484. Find Permutation

Problem Description

In this LeetCode problem, we are given a string s that represents a sequence of I (increase) and D (decrease) operations. This string s maps to a permutation perm of length n + 1, where n is the length of s. The characters in the string s determine whether the consecutive numbers in the permutation perm are in increasing (I) or decreasing (D) order.

Our goal is to reconstruct the permutation perm from the given sequence s such that the permutation is the lexicographically smallest possible. Lexicographically smallest means that if we treat each permutation as a number, we want the smallest possible number that can be formed while still satisfying the pattern of I and D from the string s.

For example, if the string s is IID, then a possible permutation that fits the pattern is 12345 where perm[0] < perm[1], perm[1] < perm[2] and perm[3] > perm[4]. However, 12345 may not be the lexicographically smallest permutation, and our task is to find the smallest one.

Intuition

The intuition behind the solution comes from the fact that we want to make the permutation as small as possible from left to right. This means that whenever we encounter an I in the string s, we should choose the smallest available number at that point to keep the permutation small. Conversely, when we encounter a D, we want to defer choosing the numbers until we reach the end of the consecutive sequence of Ds, and then fill in the numbers in reverse. By doing so, we ensure that the number before the D sequence is greater than the numbers after it, while still keeping those numbers as small as possible.

The process to arrive at this solution involves the following steps:

Fill in the array ans with the numbers from 1 to n+1 in ascending order. Assuming initially that all the following characters are 'I', this would be the lexicographically smallest permutation.
Traverse the string s and when we see a D, identify the consecutive sequence of Ds. We note where this sequence ends.
Reverse the sub-array ans[i : j + 1] where i is the start of the D sequence, and j is the index of the last D in the sequence. This places the larger numbers before the smaller numbers, which is required by the D sequence.
Continue this process until we have gone through the entire string s.

By following these steps, we construct the lexicographically smallest permutation that satisfies the given sequence s.

Learn more about Stack and Greedy patterns.

Solution Approach

The implementation of the solution follows the intuition closely, using mostly basic data structures and control flow patterns available in Python.

Here’s a step-by-step breakdown of the algorithm implemented by the Solution class:

Initialize a list called ans that contains the numbers from 1 to n + 1 (inclusive). Because the sequence must contain each number from 1 to n + 1 exactly once, this fills in the default ascending order for the "I" scenario.
Create a pointer i that will traverse the string s from the beginning.
While i is less than the length of the string s, look for consecutive "D"s. This is achieved by initializing another pointer j to i, which moves forward as long as it finds "D".
For each sequence of "D"s, reverse the corresponding subsequence in ans. The slice ans[i : j + 1] represents the numbers that need to be in descending order to satisfy this sequence of "D"s. By reversing using the [::-1] slice, we arrange them correctly while keeping them as small as possible.
Update the pointer i to continue from the end of the handled "D" sequence. The max(i + 1, j) ensures that if j has not moved (indicating that there were no "D"s), i continues to the next character. If j has moved, i skips over the "D" sequence.
After the completion of the loop, ans now represents the lexicographically smallest permutation in accordance with the pattern given by s.

This algorithm leverages Python's list indexing and slicing capabilities to reverse subarrays efficiently. The in-place reversal of subarrays helps to maintain the overall lexicographic order by ensuring that the smallest values are placed after sequences of "D"s.

Once the list ans has been fully traversed and modified, it is returned as the lexicographically smallest permutation that follows the increase-decrease pattern dictated by string s. By using this straightforward implementation, there is no need for complex data structures or additional space beyond the initial list to store the permutation. The in-place operations ensure that the space complexity stays constrained to O(n), where n is the number of elements in perm.

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Example Walkthrough

Let's illustrate the solution approach with a simple example where the string s is DID.

