Two pointers: squeezing a sorted array

Chapter 3 — Pair offers on a sorted price list

Promotions team wants pairs of distinct products whose prices sum to a campaign target (think "buy two for €50"). The price list is sorted — so you can walk two indices from both ends instead of trying every pair. That is two pointers: same catalog as before, but the question is about pairs, not a single lookup.

Pair-sum problem

Given a sorted array a and a target t, decide whether there exist two distinct indices i < j such that a[i] + a[j] == t.

public static boolean hasPairSum(int[] a, int t) {
    int left = 0, right = a.length - 1;
    while (left < right) {
        int sum = a[left] + a[right];
        if (sum == t) return true;
        if (sum < t) left++;    // too small: need a bigger number, move left right
        else         right--;   // too big: need a smaller number, move right left
    }
    return false;
}

Variations that all reuse the same idea

• Remove duplicates from a sorted array in-place (slow/fast pointers).
• Three-sum: fix i, run two pointers on the remainder.
• Container with most water: both pointers at the ends of a height array.
• Check if a string is a palindrome, ignoring case and non-letters.

Interactive — squeeze the array

Use the two-pointer lab with your own numbers; watch how each move shrinks the plausible pair space without revisiting dead combinations.

Open the two-pointer lab — drag your own sorted array →