1830. Minimum Number of Operations to Make String Sorted

Problem Description

The given problem presents you with a string s, and you are required to repeatedly perform a sequence of operations to make the string sorted in lexicographical order (i.e., dictionary order). The operations are as follows:

1. Identify the largest index i satisfying 1 <= i < s.length such that the character at s[i] is less than the character at s[i - 1].
2. Find the largest index j greater than or equal to i such that every character at position k within the range [i, j] is less than the character at s[i - 1].
3. Swap the characters at positions i - 1 and j.
4. Reverse the string segment starting from index i.

The task is then to compute how many times you must perform this sequence of operations to sort the string. However, since the number of operations could be enormous, you are asked to return this number modulo 10^9 + 7.

Intuition

The intuition behind solving this problem lies in combinatorics and understanding that each operation is essentially a step towards placing the characters in the correct order.

The idea is to iterate through the string and calculate the number of operations needed to place each character in its sorted position by considering the permutations of characters remaining to be sorted.

This approach avoids actually performing the operations on the string, which would be inefficient, and instead counts the number of operations in a combinatorial way.

Here's a step-by-step breakdown of the solution intuition:

1. Calculate factorial f and modulo multiplicative inverse g in advance for efficient computation during iteration.

2. Use a counter to keep track of the number of occurrences of each character in the string.

3. Iterate over the string:

   • Calculate the count, m, of characters less than the current character which have not already been placed correctly.

   • Determine the permutations t of the remaining characters, which would represent the number of possible strings if we were to place the current character in its correct sorted position now.

   • Adjust for the duplicate characters by multiplying with the modulo multiplicative inverse g[v] for each character count v.

   • Update the total count ans of operations needed, by adding t modulo 10^9 + 7.

   • Decrement the character count for the current character in the counter to reflect that this character has been placed in its correct position.

4. The computed ans is returned which gives us the required number of operations modulo 10^9 + 7.

5. This approach cleverly avoids actually performing the operations, which could be costly especially for long strings, by instead mathematically computing the number of steps needed to obtain the sorted string.

Learn more about Math and Combinatorics patterns.

Solution Approach

The implementation of the solution makes use of several key concepts from combinatorics, as well as Python-specific data structures and functions to efficiently perform necessary calculations.

Here's a detailed walk-through:

1. Pre-calculation of factorials (f) and their modulo multiplicative inverses (g):

   • Factorials of all possible index positions are precalculated and stored in an array f, using the modulo operation to avoid integer overflow.
   • The modulo multiplicative inverse of each factorial is calculated using pow(f[i], mod - 2, mod) due to Fermat's Little Theorem, allowing us to efficiently calculate division in modular arithmetic. These inverses are stored in the array g.

   This is a preprocessing step to speed up the calculations needed for each operation count later on.

2. Using Counter to keep track of character frequencies:

   • A Counter from the collections module is utilized to count occurrences of each character in the string. This is particularly useful for calculating the multiplicative inverse part of the permutation calculation.

3. Main algorithmic loop:

   • We loop over each character c in the string s with its index i. For each character, the goal is to find how many operations are needed to move it to its correct place given the characters that have yet to be ordered.

   • The variable m is the number of characters less than c that are currently placed to the right of their target position.

   • Compute the partial operations count t by multiplying the factorial f[n - i - 1] (representing the permutations of remaining characters) by the number m.

   • Adjust t to account for duplicate characters by multiplying it with the modulo multiplicative inverses of the counts of the remaining characters. This part is crucial because it handles the division in the permutations formula.

4. Combination of counts and modular arithmetic:

   • Each partial count t is added to the total answer ans, with modulo 10^9 + 7 being applied at every step to manage large numbers and prevent overflow.
   • After processing a character, we decrement its count in the Counter to account for having effectively placed that character in its final sorted position.

5. Final result:

   • Once the loop is done, the resulting ans variable holds the total count of operations needed to sort the string, modulo 10^9 + 7.

The use of the factorial array, multiplicative inverses, and Counter for frequency tracking together form a powerful combination that allows us to compute the total operation count efficiently without actually performing the swaps and reversals described in the problem statement.