First, we need to initialize a list ans that contains numbers from 1 to n + 1. Since the length of s is 3, n + 1 equals 4. Therefore, we initialize ans to [1, 2, 3, 4].
We then create a pointer i starting at the beginning of the string s. Initially, i = 0.
Now we start traversing the string s. At s[0], we have D, which signifies that ans[0] should be greater than ans[1]. We find the next sequence of "I" or the end of the string to determine the range to reverse. In this case, the next character at s[1] is I, signaling the end of our D sequence. So, we have a D sequence from i = 0 to j = 0.
We reverse ans[i : j + 1], but in this case, since i and j are the same, the list remains [1, 2, 3, 4].
We move to the next character in the string s and increment i to max(i + 1, j), which makes i = 1.
At s[1], we have I, so we leave ans as it is because the permutation should remain in ascending order for I.
We move i to 2 because there was no D sequence.
At s[2], we have D again, so we look for the next I or the end of the string. The end of the string comes next, so our D sequence is from i = 2 to j = 2.
We reverse ans[i : j + 1], which means reversing the slice [3, 4]. After the reversal, ans becomes [1, 2, 4, 3].
Iteration is complete as we reach the end of s.

Now, the list ans represents [1, 2, 4, 3], which is the lexicographically smallest permutation following the DID pattern of the string s. Here’s a step by step representation of ans after each iteration:

Initial ans: [1, 2, 3, 4]
After handling D at s[0]: [1, 2, 3, 4] (no change since it's a single D)
After handling I at s[1]: [1, 2, 3, 4] (no change since ans is already ascending)
After handling D at s[2]: [1, 2, 4, 3] (the last two numbers are reversed)

The final result is the permutation [1, 2, 4, 3], which satisfies the original pattern and is lexicographically the smallest possible.

Solution Implementation

1from typing import List
2
3class Solution:
4    def findPermutation(self, s: str) -> List[int]:
5        # Length of the input pattern
6        pattern_length = len(s)
7      
8        # Create an initial list of integers from 1 to n+1
9        # For the pattern 'DI', the initial list will be [1, 2, 3]
10        permutation = list(range(1, pattern_length + 2))
11      
12        # Start iterating over the pattern
13        index = 0
14        while index < pattern_length:
15            # Find the end of the current 'D' sequence 
16            sequence_end = index
17            while sequence_end < pattern_length and s[sequence_end] == 'D':
18                sequence_end += 1
19          
20            # Reverse the sub-list corresponding to the 'D' sequence found
21            # This is done because numbers must be in descending order for 'D'
22            permutation[index : sequence_end + 1] = permutation[index : sequence_end + 1][::-1]
23          
24            # Move to next starting point, one past this 'D' sequence or increment by one
25            index = max(index + 1, sequence_end)
26      
27        # Return the modified list as the resulting permutation
28        return permutation
29
30# Example usage:
31# solution = Solution()
32# print(solution.findPermutation("DID"))
33

1class Solution {
2  
3    // This function produces the permutation of numbers based on the given pattern string.
4    public int[] findPermutation(String pattern) {
5        int length = pattern.length();
6        int[] result = new int[length + 1];
7      
8        // Initialize the result array with natural numbers starting from 1 to n+1
9        for (int i = 0; i <= length; ++i) {
10            result[i] = i + 1;
11        }
12      
13        int index = 0;
14        // Traverse through the pattern string
15        while (index < length) {
16            int startIndex = index;
17            // Find the contiguous sequence of 'D's
18            while (startIndex < length && pattern.charAt(startIndex) == 'D') {
19                startIndex++;
20            }
21            // Reverse the sequence to fulfill the 'D' requirement in permutation
22            reverse(result, index, startIndex);
23            // Advance the index to the position after the reversed section or move it at least one step forward.
24            index = Math.max(index + 1, startIndex);
25        }
26      
27        return result;
28    }
29
30    // This function reverses the elements in the array between indices i and j.
31    private void reverse(int[] array, int start, int end) {
32        // Note end is decremented to swap the elements inclusively
33        for (int left = start, right = end - 1; left < right; ++left, --right) {
34            // Swap elements at indices left and right
35            int temp = array[left];
36            array[left] = array[right];
37            array[right] = temp;
38        }
39    }
40}
41

