How do you solve Build Array from Permutation on LeetCode?

Build Array from Permutation (LeetCode #1920) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem is a direct application of indexing, where the output is based on double indexing of the input array.

What is the time complexity of Build Array from Permutation?

The optimal solution for Build Array from Permutation runs in O(n) time and uses O(n) space. This complexity is optimal because we only traverse the array once, resulting in linear time complexity.

What is the brute force solution for Build Array from Permutation?

The brute force approach for Build Array from Permutation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves directly applying the permutation twice for each index in the array. This is straightforward but inefficient as it requires nested iterations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Build Array from Permutation?

Build Array from Permutation uses the following concepts: Array, Simulation. The recommended patterns to study are: Array, Index Mapping.

What are the interview tips for Build Array from Permutation?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as the smallest and largest possible inputs. Practice explaining your thought process while coding, as communication is key in interviews.

What are common mistakes when solving Build Array from Permutation?

Confusing the order of indexing, leading to incorrect results. Not initializing the output array properly, which can cause runtime errors.

#1920

Build Array from Permutation

Easy

ArraySimulationArrayIndex Mapping

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution leverages a single pass through the array to build the result, ensuring we only traverse the array once, which is efficient for larger inputs.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty array 'ans' of the same length as 'nums'.
2Step 2: For each index 'i' in 'nums', compute ans[i] = nums[nums[i]].
3Step 3: Return the 'ans' array.

solution.py4 lines

1# Full working Python code
2nums = [0, 2, 1, 5, 3, 4]
3ans = [nums[nums[i]] for i in range(len(nums))]
4print(ans)

ℹ

Complexity note: This complexity is optimal because we only traverse the array once, resulting in linear time complexity.

1The problem is a direct application of indexing, where the output is based on double indexing of the input array.
2Understanding the properties of permutations helps in recognizing that each index will map to a unique value.

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