/*
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package org.apache.lucene.search;


import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.stream.Collectors;

import org.apache.lucene.index.Impact;
import org.apache.lucene.index.Impacts;
import org.apache.lucene.index.ImpactsSource;
import org.apache.lucene.index.Term;
import org.apache.lucene.search.similarities.Similarity.SimScorer;
import org.apache.lucene.util.FixedBitSet;

/**
 * Find all slop-valid position-combinations (matches)
 * encountered while traversing/hopping the PhrasePositions.
 * <br> The sloppy frequency contribution of a match depends on the distance:
 * <br> - highest freq for distance=0 (exact match).
 * <br> - freq gets lower as distance gets higher.
 * <br>Example: for query "a b"~2, a document "x a b a y" can be matched twice:
 * once for "a b" (distance=0), and once for "b a" (distance=2).
 * <br>Possibly not all valid combinations are encountered, because for efficiency
 * we always propagate the least PhrasePosition. This allows to base on
 * PriorityQueue and move forward faster.
 * As result, for example, document "a b c b a"
 * would score differently for queries "a b c"~4 and "c b a"~4, although
 * they really are equivalent.
 * Similarly, for doc "a b c b a f g", query "c b"~2
 * would get same score as "g f"~2, although "c b"~2 could be matched twice.
 * We may want to fix this in the future (currently not, for performance reasons).
 */
final class SloppyPhraseMatcher extends PhraseMatcher {

  private final PhrasePositions[] phrasePositions;

  private final int slop;
  private final int numPostings;
  private final PhraseQueue pq; // for advancing min position
  private final boolean captureLeadMatch;

  private final DocIdSetIterator approximation;
  private final ImpactsDISI impactsApproximation;

  private int end; // current largest phrase position

  private int leadPosition;
  private int leadOffset;
  private int leadEndOffset;
  private int leadOrd;

  private boolean hasRpts; // flag indicating that there are repetitions (as checked in first candidate doc)
  private boolean checkedRpts; // flag to only check for repetitions in first candidate doc
  private boolean hasMultiTermRpts; //  
  private PhrasePositions[][] rptGroups; // in each group are PPs that repeats each other (i.e. same term), sorted by (query) offset 
  private PhrasePositions[] rptStack; // temporary stack for switching colliding repeating pps 

  private boolean positioned;
  private int matchLength;

  SloppyPhraseMatcher(PhraseQuery.PostingsAndFreq[] postings, int slop, ScoreMode scoreMode, SimScorer scorer, float matchCost, boolean captureLeadMatch) {
    super(matchCost);
    this.slop = slop;
    this.numPostings = postings.length;
    this.captureLeadMatch = captureLeadMatch;
    pq = new PhraseQueue(postings.length);
    phrasePositions = new PhrasePositions[postings.length];
    for (int i = 0; i < postings.length; ++i) {
      phrasePositions[i] = new PhrasePositions(postings[i].postings, postings[i].position, i, postings[i].terms);
    }

    approximation = ConjunctionDISI.intersectIterators(Arrays.stream(postings).map(p -> p.postings).collect(Collectors.toList()));
    // What would be a good upper bound of the sloppy frequency? A sum of the
    // sub frequencies would be correct, but it is usually so much higher than
    // the actual sloppy frequency that it doesn't help skip irrelevant
    // documents. As a consequence for now, sloppy phrase queries use dummy
    // impacts:
    final ImpactsSource impactsSource = new ImpactsSource() {
      @Override
      public Impacts getImpacts() throws IOException {
        return new Impacts() {

          @Override
          public int numLevels() {
            return 1;
          }

          @Override
          public List<Impact> getImpacts(int level) {
            return Collections.singletonList(new Impact(Integer.MAX_VALUE, 1L));
          }

          @Override
          public int getDocIdUpTo(int level) {
            return DocIdSetIterator.NO_MORE_DOCS;
          }
        };
      }

      @Override
      public void advanceShallow(int target) throws IOException {}
    };
    impactsApproximation = new ImpactsDISI(approximation, impactsSource, scorer);
  }

  @Override
  DocIdSetIterator approximation() {
    return approximation;
  }

  @Override
  ImpactsDISI impactsApproximation() {
    return impactsApproximation;
  }

  @Override
  float maxFreq() throws IOException {
    // every term position in each postings list can be at the head of at most
    // one matching phrase, so the maximum possible phrase freq is the sum of
    // the freqs of the postings lists.
    float maxFreq = 0;
    for (PhrasePositions phrasePosition : phrasePositions) {
      maxFreq += phrasePosition.postings.freq();
    }
    return maxFreq;
  }