1#include <vector>
2#include <algorithm>
3#include <numeric>
4
5class Solution {
6public:
7    // Method to find the permutation according to the input sequence.
8    vector<int> findPermutation(string sequence) {
9        int n = sequence.size(); // Size of the input string
10        // Create a vector to store the answer, initializing 
11        // it with an increasing sequence from 1 to n+1.
12        vector<int> permutation(n + 1);
13        iota(permutation.begin(), permutation.end(), 1);
14      
15        int currentIndex = 0; // Start from the beginning of the string.
16        // Iterate through the entire sequence.
17        while (currentIndex < n) {
18            int nextIndex = currentIndex;
19            // Find the sequence of 'D's in the input string.
20            while (nextIndex < n && sequence[nextIndex] == 'D') {
21                ++nextIndex; // Move to the next index if it's a 'D'.
22            }
23            // Reverse the subvector from the start of 'D's to just past the last 'D'.
24            reverse(permutation.begin() + currentIndex, permutation.begin() + nextIndex + 1);
25            // Move to the index after the sequence of 'D's or increment by one if no 'D's were found.
26            currentIndex = max(currentIndex + 1, nextIndex);
27        }
28      
29        return permutation; // Return the resulting permutation.
30    }
31};
32

1// Helper function to generate an increasing sequence array from 1 to n
2function iota(n: number): number[] {
3  return Array.from({ length: n }, (_, index) => index + 1);
4}
5
6// Helper function to reverse a subarray from start to end indices
7function reverse(array: number[], start: number, end: number): void {
8  while (start < end) {
9    [array[start], array[end]] = [array[end], array[start]]; // swap elements
10    start++;
11    end--;
12  }
13}
14
15// Function to find the permutation as per the input sequence
16function findPermutation(sequence: string): number[] {
17  const n = sequence.length; // Get the size of the input sequence
18  // Initialize the permutation array as an increasing sequence from 1 to n+1
19  let permutation = iota(n + 1);
20
21  let currentIndex = 0; // Start from the beginning of the sequence
22
23  // Iterate through the entire sequence
24  while (currentIndex < n) {
25    let nextIndex = currentIndex;
26    // Find the sequence of 'D's in the input string
27    while (nextIndex < n && sequence[nextIndex] === 'D') {
28      nextIndex++; // Advance to the next index if it's a 'D'
29    }
30    // Reverse the subarray from the start of 'D's to just past the last 'D'
31    reverse(permutation, currentIndex, nextIndex);
32    // Move to the index after the sequence of 'D's or increment by one if no 'D's were found
33    currentIndex = Math.max(currentIndex + 1, nextIndex);
34  }
35
36  return permutation; // Return the resulting permutation
37}
38

Time and Space Complexity

Time Complexity

The time complexity of the code snippet is O(n), where n is the length of the input string s. This is because the code involves iterating over each character in the input string up to twice. The while loop iterates over each character, and within this loop, there is another while loop that continues as long as the character is 'D'. However, the inner loop advances the index j every time it finds a 'D', thereby skipping that section for the outer loop. Therefore, even though there is a nested loop, each character will be looked at a constant number of times, resulting in linear time complexity overall.

Additionally, the reversal of the sublist ans[i : j + 1] takes linear time with respect to the length of the sublist, but since it's done without overlapping with other sublists, the total amount of work for all reversals summed together still does not exceed O(n).

Space Complexity

The space complexity is O(n) because we are storing the result in a list ans which contains n + 1 elements. No additional significant space is used, and the space used for the output does not count towards extra space complexity when analyzing space complexity. The in-place reversal operation does not incur additional space cost.

Learn more about how to find time and space complexity quickly using problem constraints.

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