LeetCode #2089 — EASY

Find Target Indices After Sorting Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed integer array nums and a target element target.

A target index is an index i such that nums[i] == target.

Return a list of the target indices of nums after sorting nums in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.

Example 1:

Input: nums = [1,2,5,2,3], target = 2
Output: [1,2]
Explanation: After sorting, nums is [1,2,2,3,5].
The indices where nums[i] == 2 are 1 and 2.

Example 2:

Input: nums = [1,2,5,2,3], target = 3
Output: [3]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 3 is 3.

Example 3:

Input: nums = [1,2,5,2,3], target = 5
Output: [4]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 5 is 4.

Constraints:

1 <= nums.length <= 100
1 <= nums[i], target <= 100

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums and a target element target. A target index is an index i such that nums[i] == target. Return a list of the target indices of nums after sorting nums in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[1,2,5,2,3]
2

Example 2

[1,2,5,2,3]
3

Example 3

[1,2,5,2,3]
5

Core Insight

What unlocks the optimal approach

Try "sorting" the array first.
Now find all indices in the array whose values are equal to target.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2089: Find Target Indices After Sorting Array
class Solution {
    public List<Integer> targetIndices(int[] nums, int target) {
        Arrays.sort(nums);
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == target) {
                ans.add(i);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2089: Find Target Indices After Sorting Array
func targetIndices(nums []int, target int) (ans []int) {
	sort.Ints(nums)
	for i, v := range nums {
		if v == target {
			ans = append(ans, i)
		}
	}
	return
}

# Accepted solution for LeetCode #2089: Find Target Indices After Sorting Array
class Solution:
    def targetIndices(self, nums: List[int], target: int) -> List[int]:
        nums.sort()
        return [i for i, v in enumerate(nums) if v == target]

// Accepted solution for LeetCode #2089: Find Target Indices After Sorting Array
struct Solution;

impl Solution {
    fn target_indices(mut nums: Vec<i32>, target: i32) -> Vec<i32> {
        nums.sort_unstable();
        let n = nums.len();
        let mut res = vec![];
        for i in 0..n {
            if nums[i] == target {
                res.push(i as i32);
            }
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 5, 2, 3];
    let target = 2;
    let res = vec![1, 2];
    assert_eq!(Solution::target_indices(nums, target), res);
    let nums = vec![1, 2, 5, 2, 3];
    let target = 3;
    let res = vec![3];
    assert_eq!(Solution::target_indices(nums, target), res);
    let nums = vec![1, 2, 5, 2, 3];
    let target = 5;
    let res = vec![4];
    assert_eq!(Solution::target_indices(nums, target), res);
}

// Accepted solution for LeetCode #2089: Find Target Indices After Sorting Array
function targetIndices(nums: number[], target: number): number[] {
    nums.sort((a, b) => a - b);
    let ans: number[] = [];
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i] == target) {
            ans.push(i);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.