What is the time complexity of Minimum Number of Operations to Make String Sorted?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Minimum Number of Operations to Make String Sorted?

This problem uses Math, String, Combinatorics. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Number of Operations to Make String Sorted?

Insight 1: The given operations simulate finding the lexicographically next permutation of the string. So, we need to find how many permutations are there before the sorted permutation. Insight 2: The problem can be solved by calculating the number of permutations smaller than the given string 's'. This requires calculating factorials, inverse factorials, and using a binary indexed tree (BIT) or frequency array to count the number of characters smaller than the current character.

Which companies ask LeetCode 1830. Minimum Number of Operations to Make String Sorted in interviews?

This problem is frequently asked at Samsung.

The problem asks us to find the number of operations required to sort a given string 's' using a specific set of operations. The operations involve finding the largest index 'i' where s[i] = 'i' where s[j] < s[i-1], swapping s[i-1] and s[j], and then reversing the suffix of the string starting at index 'i'. We need to repeat these operations until the string is sorted and return the number of operations modulo 10^9 + 7.

Solution in python3. Solution in java. Solution in cpp. Solution in c

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Minimum Number of Operations to Make String Sorted - LeetCode 1830 Solution

Q: How to solve Minimum Number of Operations to Make String Sorted?

The solution computes the number of permutations smaller than the given string by iterating through the string from left to right. For each character, it determines the number of characters to its right that are smaller than it. Then, it calculates the number of permutations formed by those smaller characters, taking into account the repetitions of characters. A binary indexed tree (BIT) is used to efficiently count the number of smaller characters and update their frequencies. Factorials and their modular inverses are precomputed to speed up the permutation calculation.

Minimum Number of Operations to Make String Sorted - Complete Solution Guide

Minimum Number of Operations to Make String Sorted is LeetCode problem 1830, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s ( 0-indexed ). You are asked to perform the following operation on s until you get a sorted string: Find the largest index i such that 1 <= i < s.length and s[i] < s[i - 1] . Find the largest index j such that i <= j < s.length and s[k] < s[i - 1] for all the possible values of k in the range [i, j] inclusive. Swap the two characters at indices i - 1 and j . Reverse the suffix starting at index i . Return the number of operations needed to ma

Detailed Explanation

The problem asks us to find the number of operations required to sort a given string 's' using a specific set of operations. The operations involve finding the largest index 'i' where s[i] < s[i-1], finding the largest index 'j' >= 'i' where s[j] < s[i-1], swapping s[i-1] and s[j], and then reversing the suffix of the string starting at index 'i'. We need to repeat these operations until the string is sorted and return the number of operations modulo 10^9 + 7.

Solution Approach

The solution computes the number of permutations smaller than the given string by iterating through the string from left to right. For each character, it determines the number of characters to its right that are smaller than it. Then, it calculates the number of permutations formed by those smaller characters, taking into account the repetitions of characters. A binary indexed tree (BIT) is used to efficiently count the number of smaller characters and update their frequencies. Factorials and their modular inverses are precomputed to speed up the permutation calculation.

Step-by-Step Algorithm

Step 1: Precompute factorials and inverse factorials modulo 10^9 + 7 up to the length of the string.
Step 2: Initialize a frequency array (counts) to store the count of each character in the string.
Step 3: Initialize a Binary Indexed Tree (BIT) with size 26 to store cumulative frequency of characters.
Step 4: Iterate through the string from left to right.
Step 5: For each character at index i, query the BIT to find the number of characters smaller than s[i] that appear later in the string.
Step 6: Calculate the number of permutations that can be formed using the characters to the right of s[i]. This involves multiplying the factorial of the length of the remaining string by the modular inverse of the product of the factorials of the character counts.
Step 7: Add the contribution of the current character to the total operations.
Step 8: Update the BIT and counts array to reflect the removal of the current character.
Step 9: Return the total operations modulo 10^9 + 7.

Key Insights

Insight 1: The given operations simulate finding the lexicographically next permutation of the string. So, we need to find how many permutations are there before the sorted permutation.
Insight 2: The problem can be solved by calculating the number of permutations smaller than the given string 's'. This requires calculating factorials, inverse factorials, and using a binary indexed tree (BIT) or frequency array to count the number of characters smaller than the current character.
Insight 3: Efficiently calculating factorials and their inverses modulo a large prime is crucial to avoid integer overflow and get the correct result.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Math, String, Combinatorics.

Companies

Asked at: Samsung.

