How do you solve Create Target Array in the Given Order on LeetCode?

Create Target Array in the Given Order (LeetCode #1389) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The insertion process requires careful management of indices.

What is the brute force solution for Create Target Array in the Given Order?

The brute force approach for Create Target Array in the Given Order is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves directly simulating the insertion process by iterating through the nums and index arrays. For each element, we insert the corresponding value into the target array at the specified index, shifting elements as necessary. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Create Target Array in the Given Order?

Create Target Array in the Given Order uses the following concepts: Array, Simulation. The recommended patterns to study are: Array.

What are the interview tips for Create Target Array in the Given Order?

Always consider edge cases, like inserting at the beginning or end. Explain your thought process clearly while coding. Practice simulating array manipulations to build confidence.

What are common mistakes when solving Create Target Array in the Given Order?

Not shifting elements correctly when inserting. Assuming the target array can grow without bounds.

#1389

Create Target Array in the Given Order

Easy

ArraySimulationArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal approach leverages a simple array to directly place elements in their final positions without needing to shift elements repeatedly. This is more efficient as it avoids the overhead of multiple insertions.

⚙️

Algorithm

3 steps

1Step 1: Initialize a target array of the same length as nums filled with zeros.
2Step 2: For each element in nums, place it directly in the target array at the index specified by the corresponding element in index.
3Step 3: Use a loop to shift elements to the right only when necessary, ensuring the target array is built correctly.

solution.py9 lines

1# Full working Python code
2
3def createTargetArray(nums, index):
4    target = [0] * len(nums)
5    for i in range(len(nums)):
6        for j in range(i, index[i], -1):
7            target[j] = target[j - 1]
8        target[index[i]] = nums[i]
9    return target

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loop structure, but we avoid the overhead of dynamic array resizing. The space complexity is O(n) because we create a new target array.

1The insertion process requires careful management of indices.
2Understanding how to manipulate arrays is crucial for this problem.

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