  @Override
  public void reset() throws IOException {
    this.positioned = initPhrasePositions();
    this.matchLength = Integer.MAX_VALUE;
    this.leadPosition = Integer.MAX_VALUE;
  }

  @Override
  float sloppyWeight() {
    return 1f / (1f + matchLength);
  }

  @Override
  public boolean nextMatch() throws IOException {
    if (!positioned) {
      return false;
    }
    PhrasePositions pp = pq.pop();
    assert pp != null;  // if the pq is not full, then positioned == false
    captureLead(pp);
    matchLength = end - pp.position;
    int next = pq.top().position; 
    while (advancePP(pp)) {
      if (hasRpts && !advanceRpts(pp)) {
        break; // pps exhausted
      }
      if (pp.position > next) { // done minimizing current match-length
        pq.add(pp);
        if (matchLength <= slop) {
          return true;
        }
        pp = pq.pop();
        next = pq.top().position;
        assert pp != null;  // if the pq is not full, then positioned == false
        matchLength = end - pp.position;
      } else {
        int matchLength2 = end - pp.position;
        if (matchLength2 < matchLength) {
          matchLength = matchLength2;
        }
      }
      captureLead(pp);
    }
    positioned = false;
    return matchLength <= slop;
  }

  private void captureLead(PhrasePositions pp) throws IOException {
    if (captureLeadMatch == false) {
      return;
    }
    leadOrd = pp.ord;
    leadPosition = pp.position + pp.offset;
    leadOffset = pp.postings.startOffset();
    leadEndOffset = pp.postings.endOffset();
  }

  @Override
  public int startPosition() {
    // when a match is detected, the top postings is advanced until it has moved
    // beyond its successor, to ensure that the match is of minimal width.  This
    // means that we need to record the lead position before it is advanced.
    // However, the priority queue doesn't guarantee that the top postings is in fact the
    // earliest in the list, so we need to cycle through all terms to check.
    // this is slow, but Matches is slow anyway...
    int leadPosition = this.leadPosition;
    for (PhrasePositions pp : phrasePositions) {
      leadPosition = Math.min(leadPosition, pp.position + pp.offset);
    }
    return leadPosition;
  }

  @Override
  public int endPosition() {
    int endPosition = leadPosition;
    for (PhrasePositions pp : phrasePositions) {
      if (pp.ord != leadOrd) {
        endPosition = Math.max(endPosition, pp.position + pp.offset);
      }
    }
    return endPosition;
  }

  @Override
  public int startOffset() throws IOException {
    // when a match is detected, the top postings is advanced until it has moved
    // beyond its successor, to ensure that the match is of minimal width.  This
    // means that we need to record the lead offset before it is advanced.
    // However, the priority queue doesn't guarantee that the top postings is in fact the
    // earliest in the list, so we need to cycle through all terms to check
    // this is slow, but Matches is slow anyway...
    int leadOffset = this.leadOffset;
    for (PhrasePositions pp : phrasePositions) {
      leadOffset = Math.min(leadOffset, pp.postings.startOffset());
    }
    return leadOffset;
  }

  @Override
  public int endOffset() throws IOException {
    int endOffset = leadEndOffset;
    for (PhrasePositions pp : phrasePositions) {
      if (pp.ord != leadOrd) {
        endOffset = Math.max(endOffset, pp.postings.endOffset());
      }
    }
    return endOffset;
  }

  /** advance a PhrasePosition and update 'end', return false if exhausted */
  private boolean advancePP(PhrasePositions pp) throws IOException {
    if (!pp.nextPosition()) {
      return false;
    }
    if (pp.position > end) {
      end = pp.position;
    }
    return true;
  }
  
