What is the time complexity of Create Sorted Array through Instructions?

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

What data structures are used to solve Create Sorted Array through Instructions?

This problem uses Array, Binary Search, Divide and Conquer, Binary Indexed Tree, Segment Tree, Merge Sort, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Create Sorted Array through Instructions?

Insight 1: The core idea is to efficiently count the number of elements less than and greater than the current element being inserted. Insight 2: A Binary Indexed Tree (BIT), also known as a Fenwick Tree, is the perfect data structure for efficiently counting the number of elements less than or equal to a given value in a dynamic array.

Which companies ask LeetCode 1649. Create Sorted Array through Instructions in interviews?

This problem is frequently asked at Akuna Capital.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Create Sorted Array through Instructions - LeetCode 1649 Solution

Create Sorted Array through Instructions - Complete Solution Guide

Create Sorted Array through Instructions is LeetCode problem 1649, 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

Given an integer array instructions , you are asked to create a sorted array from the elements in instructions . You start with an empty container nums . For each element from left to right in instructions , insert it into nums . The cost of each insertion is the minimum of the following: The number of elements currently in nums that are strictly less than instructions[i] . The number of elements currently in nums that are strictly greater than instructions[i] . For example, if inserting element

Detailed Explanation

The problem requires you to simulate building a sorted array `nums` by inserting elements from the `instructions` array one by one. The 'cost' of inserting each element is the minimum between the number of elements in `nums` that are strictly less than the element being inserted and the number of elements in `nums` that are strictly greater than the element. The goal is to calculate the total cost of inserting all elements, modulo 10^9 + 7.

Solution Approach

The solution uses a Binary Indexed Tree (BIT) to track the elements that have been inserted into the sorted array. For each element in the `instructions` array, it queries the BIT to find the number of elements less than and greater than the current element. The minimum of these two counts is added to the total cost. After calculating the cost, the BIT is updated to reflect the insertion of the current element.

Step-by-Step Algorithm

Step 1: Initialize a Binary Indexed Tree (BIT) of size 100002 (1-indexed to handle values from 1 to 100001).
Step 2: Iterate through the `instructions` array.
Step 3: For each element `num` in `instructions`, query the BIT to find the number of elements strictly less than `num` (i.e., `query(num - 1)`).
Step 4: Calculate the number of elements strictly greater than `num`. This is equal to `i - query(num)`, where `i` is the current index in the `instructions` array.
Step 5: Add the minimum of the 'less count' and 'greater count' to the `total_cost`.
Step 6: Update the BIT to reflect the insertion of `num` (i.e., `update(num)`).
Step 7: After processing all elements in `instructions`, return the `total_cost` modulo 10^9 + 7.

Key Insights

Insight 1: The core idea is to efficiently count the number of elements less than and greater than the current element being inserted.
Insight 2: A Binary Indexed Tree (BIT), also known as a Fenwick Tree, is the perfect data structure for efficiently counting the number of elements less than or equal to a given value in a dynamic array.
Insight 3: The problem's constraints (1 <= instructions[i] <= 10^5) are important, as they allow us to use a BIT of a fixed size without needing to dynamically resize or use a more complex data structure.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(m)

Topics

This problem involves: Array, Binary Search, Divide and Conquer, Binary Indexed Tree, Segment Tree, Merge Sort, Ordered Set.

Companies

Asked at: Akuna Capital.

Create Sorted Array through Instructions | LeetCode 1649

Hard

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set

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

Solution Code

Available in:

class Solution {
    public int createSortedArray(int[] instructions) {
        int MOD = 1_000_000_007;
        
        // Max value in instructions is 10^5.
        // We use a Fenwick Tree (BIT) for values from 1 to 10^5.
        // The BIT will be 1-indexed, so size needs to be max_val + 2.
        int size = 100002;
        int[] bit = new int[size];

        // No need for static keyword since it's an inner class in Java
        class BIT {
            public void update(int index) {
                /**Increments the count for the number at `index`*/
                while (index < size) {
                    bit[index] += 1;
                    index += index & -index;
                }
            }

            public int query(int index) {
                /**Returns the count of numbers less than or equal to `index`*/
                int res = 0;
                while (index > 0) {
                    res += bit[index];
                    index -= index & -index;
                }
                return res;
            }
        }
        
