Coding Interview Patterns

Two Pointers & Sliding Window

Two indices walking across an array or string — from opposite ends, or as a window that grows and shrinks — turn a large class of O(n²) brute-force scans into O(n).

Opposite-direction pointers on sorted data: two-sum on a sorted array, container-with-most-water
Same-direction pointers as a sliding window: longest substring without repeating characters, smallest subarray with a given sum
The window expands to explore and shrinks to restore a violated invariant — it never restarts from scratch, which is exactly what makes it O(n) instead of O(n²)
Recognize the shape: "contiguous subarray/substring satisfying some condition" is the sliding-window tell
Opposite-end two pointers usually require sorted input first — sorting costs O(n log n), which then dominates the two-pointer O(n) scan

Longest substring without repeating characters

static int lengthOfLongestSubstring(String s) {
    Map<Character, Integer> lastSeen = new HashMap<>();
    int longest = 0, windowStart = 0;
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (lastSeen.containsKey(c) && lastSeen.get(c) >= windowStart) {
            windowStart = lastSeen.get(c) + 1;   // jump the window start past the duplicate
        }
        lastSeen.put(c, i);
        longest = Math.max(longest, i - windowStart + 1);
    }
    return longest;
}

Two-pointer variants
Shape	Pointer movement	Example
Opposite ends	lo++ or hi-- based on a comparison	two-sum on sorted array, container with most water
Fixed-size window	both ends move together, one step at a time	max sum of every k-length subarray
Variable-size window	right expands, left catches up when invalid	longest substring without repeats, minimum window substring