Contains Nearby Duplicate

L3 Easy Arrays & Hashing

Concept

Storing "last seen index" inside a dictionary turns a closeness question into constant-time lookups as you sweep the array.

Given nums and a distance k, return true if the same integer appears at two different indices that are at most k apart. Return false otherwise.

Examples

▸ nums = [1, 2, 3, 1], k = 3

→ true

▸ nums = [1, 0, 1, 1], k = 1

→ true

▸ nums = [1, 2, 3, 1, 2, 3], k = 2

→ false

Progressive Hints

Hint 1 · Nudge

Remember where you last saw each value, not just that you saw it.

Hint 2 · Plan

Keep a map from value to its most recent index. For each position: if the value was seen before and the gap between the two indexes is at most k, return true. Always record the current index for that value as you go.

Hint 3 · Approach

last = map from value to index. For each i, n: if n is in last and i - last[n] <= k, return true. Then last[n] = i. Return false.

Output

// Run your code to see the output here.