HashMap & HashSet Problems

Sort, fix one element, then two-pointer the rest inward, skipping duplicates to keep triples unique.

Sort first, fix one element, then run opposite-end two pointers on the rest, skipping duplicates, to collect all triplets in O(n squared).

Track each key's frequency, group keys into frequency buckets of insertion order, and keep a minimum-frequency pointer so the least-used, oldest-tied entry is evicted in constant time.

Add each element to a HashSet; the first time add returns false you have a duplicate.

Follow values as next-index pointers; the duplicate creates a cycle, and Floyd's algorithm finds the cycle's entry, which is the duplicate.

Traverse the graph keeping a HashMap from original to clone; create each clone once and wire neighbors through the map to handle cycles and shared nodes.

Sort, then at each level skip a value equal to its predecessor to drop duplicate subsets.

Sort, then skip a duplicate value whose equal predecessor is unused at this branch to avoid repeated permutations.

Advance index by one (use-once) and skip same-level duplicates to get unique combinations from a multiset.

Pick the pattern that matches the problem – sliding window for subarrays, HashMap for lookups, DFS/BFS for graphs and trees, DP for optimization – and always state the time and space complexity.

Put everything in a set, start counting only from numbers whose predecessor is absent, and expand forward.

Place each value at its home index; the first time the home already holds that same value, you have found the duplicate.

Place v at index v-1; every stranded value where value != index+1 is a duplicate.

After cyclic placement, the one mismatched slot gives the duplicate (its value) and the missing (its index+1) together.

In sorted data, duplicates are neighbours, so one slow-fast sweep keeps only the first of each run.

A HashMap turns a key into a bucket index so you can fetch the value in one jump instead of searching.

For each value, look up target minus value in a HashMap of seen values; a hit is the answer, otherwise record the value and move on.