# Two Pointers

Used with certain patterns revolving around arrays, often times sorted, or in combination with a sliding window. Also to check for a palindrome, or reversing.
The idea here is to iterate two different parts of the array simultaneously to get the answer faster.

## Use Cases

• Searching: Finding elements or patterns
• Sorting: Rearranging elements in specific order
• Sliding Window

## Steps

• Init the pointers
• Loop through the DS according to requirements
• Update the pointers based on requirements
• Return when pointers or satisfy problem's logic

## Two Sum

Given an array of sorted numbers and a target sum, find a pair in the array whose sum is equal to the given target.

We could BF it by looking at each element and compare it with the rest, but that would put our Big O into N².

## How to get from O(N²) to O(N)?

As we know that our input array is sorted we can introduce a L and R Pointer and use them to move through the array. If arr[L] + arr[R] > target then reduce R pointer, otherwise increase L pointer.

## What is the difference between a Two Pointer and a Sliding Window approach?

The Sliding Window approach cares about a subset of an array between Two Pointers, whereas the Two Pointers approach cares about the values at those pointers.

## What loop are we using?

While loop as the two Pointers, L and R, move towards each other and.

``````while (L < R) { }
``````

## Problem Statement

Given an array of sorted numbers and a target sum, find a pair in the array whose sum is equal to the given target.

Write a function to return the indices of the two numbers (i.e. the pair) such that they add up to the given target.

`Input: [2, 5, 9, 11], target=11 `Output: [0, 2]
`Explanation: The numbers at index 0 and 2 add up to 11: 2+9=11

``````const targetSum = (arr, target) => {
let left = 0,
right = arr.length - 1

while (left < right) {

// can be improved by making variable sum
if (arr[left] + arr[right] === target) {
return [left, right]
}

if (arr[left] + arr[right] > target) {
right--
} else {
left++
}
}
return [-1, -1]
}
``````

## Alternate Approach

This can also be solved with a hashmap, which also works with non-sorted arrays.

LC