CPMagical CP Library

Problem solution · Java

Groups of Strings

Groups of Strings: a Java solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Disjoint set union
Source
walkccc LeetCode Solutions
Length
81 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Disjoint set union

For Groups of Strings, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 81 lines of Java from the credited upstream file 2157.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 5 loop blocks detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeGroups of Strings · JavaJava
Use this to learn the idea, then write your own version.
class UnionFind {  public UnionFind(int n) {    count = n;    id = new int[n];    sz = new int[n];    for (int i = 0; i < n; ++i)      id[i] = i;    for (int i = 0; i < n; ++i)      sz[i] = 1;  }   public void unionBySize(int u, int v) {    final int i = find(u);    final int j = find(v);    if (i == j)      return;    if (sz[i] < sz[j]) {      sz[j] += sz[i];      id[i] = j;    } else {      sz[i] += sz[j];      id[j] = i;    }    --count;  }   public int getCount() {    return count;  }   public int getMaxSize() {    return Arrays.stream(sz).max().getAsInt();  }   private int count;  private int[] id;  private int[] sz;   private int find(int u) {    return id[u] == u ? u : (id[u] = find(id[u]));  }} class Solution {  public int[] groupStrings(String[] words) {    UnionFind uf = new UnionFind(words.length);    Map<Integer, Integer> maskToIndex = new HashMap<>();    Map<Integer, Integer> deletedMaskToIndex = new HashMap<>();     for (int i = 0; i < words.length; ++i) {      final int mask = getMask(words[i]);      for (int j = 0; j < 26; ++j)        if ((mask >> j & 1) == 1) {          // Going to delete this bit.          final int m = mask ^ 1 << j;          if (maskToIndex.containsKey(m))            uf.unionBySize(i, maskToIndex.get(m));          if (deletedMaskToIndex.containsKey(m))            uf.unionBySize(i, deletedMaskToIndex.get(m));          else            deletedMaskToIndex.put(m, i);        } else {          // Going to add this bit.          final int m = mask | 1 << j;          if (maskToIndex.containsKey(m))            uf.unionBySize(i, maskToIndex.get(m));        }      maskToIndex.put(mask, i);    }     return new int[] {uf.getCount(), uf.getMaxSize()};  }   private int getMask(final String s) {    int mask = 0;    for (final char c : s.toCharArray())      mask |= 1 << c - 'a';    return mask;  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