EasyTwo Pointers

Remove Element

Explanation & Solution

Description

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place and return the new length. The order of the elements may be changed. It does not matter what you leave beyond the new length.

Input: nums = [3, 2, 2, 3], val = 3

Output: 2

Explanation: The function returns 2, with the first two elements of nums being [2, 2]. The value 3 has been removed.

Constraints

0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100

Approach

Two Pointers pattern

1. Initialize the Slow Pointer

Set k = 0 to track the write position for non-val elements
k will also represent the final length of the modified array

2. Scan Through the Array

Use a fast pointer i to iterate from 0 to nums.length - 1
Each element is checked exactly once

3. Keep Non-Matching Elements

If nums[i] !== val, the element should be kept
Copy nums[i] into nums[k] and increment k
This shifts all non-val elements to the front of the array

4. Skip Matching Elements

If nums[i] === val, do nothing
The fast pointer i advances, but k stays in place
This effectively overwrites val elements with the next valid one

5. Return the New Length

After the loop, k equals the number of elements that are not val
The first k elements of nums contain all the kept values

Key Insight

The two-pointer technique uses a slow pointer k (write position) and a fast pointer i (read position)
Every non-val element is written to the front, preserving them while discarding matches

Time

O(n)single pass through the array

Space

O(1)in-place modification, no extra storage

Visualization

Input:

[3, 2, 2, 3], val = 3

—

Press ▶ or use ← → to step through

Pointer iPointer jProcessedDone

7 steps

Solution Code

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