How do you solve Find Target Indices After Sorting Array on LeetCode?

Find Target Indices After Sorting Array (LeetCode #2089) 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: Sorting is essential to find target indices efficiently.

What is the time complexity of Find Target Indices After Sorting Array?

The optimal solution for Find Target Indices After Sorting Array runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, and the space complexity is O(n) for storing the sorted array and the indices.

What is the brute force solution for Find Target Indices After Sorting Array?

The brute force approach for Find Target Indices After Sorting Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves sorting the array first and then scanning through the sorted array to find the indices of the target. This method is straightforward but not the most efficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find Target Indices After Sorting Array?

Find Target Indices After Sorting Array uses the following concepts: Array, Binary Search, Sorting. The recommended patterns to study are: Sorting, Array.

What are the interview tips for Find Target Indices After Sorting Array?

Always clarify if you need to return original or sorted indices. Discuss your thought process while coding to show your understanding. Consider edge cases, such as when the target is not present.

What are common mistakes when solving Find Target Indices After Sorting Array?

Not considering the original indices after sorting. Overlooking the need to return sorted indices.

#2089

Find Target Indices After Sorting Array

Easy

ArrayBinary SearchSortingSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution involves sorting the array and then using a single pass to find the indices of the target. This is efficient because we only need to sort once and can gather indices in a single traversal.

⚙️

Algorithm

5 steps

1Step 1: Sort the array nums.
2Step 2: Initialize an empty list to store the target indices.
3Step 3: Iterate through the sorted array and check if each element equals the target.
4Step 4: If it does, append the current index to the list.
5Step 5: Return the list of target indices.

solution.py5 lines

1# Full working Python code
2
3def target_indices(nums, target):
4    sorted_nums = sorted(nums)
5    return [i for i, num in enumerate(sorted_nums) if num == target]

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, and the space complexity is O(n) for storing the sorted array and the indices.

1Sorting is essential to find target indices efficiently.
2Using a single pass after sorting can optimize the search for indices.

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