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Problem solution · Java

Frequencies of Shortest Supersequences

Frequencies of Shortest Supersequences: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
141 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

Hash-based lookup

For Frequencies of Shortest Supersequences, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 141 lines of Java from the credited upstream file 3435.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 16 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeFrequencies of Shortest Supersequences · JavaJava
Use this to learn the idea, then write your own version.
enum State { INIT, VISITING, VISITED } class Solution {  public List<List<Integer>> supersequences(String[] words) {    List<List<Integer>> ans = new ArrayList<>();    List<int[]> edges = getEdges(words);    List<Integer> nodes = getNodes(edges);    int[] letterToIndex = getLetterToIndex(nodes);    List<Integer>[] graph = new List[nodes.size()];     for (int i = 0; i < nodes.size(); i++)      graph[i] = new ArrayList<>();     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      graph[letterToIndex[u]].add(letterToIndex[v]);    }     for (List<Integer> doubledSubset : getMinimumSubsets(graph)) {      int[] freq = new int[26];      for (final int letter : nodes)        freq[letter] = 1;      for (final int index : doubledSubset)        freq[nodes.get(index)] = 2;      ans.add(Arrays.stream(freq).boxed().collect(Collectors.toList()));    }     return ans;  }   // Returns a list of the minimum subsets of nodes that do not create a cycle  // when skipped.  private List<List<Integer>> getMinimumSubsets(List<Integer>[] graph) {    final int n = graph.length;    List<List<Integer>> res = new ArrayList<>();     for (int subsetSize = 0; subsetSize <= n; ++subsetSize) {      boolean[] combination = new boolean[n];      Arrays.fill(combination, n - subsetSize, n, true);      do {        List<Integer> doubledSubset = new ArrayList<>();        for (int i = 0; i < n; i++)          if (combination[i])            doubledSubset.add(i);        if (!hasCycleSkipping(graph, new HashSet<>(doubledSubset)))          res.add(doubledSubset);      } while (nextPermutation(combination));      if (!res.isEmpty())        return res;    }    return res;  }   // Returns true if there is a cycle in the `graph` when skipping any edges  // whose both endpoints are in `doubledSubset`.  private boolean hasCycleSkipping(List<Integer>[] graph, Set<Integer> doubledSubset) {    State[] states = new State[graph.length];    for (int i = 0; i < graph.length; ++i)      if (hasCycle(graph, i, states, doubledSubset))        return true;    return false;  }   private boolean hasCycle(List<Integer>[] graph, int u, State[] states,                           Set<Integer> doubledSubset) {    if (states[u] == State.VISITING)      return true;    if (states[u] == State.VISITED)      return false;    states[u] = State.VISITING;    if (!doubledSubset.contains(u))      for (final int v : graph[u])        if (!doubledSubset.contains(v) && hasCycle(graph, v, states, doubledSubset))          return true;    states[u] = State.VISITED;    return false;  }   private List<int[]> getEdges(String[] words) {    List<int[]> edges = new ArrayList<>();    for (final String word : words)      edges.add(new int[] {word.charAt(0) - 'a', word.charAt(1) - 'a'});    return edges;  }   private List<Integer> getNodes(List<int[]> edges) {    TreeSet<Integer> nodes = new TreeSet<>();    for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      nodes.add(u);      nodes.add(v);    }    return new ArrayList<>(nodes);  }   private int[] getLetterToIndex(List<Integer> nodes) {    int[] letterToIndex = new int[26];    for (int i = 0; i < nodes.size(); ++i)      letterToIndex[nodes.get(i)] = i;    return letterToIndex;  }   private boolean nextPermutation(boolean[] arr) {    final int n = arr.length;     // From back to front, find the first false followed by true    int i;    for (i = n - 2; i >= 0; --i)      if (!arr[i] && arr[i + 1])        break;     // If no such pair found, we've reached the last permutation    if (i < 0)      return false;     // From back to front, find the first true to swap with arr[i].    for (int j = n - 1; j > i; --j)      if (arr[j] && !arr[i]) {        swap(arr, i, j);        break;      }     // Reverse arr[i + 1..n - 1].    reverse(arr, i + 1, n - 1);    return true;  }   private void reverse(boolean[] arr, int l, int r) {    while (l < r)      swap(arr, l++, r--);  }   private void swap(boolean[] arr, int i, int j) {    boolean temp = arr[i];    arr[i] = arr[j];    arr[j] = temp;  }}

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