring 'ab' appears twice in the string and satisfies the condition of having 2 unique letters and a size of 2.
• Input: Input: s = "xxxyyy", maxLetters = 1, minSize = 2, maxSize = 3
• Explanation: The substring 'xx' appears twice and satisfies the condition of having 1 unique letter and a size of 2.
Approach: The problem can be solved by generating all possible substrings of valid lengths and counting their occurrences. We need to ensure that only valid substrings are considered.
Observations:
• We need to generate substrings of different lengths between `minSize` and `maxSize`.
• The number of unique characters in each substring must be tracked to check if it is valid.
• A sliding window approach or hashing can be used to efficiently count occurrences of substrings while ensuring they meet the conditions.
Steps:
• Iterate through all possible substring lengths from `minSize` to `maxSize`.
• For each substring, track the unique characters using a set or hash map.
• If the number of unique characters is less than or equal to `maxLetters`, update the count for that substring.
• After processing all substrings, return the maximum count of any valid substring.
Empty Inputs:
• The string is non-empty, as per the problem constraints.
Large Inputs:
• The solution must efficiently handle strings up to length 10^5.
Special Values:
• If `maxLetters` is larger than the number of unique letters in the string, all substrings are valid.
Constraints:
• The solution should be optimized to run within time limits for large input sizes.
bool isPossibleDivide(vector<int>& nums, int k) {
map<int, int> cnt;
int n = nums.size();
for(int num : nums)
cnt[num]++;
for(auto it : cnt) {
int frq = it.second;
if(frq > 0)
for(int i = 0; i < k; i++) {
if(cnt[it.first + i] < frq) return false;
else cnt[it.first + i] -= frq;
}
}
return true;
}
1 : Function Definition
bool isPossibleDivide(vector<int>& nums, int k) {
This is the function header, where the function is defined with two parameters: 'nums' (a vector of integers) and 'k' (the size of each group).
2 : Map Initialization
map<int, int> cnt;
This creates a map to store the frequency of each integer in the input array 'nums'. The key is the integer, and the value is its frequency.
3 : Array Length
int n = nums.size();
This stores the size of the input array 'nums' in the variable 'n'.
4 : Iterate Through Numbers
for(int num : nums)
This loop iterates through each integer in the input array 'nums'.
5 : Update Frequency Map
cnt[num]++;
This increments the frequency of the current number in the map 'cnt'.
6 : Frequency Map Iteration
Now, we begin iterating through the frequency map to check if it is possible to form groups.
7 : Iterate Through Frequency Map
for(auto it : cnt) {
This loop iterates through each key-value pair in the frequency map, where 'it.first' is the number and 'it.second' is its frequency.
8 : Get Frequency of Current Number
int frq = it.second;
Extracts the frequency of the current number from the map.
9 : Frequency Check
if(frq > 0)
Checks if the current frequency is greater than 0 before attempting to form a group.
10 : Loop for Consecutive Numbers
for(int i = 0; i < k; i++) {
This inner loop iterates over the next 'k' consecutive numbers starting from the current number.
11 : Condition for Group Formation
Here, we check if the group of consecutive numbers can be formed.
12 : Check for Sufficient Frequency
if(cnt[it.first + i] < frq) return false;
If the frequency of any of the consecutive numbers is less than the required frequency, it returns false as the group cannot be formed.
13 : Update Frequency for Used Numbers
else cnt[it.first + i] -= frq;
If the consecutive number can be used, its frequency is decreased by 'frq'.
14 : Return Statement
return true;
If the loop completes without returning false, it means it's possible to divide the array into consecutive groups of size 'k', so we return true.
Best Case: O(n) - In the best case, no substrings need to be checked or all substrings are valid.
Average Case: O(n^2) - Substring generation and uniqueness checks dominate the complexity.
Worst Case: O(n^2) - In the worst case, we need to check all possible substrings for validity.
Description: The time complexity is O(n^2) due to the need to generate and check all substrings.
Best Case: O(n) - Space complexity remains O(n) even in the best case.
Worst Case: O(n) - Space complexity is O(n) due to storage of substring counts and temporary sets.
Description: The space complexity is O(n) due to the temporary storage of unique substrings.
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