Two pointers (opposite ends)

O(n)

Two indices start at the two ends of a sorted array and walk toward each other, deciding at each step which side to move.

Signals

sorted array (or you are allowed to sort)find a pair/triple with a target sumsqueeze from both ends (container, area)palindrome check

Template

function twoPointers(arr, target) {
    let lo = 0;
    let hi = arr.length - 1;
    while (lo < hi) {
        const sum = arr[lo] + arr[hi];
        if (sum === target) return [lo, hi];
        if (sum < target) lo++;
        else hi--;
    }
    return null;
}

Looks similar, isn't

Sliding window: A window keeps a contiguous span and grows/shrinks it under a condition. Opposite pointers converge to a single answer (a pair, an area) and never keep a span.
Hash set / map: Hashing finds a complement in unsorted data in O(n) time but O(n) space. Reach for two pointers only when the array is sorted and O(1) extra space matters.

n up to 1e5..1e6 and the array is sorted -> O(n): one linear pass, each pointer moves at most n steps. O(1) extra space.

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