713590. Minimum Days
Minimum Days Slug: minimum-days--170646 Difficulty: Medium Id: 713590 Topic Tags: Strings, permutation, Data Structures Company Tags: None
Summary
The Minimum Days problem involves finding the minimum number of days required to make a given string of characters unique. This is achieved by rearranging the characters in the string to form all possible permutations and then counting the number of days it takes for each character to appear once. The key concepts involved are permutation, strings, and data structures.
Detailed Explanation
The approach to solving this problem involves generating all possible permutations of the given string using a backtracking algorithm. Each permutation is checked to see if it contains any duplicate characters. If no duplicates are found, then the number of days required for that permutation to appear in the sequence increases by one. The overall minimum number of days required is the maximum number of days found across all permutations.
Here's a step-by-step breakdown of the solution:
- Initialize an integer variable
min_daysto store the minimum number of days required. - Define a recursive function
permutationsthat generates all possible permutations of the given string and checks for duplicates. - Inside the
permutationsfunction:- For each character in the permutation, check if it's already been seen before.
- If not, increment a counter
days.
- Compare the current
dayswith the storedmin_days. Updatemin_daysif necessary. - Return the minimum number of days required.
Time complexity: O(n!), where n is the length of the string, because we're generating all possible permutations of the string. Space complexity: O(n), as we need to store the current permutation and other variables.
Optimized Solutions
Java
import java.util.*;
public class MinimumDays {
public static int minDays(String str) {
int n = str.length();
char[] chars = str.toCharArray();
Arrays.sort(chars);
int days = 1;
for (int i = 0; i < n; i++) {
if (chars[i] == chars[i - 1]) {
continue;
}
days++;
}
return days;
}
}