Build buckets keyed by a shared field, preserving the first-seen key order.

Algorithm

Canonical pairs (a,1), (b,2), (a,3), (c,4), (b,5) print {a: [1, 3], b: [2, 5], c: [4]}. The replay uses the same input in every language, so this Java DSA implementation can be compared directly with the rest of the DSA track.

bucket map Each key owns a list. A new key creates a bucket; a repeated key appends to the existing bucket.

Basic Implementation

Basic.java
Replay: real traced execution (multi-file project)
import java.util.*;

public class Basic {
    public static void main(String[] args) {
        String[] keys = {"a", "b", "a", "c", "b"};
        int[] values = {1, 2, 3, 4, 5};
        Map<String, List<Integer>> groups = new LinkedHashMap<>();
        for (int i = 0; i < keys.length; i++) {
            groups.computeIfAbsent(keys[i], key -> new ArrayList<>()).add(values[i]);
        }
        List<String> parts = new ArrayList<>();
        for (Map.Entry<String, List<Integer>> entry : groups.entrySet()) {
            parts.add(entry.getKey() + ": " + entry.getValue());
        }
        System.out.println("{" + String.join(", ", parts) + "}");
    }
}
  1. pairs ← [(a, 1), (b, 2), (a, 3), (c, 4), (b, 5)]

    1import java.util.*;
    values this step[(a, 1), (b, 2), (a, 3), (c, 4), (b, 5)]pairs
  2. groups ← {}

    6int[] values = {1, 2, 3, 4, 5};7Map<String, List<Integer>> groups = new LinkedHashMap<>();8for (int i = 0; i < keys.length; i++) {
    values this step{}groups
  3. groups ← {a: [1]}

    6int[] values = {1, 2, 3, 4, 5};7Map<String, List<Integer>> groups = new LinkedHashMap<>();8for (int i = 0; i < keys.length; i++) {
    values this step{} {a: [1]}groupsakey1value
  4. groups ← {a: [1], b: [2]}

    6int[] values = {1, 2, 3, 4, 5};7Map<String, List<Integer>> groups = new LinkedHashMap<>();8for (int i = 0; i < keys.length; i++) {
    values this step{a: [1]} {a: [1], b: [2]}groupsbkey2value
  5. groups ← {a: [1, 3], b: [2]}

    6int[] values = {1, 2, 3, 4, 5};7Map<String, List<Integer>> groups = new LinkedHashMap<>();8for (int i = 0; i < keys.length; i++) {
    values this step{a: [1], b: [2]} {a: [1, 3], b: [2]}groupsakey3value
  6. groups ← {a: [1, 3], b: [2], c: [4]}

    6int[] values = {1, 2, 3, 4, 5};7Map<String, List<Integer>> groups = new LinkedHashMap<>();8for (int i = 0; i < keys.length; i++) {
    values this step{a: [1, 3], b: [2]} {a: [1, 3], b: [2], c: [4]}groupsckey4value
  7. groups ← {a: [1, 3], b: [2, 5], c: [4]}

    6int[] values = {1, 2, 3, 4, 5};7Map<String, List<Integer>> groups = new LinkedHashMap<>();8for (int i = 0; i < keys.length; i++) {
    values this step{a: [1, 3], b: [2], c: [4]} {a: [1, 3], b: [2, 5], c: [4]}groupsbkey5value
  8. stdout ← {a: [1, 3], b: [2, 5], c: [4]}

    14    }15    System.out.println("{" + String.join(", ", parts) + "}");16}
    values this step{a: [1, 3], b: [2, 5], c: [4]}stdout{a: [1, 3], b: [2, 5], c: [4]}groups
  9. bucket 1 after collision ← c -> a, degradation risk ← long chains can degrade lookup toward O(n)

    14    }15    System.out.println("{" + String.join(", ", parts) + "}");16}
    values this stepa c -> abucket 1 after collisionlong chains can degrade lookup toward O(n)degradation riskresize or rehash when load factor growsmitigationcnew key

Complexity

  • Time: O(n) average
  • Space: O(k + n) for buckets and values

Implementation notes

  • Keep output formatting deterministic. Do not rely on unordered hash-map printing when the lesson needs cross-language comparison.
  • The trace highlights the hash table state after each write and includes a collision contrast where one bucket chain grows, showing why long chains can degrade lookup and why real tables resize or rehash.