LeetCode 484: Find Permutation Solution
Master LeetCode problem 484 (Find Permutation), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation:
This problem asks us to find a permutation that is lexicographically smaller than the given string. We can solve this problem by observing that we can always construct a lexicographically smaller permutation by starting with a sorted sequence of integers from 1 to n and then reversing a subarray within this sequence.
Here is the step-by-step algorithm:
- Initialize a list containing integers from 1 to n in ascending order.
- For each character in the input string, if it is 'D', reverse a subarray starting from the current index in the list.
- Return the list as the final permutation.
Time complexity: O(n) Space complexity: O(n)
:
Solution Code
class Solution {
public int[] findPermutation(String s) {
int n = s.length() + 1;
int[] result = new int[n];
for (int i = 0; i < n; i++) {
result[i] = i + 1;
}
for (int i = 0; i < n - 1; i++) {
if (s.charAt(i) == 'D') {
int start = i;
while (i < n - 1 && s.charAt(i) == 'D') {
i++;
}
reverse(result, start, i);
}
}
return result;
}
private void reverse(int[] arr, int start, int end) {
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
}Try It Yourself
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