LeetCode 2272: Substring With Largest Variance
LeetCode 2272 Solution Explanation
Explanation
To solve this problem, we can iterate through all substrings of the given string s
and calculate the variance for each substring. For each substring, we count the occurrences of each character and find the difference between the maximum and minimum occurrence. We track the maximum variance seen so far and return it at the end.
- We use two pointers to define the substring window.
- We maintain a hashmap to keep track of the count of characters within the window.
- We slide the window to cover all possible substrings and calculate the variance for each.
Time Complexity: O(N^2) where N is the length of the input string s
.
Space Complexity: O(26) - since we are using a hashmap to store the count of characters.
LeetCode 2272 Solutions in Java, C++, Python
class Solution {
public int largestVariance(String s) {
int maxVariance = 0;
for (int i = 0; i < s.length(); i++) {
int[] count = new int[26];
for (int j = i; j < s.length(); j++) {
count[s.charAt(j) - 'a']++;
int maxCount = 0, minCount = Integer.MAX_VALUE;
for (int c : count) {
if (c > 0) {
maxCount = Math.max(maxCount, c);
minCount = Math.min(minCount, c);
}
}
maxVariance = Math.max(maxVariance, maxCount - minCount);
}
}
return maxVariance;
}
}
Interactive Code Editor for LeetCode 2272
Improve Your LeetCode 2272 Solution
Use the editor below to refine the provided solution for LeetCode 2272. Select a programming language and try the following:
- Add import statements if required.
- Optimize the code for better time or space complexity.
- Add test cases to validate edge cases and common scenarios.
- Handle error conditions or invalid inputs gracefully.
- Experiment with alternative approaches to deepen your understanding.
Click "Run Code" to execute your solution and view the output. If errors occur, check the line numbers and debug accordingly. Resize the editor by dragging its bottom edge.
Loading editor...