        BIT bitInstance = new BIT();

        int total_cost = 0;
        for (int i = 0; i < instructions.length; i++) {
            int num = instructions[i];
            // Count of elements strictly less than num
            int less_count = bitInstance.query(num - 1);
            
            // Count of elements strictly greater than num
            // i is the number of elements already inserted.
            // query(num) is the count of elements <= num.
            // So, i - query(num) gives elements > num.
            int greater_count = i - bitInstance.query(num);
            
            total_cost += Math.min(less_count, greater_count);
            
            // Add the current number to our data structure
            bitInstance.update(num);
        }
        
        return total_cost % MOD;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1649: Create Sorted Array through Instructions - Detailed explanation and solution approaches.

Topics

Array Binary Search Divide and Conquer Binary Indexed Tree Segment Tree Merge Sort Ordered Set

Companies

Akuna Capital

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1649

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

7 steps

1
Step 1: Initialize a Binary Indexed Tree (BIT) of size 100002 (1-indexed to handle values from 1 to 100001).
2
Step 2: Iterate through the `instructions` array.
3
Step 3: For each element `num` in `instructions`, query the BIT to find the number of elements strictly less than `num` (i.e., `query(num - 1)`).
4
Step 4: Calculate the number of elements strictly greater than `num`. This is equal to `i - query(num)`, where `i` is the current index in the `instructions` array.
5
Step 5: Add the minimum of the 'less count' and 'greater count' to the `total_cost`.
6
Step 6: Update the BIT to reflect the insertion of `num` (i.e., `update(num)`).
7
Step 7: After processing all elements in `instructions`, return the `total_cost` 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(m)

📝

Why?

The time complexity is determined by the number of iterations through the instructions (n) multiplied by the time complexity of the BIT operations (log m). The space complexity is determined by the size of the BIT, which is dependent on the maximum value in the input array (m), but independent of the number of instructions (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

Ordered Set (Self-Balancing BST)

Instead of a BIT, an ordered set (like `std::set` in C++ or `TreeSet` in Java) can be used. We can insert elements into the ordered set and use `lower_bound` or equivalent to find the number of elements less than and greater than the target.

Time: O(n log n), where n is the length of the instructions array. Inserting and querying elements in a balanced BST takes O(log n) on average, and we perform these operations n times.Space: O(n) to store the ordered set.

Trade-offs: The ordered set approach has a time complexity similar to the BIT approach, but it might have slightly higher constant factors. It has higher space complexity compared to the BIT, as the space used depends on the number of instructions rather than the maximum value in the instructions.

Merge Sort

Merge Sort can be used but the constant factor will be very large and will result in TLE for large inputs

Time: O(n^2)Space: O(n)

Trade-offs: Merge sort will result in TLE, and should not be used.

✨

Key Insights

1Insight 1: The core idea is to efficiently count the number of elements less than and greater than the current element being inserted.
2Insight 2: A Binary Indexed Tree (BIT), also known as a Fenwick Tree, is the perfect data structure for efficiently counting the number of elements less than or equal to a given value in a dynamic array.
3Insight 3: The problem's constraints (1 <= instructions[i] <= 10^5) are important, as they allow us to use a BIT of a fixed size without needing to dynamically resize or use a more complex data structure.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Duplicate numbers in the `instructions` array. The BIT correctly handles duplicates because the `query` function counts the number of elements less than or *equal to* a given value, and the `update` function increments the count for that value.
Edge case 2: Empty `instructions` array. The solution correctly returns 0 because the loop will not execute, and the initial `total_cost` is 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of elements greater than the current element. Remember that `i - query(num)` gives the number of elements *strictly greater than* `num` because `query(num)` gives the number of elements less than or *equal to* `num`.

✗Common mistake 2: Not taking the modulo at the end of the calculation. The problem statement specifies that the answer should be returned modulo 10^9 + 7. Failing to do so can lead to incorrect results due to integer overflow.

✅

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: Calculating the cost of inserting data into a sorted data stream, where the cost depends on the position of the new data point relative to existing data points. For example, optimizing placement of advertisements based on user preferences.

🏦Real-world scenario 2: Rank aggregation systems. Consider ranking different search results based on multiple criteria. The cost to insert a new ranking can be derived from the algorithm.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: June 30, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Create Sorted Array through Instructions"?

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

🤖 Need help with this problem?