LeetCode 2910: Minimum Number of Groups to Create a Valid Assignment
LeetCode 2910 Solution Explanation
Explanation:
To solve this problem, we can use a greedy approach. We will iterate through the balls and keep track of the count of each ball number. We will then sort the counts in descending order and assign the balls to boxes in a way that the difference in the number of balls in each box does not exceed one. We will keep track of the number of boxes used and return this count as the result.
Algorithm:
- Create a HashMap to store the count of each ball number.
- Iterate through the balls and update the count in the HashMap.
- Sort the counts of ball numbers in descending order.
- Initialize a variable
boxesto keep track of the number of boxes used. - Iterate through the sorted counts:
- Assign the current count of balls to a new box.
- Update the number of boxes used.
- Decrement the count if the count is greater than 1.
- Return the final number of boxes used.
Time Complexity: O(n log n) where n is the number of balls.
Space Complexity: O(n) for the HashMap to store counts.
:
LeetCode 2910 Solutions in Java, C++, Python
class Solution {
public int minNumberOfFrogs(String croakOfFrogs) {
Map<Character, Integer> count = new HashMap<>();
int frogs = 0, maxFrogs = 0;
for (char ch : croakOfFrogs.toCharArray()) {
if (ch == 'c') {
count.put('c', count.getOrDefault('c', 0) + 1);
maxFrogs = Math.max(maxFrogs, ++frogs);
continue;
}
char prev = (char) (ch - 1);
if (count.getOrDefault(prev, 0) == 0) {
return -1;
}
count.put(prev, count.get(prev) - 1);
if (ch == 'k') {
frogs--;
} else {
count.put(ch, count.getOrDefault(ch, 0) + 1);
}
}
return frogs == 0 ? maxFrogs : -1;
}
}Interactive Code Editor for LeetCode 2910
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