First and Last Occurrence of Element in DSA Typescript - Execution Trace

Concept Flow - First and Last Occurrence of Element

Start

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Initialize first = -1, last = -1

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Traverse array from i=0 to end

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Check if array[i

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If first == -1, set first = i

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Set last = i

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Continue traversal

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End traversal

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Return first and last indices

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Done

We scan the array once, updating first and last positions when we find the target element.

Execution Sample

DSA Typescript

function findFirstLast(arr: number[], target: number): [number, number] {
  let first = -1;
  let last = -1;
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === target) {
      if (first === -1) first = i;
      last = i;
    }
  }
  return [first, last];
}

This code finds the first and last positions of the target number in the array.

Execution Table

Step	Operation	Index i	arr[i]	Condition arr[i] == target	first	last	Visual State
1	Initialize first and last	-	-	-	-1	-1	first = -1, last = -1
2	Check element	0	5	No	-1	-1	No change
3	Check element	1	3	Yes	1	1	first = 1, last = 1
4	Check element	2	7	No	1	1	No change
5	Check element	3	3	Yes	1	3	last updated to 3
6	Check element	4	9	No	1	3	No change
7	Check element	5	3	Yes	1	5	last updated to 5
8	End traversal	-	-	-	1	5	Return [1, 5]

💡 Reached end of array, traversal stops

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
first	-1	-1	1	1	1	1	1	1
last	-1	-1	1	1	3	3	5	5
i	-	0	1	2	3	4	5	-

Key Moments - 3 Insights

Why do we check if first == -1 before setting it?

Because first should store the earliest index where target appears. We only set it once at the first occurrence (see Step 3 in execution_table).

Why do we update last every time we find the target?

Because last should always point to the most recent occurrence of the target (see Steps 3, 5, 7 in execution_table).

What if the target is not in the array at all?

Then first and last remain -1, indicating target not found (see initial values in Step 1 and no updates in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'first' after Step 5?

A 1

B 3

C -1

D 5

💡 Hint

Check the 'first' column in row for Step 5 in execution_table.

At which step does the condition arr[i] == target become false for the first time?

A Step 3

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'Condition arr[i] == target' column in execution_table rows.

If the target was 7 instead of 3, what would be the final value of 'last'?

A -1

B 2

C 5

D 3

💡 Hint

Check where 7 appears in the array and see last updates in execution_table.

Concept Snapshot

Find first and last occurrence of target in array:
- Initialize first and last to -1
- Traverse array once
- If arr[i] == target:
  - If first == -1, set first = i
  - Update last = i
- Return [first, last]
- If target not found, returns [-1, -1]

Full Transcript

This concept finds the first and last positions of a target element in an array by scanning it once. We start with first and last set to -1, meaning not found. As we move through the array, when we find the target for the first time, we set first to that index. Every time we find the target, we update last to that index. After scanning all elements, we return the two indices. If the target is not found, both remain -1. The execution table shows each step with index, element checked, condition result, and updates to first and last. This helps visualize how the pointers move and how the final result is formed.