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easyBasicsSometimes asked

First and Last Occurrence

Basics · 6 min

Given an integer array nums and a target, return the first and last indices where target occurs, as [first, last], or [-1, -1] if absent.

In plain English

Scan once, recording the first match index and updating the last match index on every hit.

Examples

Example 1

Input: nums = [1, 2, 2, 3, 2], target = 2
Output: [1, 4]
Note: First at 1, last at 4.

Example 2

Input: nums = [1, 2, 3], target = 5
Output: [-1, -1]
Note: Absent.

Example 3

Input: nums = [4], target = 4
Output: [0, 0]
Note: Single hit.

Constraints

0 ≤ nums.length ≤ 10^5

Clarifying questions2First question: Is the array sorted?Explore questions

Q1.Is the array sorted?
If so I'd use two binary searches for O(log n).
Q2.Return value when absent?
[-1, -1].

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Track `first = -1` and `last = -1`. On each match, set `first` only if it's still -1, and always update `last`. O(n). (Sorted → two binary searches, O(log n).)

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you capture first/last in one pass and handle the absent case.

Where candidates slip

Overwriting first, or wrong absent sentinel.

How to stand out

Give the O(log n) two-binary-search solution for sorted input (LC34).

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Single-pass first/last

TimeO(n)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Can one pass capture both the earliest and latest match?

2Hint 2Explore hint

Set first only once; always update last.

3Hint 3Explore hint

What do you return when the target is absent?

A · Core idea

Return the first and last indices of a target. A single pass suffices: record first only the first time you match, and overwrite last on every match.

Why guard `first` but not `last`. The earliest index should be set once and never changed (guard it with the first == -1 check). The latest index is simply whatever the final match was, so you update it unconditionally on each hit.

Why both default to -1. If the target never appears, both stay -1, which is the agreed 'absent' signal. Returning [0, 0] for an absent target is the classic bug.

The sorted upgrade. On a sorted array, the matches form a contiguous block, so two binary searches (leftmost and rightmost index) find both ends in O(log n) — the standard follow-up (LeetCode 34).

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

nums = [1, 2, 2, 3, 2], target = 2

i=1

match, first=-1→1

Statefirst=1,last=1

Final result[1, 4]

C · Code with teaching comments

solution.javaJava

Single-pass first/last — optimal

2 teaching notes

1
public int[] firstLast(int[] nums, int target) {
2
    int first = -1, last = -1;
3
    for (int i = 0; i < nums.length; i++) {
4
        if (nums[i] == target) {
5
            // Record the first match only once.
6
            if (first == -1) first = i;
7
            // Always overwrite with the most recent match.
8
            last = i;
9
        }
10
    }
11
    return new int[]{first, last};
12
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.12 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

On a sorted array, two binary searches (leftmost and rightmost) give O(log n) — the classic follow-up.

ComplexityO(n) time, O(1) space — one linear pass with two index variables; O(log n) via two binary searches if sorted.

E · Edge cases and pitfalls4nums = [1, 2, 3], target = 5 — Absent → [-1, -1].Explore edge cases

Edge cases handled

nums = [1, 2, 3], target = 5Absent → [-1, -1].
nums = [4], target = 4Single hit → [0, 0].

Pitfalls

Overwriting `first` on every match (loses the earliest).
Returning [0, 0] instead of [-1, -1] when absent.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic int[] firstLast(int[] nums, int target) { int first = -1, last = -1; for (int i = 0; i < nums.length; i++) { if (nums[i] == target) { if (first == -1) first = i;Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Overwriting `first` on every match (loses the earliest).
Returning [0, 0] instead of [-1, -1] when absent.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Single-pass first/last

Code pattern to memorize

first=-1,last=-1; on match: if first<0 first=i; last=i

Rules to remember

1Guard first with -1.
2Always update last.
3Both -1 if absent.