Example Walkthrough

Let's consider an example with the string s = "bca" and walk through the solution approach.

1. Pre-calculation of factorials (f) and their modulo multiplicative inverses (g):

   For a string of length 3, we would calculate factorials for 0, 1, 2, 3 and their inverses accordingly. Let's assume we have computed them as follows (using a modulus mod = 10^9 + 7):

   • f = [1, 1, 2, 6] (factorials modulo mod)
   • g = [1, 1, 500000004, 166666668] (inverses modulo mod)

2. Using Counter to keep track of character frequencies:

   By counting the occurrences of each character, we get:

   • Counter = {'b': 1, 'c': 1, 'a': 1}

3. Main algorithmic loop:

   Now, we iterate over the string:

   • For character b at index 0:

     • m is the count of characters less than b to its right, which is 1 ('a').
     • Calculate permutation count t as m * f[n - i - 1], which gives 1 * f[2]
     • Correct for any duplicates, which isn't necessary here as all counts are 1.
     • t is now 1 * 2 = 2.
     • Add t to the total operations count, ans += 2 % mod.

   • For character c at index 1:

     • m is the count of characters less than c to its right, which is 1 ('a').
     • Calculate permutation count t as m * f[n - i - 1], which gives 1 * f[1]
     • Correct for any duplicates, which again isn't needed.
     • t is 1 * 1 = 1.
     • Add t to the total operations count, ans += 1 % mod.

   • For the last character at index 2, no operation is needed as it's already in the last position, so m = 0.

4. Combination of counts and modular arithmetic:

   • By summing the partial counts (while taking mod at each step):
     • ans = (0 + 2 + 1) % mod
     • ans = 3 % mod, which means 3 operations are required to sort "bca" into "abc".

5. Final result:

   Thus, the total count of operations needed to sort the string "bca" into "abc" is 3, and considering the modulus, the answer remains 3, representing the result modulo 10^9 + 7.

Note that in a longer string, the factors and the multiplicative inverses would allow us to handle the calculations without actually having to perform the swaps and reverses as the string's sorting operations. The example uses small factorials and inverses for simplicity, and in practice, these would be precomputed for efficient computation in the main loop for larger strings.

Solution Implementation

Time and Space Complexity

The given Python code is designed to compute the number of permutations that are lexicographically smaller than the input string s, using modular arithmetic with a modulus of mod = 10**9 + 7.

Time Complexity

The time complexity can be analyzed as follows:

1. The initialization of lists f and g has a complexity of O(n), where n is the length of f and g.

2. The loop to compute f and g also runs in O(n), as it iterates once for each of the n elements, and each operation within the loop (multiplication, modulo, and power operation) is O(1) in time complexity (since the modular exponentiation is computed in logarithmic time relative to the exponent).

3. Inside the makeStringSorted method, there is a loop that goes over each character of the string s. The length of s could be n at most, so this loop could iterate n times.

4. Inside this loop, there is a sum operation which in the worst case can iterate over 26 characters (all lowercase English letters), so this part is O(1) as the number of lowercase English letters is constant.

5. Computation of t inside the loop is O(1) as it involves a single multiplication and module operation.

6. The most expensive operation inside the loop is the subsequent loop which computes the multiplication of t with g[v] for each value v in cnt values. Since there can be at most 26 unique characters, this loop runs in O(1) time complexity.

7. Updating the cnt dictionary is O(1) since dictionaries in Python have average-case constant time insertions, deletions, and lookups.

Considering these steps, each iteration of the loop is O(1). Therefore, the overall time complexity for the makeStringSorted function is O(n).

Space Complexity

The space complexity can be analyzed as follows:

1. The f and g arrays each require O(n) space to store the factorial and the modular multiplicative inverses, respectively.

2. The Counter instance cnt will store at most 26 unique characters with their counts, which is O(1).

3. Temporary variables used within the for loops such as i, c, m, t take O(1) space.

Therefore, the dominant term in the space complexity is the storage for f and g, which results in an overall space complexity of O(n).

To summarize, the time complexity of the code is O(n), and the space complexity is O(n).

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