  /** pp was just advanced. If that caused a repeater collision, resolve by advancing the lesser
   * of the two colliding pps. Note that there can only be one collision, as by the initialization
   * there were no collisions before pp was advanced.  */
  private boolean advanceRpts(PhrasePositions pp) throws IOException {
    if (pp.rptGroup < 0) {
      return true; // not a repeater
    }
    PhrasePositions[] rg = rptGroups[pp.rptGroup];
    FixedBitSet bits = new FixedBitSet(rg.length); // for re-queuing after collisions are resolved
    int k0 = pp.rptInd;
    int k;
    while((k=collide(pp)) >= 0) {
      pp = lesser(pp, rg[k]); // always advance the lesser of the (only) two colliding pps
      if (!advancePP(pp)) {
        return false; // exhausted
      }
      if (k != k0) { // careful: mark only those currently in the queue
        bits = FixedBitSet.ensureCapacity(bits, k);
        bits.set(k); // mark that pp2 need to be re-queued
      }
    }
    // collisions resolved, now re-queue
    // empty (partially) the queue until seeing all pps advanced for resolving collisions
    int n = 0;
    // TODO would be good if we can avoid calling cardinality() in each iteration!
    int numBits = bits.length(); // larges bit we set
    while (bits.cardinality() > 0) {
      PhrasePositions pp2 = pq.pop();
      rptStack[n++] = pp2;
      if (pp2.rptGroup >= 0 
          && pp2.rptInd < numBits  // this bit may not have been set
          && bits.get(pp2.rptInd)) {
        bits.clear(pp2.rptInd);
      }
    }
    // add back to queue
    for (int i=n-1; i>=0; i--) {
      pq.add(rptStack[i]);
    }
    return true;
  }

  /** compare two pps, but only by position and offset */
  private PhrasePositions lesser(PhrasePositions pp, PhrasePositions pp2) {
    if (pp.position < pp2.position ||
        (pp.position == pp2.position && pp.offset < pp2.offset)) {
      return pp;
    }
    return pp2;
  }

  /** index of a pp2 colliding with pp, or -1 if none */
  private int collide(PhrasePositions pp) {
    int tpPos = tpPos(pp);
    PhrasePositions[] rg = rptGroups[pp.rptGroup];
    for (int i=0; i<rg.length; i++) {
      PhrasePositions pp2 = rg[i];
      if (pp2 != pp && tpPos(pp2) == tpPos) {
        return pp2.rptInd;
      }
    }
    return -1;
  }

  /**
   * Initialize PhrasePositions in place.
   * A one time initialization for this scorer (on first doc matching all terms):
   * <ul>
   *  <li>Check if there are repetitions
   *  <li>If there are, find groups of repetitions.
   * </ul>
   * Examples:
   * <ol>
   *  <li>no repetitions: <b>"ho my"~2</b>
   *  <li>repetitions: <b>"ho my my"~2</b>
   *  <li>repetitions: <b>"my ho my"~2</b>
   * </ol>
   * @return false if PPs are exhausted (and so current doc will not be a match) 
   */
  private boolean initPhrasePositions() throws IOException {
    end = Integer.MIN_VALUE;
    if (!checkedRpts) {
      return initFirstTime();
    }
    if (!hasRpts) {
      initSimple();
      return true; // PPs available
    }
    return initComplex();
  }
  
  /** no repeats: simplest case, and most common. It is important to keep this piece of the code simple and efficient */
  private void initSimple() throws IOException {
    //System.err.println("initSimple: doc: "+min.doc);
    pq.clear();
    // position pps and build queue from list
    for (PhrasePositions pp : phrasePositions) {
      pp.firstPosition();
      if (pp.position > end) {
        end = pp.position;
      }
      pq.add(pp);
    }
  }
  
  /** with repeats: not so simple. */
  private boolean initComplex() throws IOException {
    //System.err.println("initComplex: doc: "+min.doc);
    placeFirstPositions();
    if (!advanceRepeatGroups()) {
      return false; // PPs exhausted
    }
    fillQueue();
    return true; // PPs available
  }

  /** move all PPs to their first position */
  private void placeFirstPositions() throws IOException {
    for (PhrasePositions pp : phrasePositions) {
      pp.firstPosition();
    }
  }

  /** Fill the queue (all pps are already placed */
  private void fillQueue() {
    pq.clear();
    for (PhrasePositions pp : phrasePositions) {  // iterate cyclic list: done once handled max
      if (pp.position > end) {
        end = pp.position;
      }
      pq.add(pp);
    }
  }

  /** At initialization (each doc), each repetition group is sorted by (query) offset.
   * This provides the start condition: no collisions.
   * <p>Case 1: no multi-term repeats<br>
   * It is sufficient to advance each pp in the group by one less than its group index.
   * So lesser pp is not advanced, 2nd one advance once, 3rd one advanced twice, etc.
   * <p>Case 2: multi-term repeats<br>
   * 
   * @return false if PPs are exhausted. 
   */
  private boolean advanceRepeatGroups() throws IOException {
    for (PhrasePositions[] rg: rptGroups) { 
      if (hasMultiTermRpts) {
        // more involved, some may not collide
        int incr;
        for (int i=0; i<rg.length; i+=incr) {
          incr = 1;
          PhrasePositions pp = rg[i];
          int k;
          while((k=collide(pp)) >= 0) {
            PhrasePositions pp2 = lesser(pp, rg[k]);
            if (!advancePP(pp2)) {  // at initialization always advance pp with higher offset
              return false; // exhausted
            }
            if (pp2.rptInd < i) { // should not happen?
              incr = 0;
              break;
            }
          }
        }
      } else {
        // simpler, we know exactly how much to advance
        for (int j=1; j<rg.length; j++) {
          for (int k=0; k<j; k++) {
            if (!rg[j].nextPosition()) {
              return false; // PPs exhausted
            }
          }
        }
      }
    }
    return true; // PPs available
  }
  