Use this pattern when

Range of a value's positions

Choose another pattern when

Sorted — binary-search both ends

Interview takeaway

One scan can capture both the first and last occurrence.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Single-pass first/last. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Find First and Last Position (sorted)
Coming soon
Two binary searches on a sorted array.
Count Occurrences
last − first + 1 gives the count on sorted data.

easyBasicsSometimes asked

First and Last Occurrence

Basics · 6 min

Given an integer array nums and a target, return the first and last indices where target occurs, as [first, last], or [-1, -1] if absent.

In plain English

Scan once, recording the first match index and updating the last match index on every hit.

Examples

Example 1

Input: nums = [1, 2, 2, 3, 2], target = 2
Output: [1, 4]
Note: First at 1, last at 4.

Example 2

Input: nums = [1, 2, 3], target = 5
Output: [-1, -1]
Note: Absent.

Example 3

Input: nums = [4], target = 4
Output: [0, 0]
Note: Single hit.

Constraints

0 ≤ nums.length ≤ 10^5

Clarifying questions2First question: Is the array sorted?Explore questions

Q1.Is the array sorted?
If so I'd use two binary searches for O(log n).
Q2.Return value when absent?
[-1, -1].

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Track `first = -1` and `last = -1`. On each match, set `first` only if it's still -1, and always update `last`. O(n). (Sorted → two binary searches, O(log n).)

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you capture first/last in one pass and handle the absent case.

Where candidates slip

Overwriting first, or wrong absent sentinel.

How to stand out

Give the O(log n) two-binary-search solution for sorted input (LC34).

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Single-pass first/last

TimeO(n)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Can one pass capture both the earliest and latest match?

2Hint 2Explore hint

Set first only once; always update last.

3Hint 3Explore hint

What do you return when the target is absent?

A · Core idea

Return the first and last indices of a target. A single pass suffices: record first only the first time you match, and overwrite last on every match.

Why both default to -1. If the target never appears, both stay -1, which is the agreed 'absent' signal. Returning [0, 0] for an absent target is the classic bug.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

nums = [1, 2, 2, 3, 2], target = 2

i=1

match, first=-1→1

Statefirst=1,last=1

Final result[1, 4]

C · Code with teaching comments

solution.javaJava

Single-pass first/last — optimal

2 teaching notes

1
public int[] firstLast(int[] nums, int target) {
2
    int first = -1, last = -1;
3
    for (int i = 0; i < nums.length; i++) {
4
        if (nums[i] == target) {
5
            // Record the first match only once.
6
            if (first == -1) first = i;
7
            // Always overwrite with the most recent match.
8
            last = i;
9
        }
10
    }
11
    return new int[]{first, last};
12
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.12 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

On a sorted array, two binary searches (leftmost and rightmost) give O(log n) — the classic follow-up.

ComplexityO(n) time, O(1) space — one linear pass with two index variables; O(log n) via two binary searches if sorted.

E · Edge cases and pitfalls4nums = [1, 2, 3], target = 5 — Absent → [-1, -1].Explore edge cases

Edge cases handled

nums = [1, 2, 3], target = 5Absent → [-1, -1].
nums = [4], target = 4Single hit → [0, 0].

Pitfalls

Overwriting `first` on every match (loses the earliest).
Returning [0, 0] instead of [-1, -1] when absent.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Overwriting `first` on every match (loses the earliest).
Returning [0, 0] instead of [-1, -1] when absent.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Single-pass first/last

Code pattern to memorize

first=-1,last=-1; on match: if first<0 first=i; last=i

Rules to remember

1Guard first with -1.
2Always update last.
3Both -1 if absent.

Use this pattern when

Range of a value's positions

Choose another pattern when

Sorted — binary-search both ends

Interview takeaway

One scan can capture both the first and last occurrence.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Single-pass first/last. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Find First and Last Position (sorted)
Coming soon
Two binary searches on a sorted array.
Count Occurrences
last − first + 1 gives the count on sorted data.