Minimum Number of Operations to Make String Sorted | LeetCode 1830

Hard

Math String Combinatorics

Time: O(n log n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int makeStringSorted(String s) {
        int MOD = (int)1e9 + 7;
        
        int n = s.length();
        
        int[] tree = new int[27];
        
        
        
        
        
        int[] fact = new int[n + 1];
        int[] inv_fact = new int[n + 1];
        fact[0] = 1;
        inv_fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = (int)((1L * fact[i - 1] * i) % MOD);
        }
        
        inv_fact[n] = power(fact[n], MOD - 2, MOD);
        for (int i = n - 1; i >= 0; i--) {
            inv_fact[i] = (int)((1L * inv_fact[i + 1] * (i + 1)) % MOD);
        }

        int[] counts = new int[26];
        for (char charr : s.toCharArray()) {
            counts[charr - 'a']++;
        }
        
        for (int i = 0; i < 26; i++) {
            if (counts[i] > 0) {
                update(tree, i, counts[i]);
            }
        }

        long denominator_inv = 1;
        for (int count : counts) {
            denominator_inv = (denominator_inv * inv_fact[count]) % MOD;
        }

        long total_ops = 0;
        
        for (int i = 0; i < n; i++) {
            int m = n - i;
            int char_code = s.charAt(i) - 'a';
            
            int n_smaller = query(tree, char_code - 1);
            
            long perms = ((1L * fact[m - 1] * denominator_inv) % MOD);
            long contribution = ((perms * n_smaller) % MOD);
            total_ops = (total_ops + contribution) % MOD;
            
            if (counts[char_code] > 0) {
                denominator_inv = (denominator_inv * counts[char_code]) % MOD;
                update(tree, char_code, -1);
                counts[char_code]--;
            } else {
                // This case indicates an issue or the string is fully processed.
                // In this problem context, it shouldn't be reached before the loop ends.
                break;
            }

        }

        return (int)total_ops;
    }

    private void update(int[] tree, int idx, int delta) {
        int i = idx + 1;
        while (i < 27) {
            tree[i] += delta;
            i += i & (-i);
        }
    }

    private int query(int[] tree, int idx) {
        if (idx < 0) {
            return 0;
        }
        int s = 0;
        int i = idx + 1;
        while (i > 0) {
            s += tree[i];
            i -= i & (-i);
        }
        return s;
    }
    
    private int power(int base, int exp, int mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if ((exp & 1) == 1) res = (res * base) % mod;
            exp >>= 1;
            base = (base * base) % mod;
        }
        return (int)res;
    }
}

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Problem Info

Description

Problem 1830: Minimum Number of Operations to Make String Sorted - Detailed explanation and solution approaches.

Topics

Math String Combinatorics

Companies

Samsung

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1830

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Precompute factorials and inverse factorials modulo 10^9 + 7 up to the length of the string.
2
Step 2: Initialize a frequency array (counts) to store the count of each character in the string.
3
Step 3: Initialize a Binary Indexed Tree (BIT) with size 26 to store cumulative frequency of characters.
4
Step 4: Iterate through the string from left to right.
5
Step 5: For each character at index i, query the BIT to find the number of characters smaller than s[i] that appear later in the string.
6
Step 6: Calculate the number of permutations that can be formed using the characters to the right of s[i]. This involves multiplying the factorial of the length of the remaining string by the modular inverse of the product of the factorials of the character counts.
7
Step 7: Add the contribution of the current character to the total operations.
8
Step 8: Update the BIT and counts array to reflect the removal of the current character.
9
Step 9: Return the total operations modulo 10^9 + 7.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the loop through the string and the BIT operations. Factorial precomputation and initial counts are O(n), but those are performed only once. Each BIT operation takes O(log n). Thus, processing each character takes O(log n), which is repeated n times so O(n log n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Frequency Array for Querying

Instead of using a BIT, we can use a frequency array of size 26 to store the frequency of characters. To query the number of characters smaller than the current character, we iterate through the frequency array from 0 to s[i] - 'a' and sum up the frequencies.

Time: O(n * 26) = O(n), since the inner loop has constant iterations.Space: O(n), to store the factorial and the inverse factorial tables.

Trade-offs: This approach is slightly simpler to implement, but for a larger alphabet size the BIT approach provides better performance as queries in BIT can be faster than iterating on the array, and BIT allows quick updates too.

✨

Key Insights

1Insight 1: The given operations simulate finding the lexicographically next permutation of the string. So, we need to find how many permutations are there before the sorted permutation.
2Insight 2: The problem can be solved by calculating the number of permutations smaller than the given string 's'. This requires calculating factorials, inverse factorials, and using a binary indexed tree (BIT) or frequency array to count the number of characters smaller than the current character.
3Insight 3: Efficiently calculating factorials and their inverses modulo a large prime is crucial to avoid integer overflow and get the correct result.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty or single-character string. The code implicitly handles these cases as the loop will not execute or execute only once, leading to 0 operations for single-character which is correct.
Edge case 2: String already sorted. The solution correctly calculates 0 operations.
Edge case 3: String with repeating characters. The solution correctly handles this by calculating the number of distinct permutations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo operation at each step during factorial and inverse factorial calculation can lead to integer overflow and incorrect results.

✗Common mistake 2: Forgetting to calculate the modular inverse correctly using Fermat's Little Theorem can result in incorrect permutations.

✗Common mistake 3: Incorrectly updating the frequencies or BIT after processing each character can lead to miscounting the number of smaller permutations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Ranking permutations or combinations. This problem is related to determining the rank of a given permutation in the lexicographical order of all possible permutations. This can be used in data compression or indexing.

🏦Real-world scenario 2: Evaluating the probability of certain event orderings. The permutation logic and counting can be adapted to calculate probabilities in simulations or statistical analysis where the order of events is important.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Number of Operations to Make String Sorted"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?