How do you solve Create Sorted Array through Instructions on LeetCode?

Create Sorted Array through Instructions (LeetCode #1649) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using data structures like BIT can significantly reduce the time complexity of counting operations.

What is the brute force solution for Create Sorted Array through Instructions?

The brute force approach for Create Sorted Array through Instructions is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves inserting each element into the sorted array and counting the elements less than and greater than the current element. This is simple but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Create Sorted Array through Instructions?

Always discuss your thought process and approach before jumping into coding. Consider edge cases and how they might affect your solution. Practice implementing data structures like BIT or Segment Trees, as they are commonly used in optimization problems.

What are common mistakes when solving Create Sorted Array through Instructions?

Not considering edge cases where elements are repeated, leading to incorrect counts. Failing to optimize the insertion process, resulting in timeouts for larger inputs.

#1649

Create Sorted Array through Instructions

Hard

ArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetBinary Indexed TreeSegment TreeCounting Inversions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

💡

Intuition

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

Using a Binary Indexed Tree (BIT) allows us to efficiently count the number of elements less than and greater than the current element in logarithmic time, significantly speeding up the process.

⚙️

Algorithm

5 steps

1Step 1: Initialize a Binary Indexed Tree (BIT) to keep track of counts of elements.
2Step 2: For each element in instructions, use BIT to find the count of elements less than and greater than the current element.
3Step 3: Calculate the cost as the minimum of these two counts and add it to total_cost.
4Step 4: Update the BIT with the current element to include it in future counts.
5Step 5: Return total_cost after processing all elements.

solution.py29 lines

1# Full working Python code
2
3class BIT:
4    def __init__(self, size):
5        self.size = size
6        self.tree = [0] * (size + 1)
7
8    def update(self, index, value):
9        while index <= self.size:
10            self.tree[index] += value
11            index += index & -index
12
13    def query(self, index):
14        total = 0
15        while index > 0:
16            total += self.tree[index]
17            index -= index & -index
18        return total
19
20def createSortedArray(instructions):
21    max_val = max(instructions)
22    bit = BIT(max_val)
23    total_cost = 0
24    for instruction in instructions:
25        less_count = bit.query(instruction - 1)
26        greater_count = len(instructions) - bit.query(instruction)
27        total_cost += min(less_count, greater_count)
28        bit.update(instruction, 1)
29    return total_cost % (10**9 + 7)

ℹ

Complexity note: The complexity is reduced to O(n log n) because each insertion and query operation in the BIT takes logarithmic time, making it efficient for large inputs.

1Using data structures like BIT can significantly reduce the time complexity of counting operations.
2Understanding how to maintain sorted order efficiently is crucial for problems involving dynamic arrays.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.