  /** initialize with checking for repeats. Heavy work, but done only for the first candidate doc.<p>
   * If there are repetitions, check if multi-term postings (MTP) are involved.<p>
   * Without MTP, once PPs are placed in the first candidate doc, repeats (and groups) are visible.<br>
   * With MTP, a more complex check is needed, up-front, as there may be "hidden collisions".<br>
   * For example P1 has {A,B}, P1 has {B,C}, and the first doc is: "A C B". At start, P1 would point
   * to "A", p2 to "C", and it will not be identified that P1 and P2 are repetitions of each other.<p>
   * The more complex initialization has two parts:<br>
   * (1) identification of repetition groups.<br>
   * (2) advancing repeat groups at the start of the doc.<br>
   * For (1), a possible solution is to just create a single repetition group, 
   * made of all repeating pps. But this would slow down the check for collisions, 
   * as all pps would need to be checked. Instead, we compute "connected regions" 
   * on the bipartite graph of postings and terms.  
   */
  private boolean initFirstTime() throws IOException {
    //System.err.println("initFirstTime: doc: "+min.doc);
    checkedRpts = true;
    placeFirstPositions();

    LinkedHashMap<Term,Integer> rptTerms = repeatingTerms(); 
    hasRpts = !rptTerms.isEmpty();

    if (hasRpts) {
      rptStack = new PhrasePositions[numPostings]; // needed with repetitions
      ArrayList<ArrayList<PhrasePositions>> rgs = gatherRptGroups(rptTerms);
      sortRptGroups(rgs);
      if (!advanceRepeatGroups()) {
        return false; // PPs exhausted
      }
    }
    
    fillQueue();
    return true; // PPs available
  }

  /** sort each repetition group by (query) offset. 
   * Done only once (at first doc) and allows to initialize faster for each doc. */
  private void sortRptGroups(ArrayList<ArrayList<PhrasePositions>> rgs) {
    rptGroups = new PhrasePositions[rgs.size()][];
    Comparator<PhrasePositions> cmprtr = new Comparator<PhrasePositions>() {
      @Override
      public int compare(PhrasePositions pp1, PhrasePositions pp2) {
        return pp1.offset - pp2.offset;
      }
    };
    for (int i=0; i<rptGroups.length; i++) {
      PhrasePositions[] rg = rgs.get(i).toArray(new PhrasePositions[0]);
      Arrays.sort(rg, cmprtr);
      rptGroups[i] = rg;
      for (int j=0; j<rg.length; j++) {
        rg[j].rptInd = j; // we use this index for efficient re-queuing
      }
    }
  }

  /** Detect repetition groups. Done once - for first doc */
  private ArrayList<ArrayList<PhrasePositions>> gatherRptGroups(LinkedHashMap<Term,Integer> rptTerms) throws IOException {
    PhrasePositions[] rpp = repeatingPPs(rptTerms); 
    ArrayList<ArrayList<PhrasePositions>> res = new ArrayList<>();
    if (!hasMultiTermRpts) {
      // simpler - no multi-terms - can base on positions in first doc
      for (int i=0; i<rpp.length; i++) {
        PhrasePositions pp = rpp[i];
        if (pp.rptGroup >=0) continue; // already marked as a repetition
        int tpPos = tpPos(pp);
        for (int j=i+1; j<rpp.length; j++) {
          PhrasePositions pp2 = rpp[j];
          if (
              pp2.rptGroup >=0        // already marked as a repetition
              || pp2.offset == pp.offset // not a repetition: two PPs are originally in same offset in the query! 
              || tpPos(pp2) != tpPos) {  // not a repetition
            continue; 
          }
          // a repetition
          int g = pp.rptGroup;
          if (g < 0) {
            g = res.size();
            pp.rptGroup = g;  
            ArrayList<PhrasePositions> rl = new ArrayList<>(2);
            rl.add(pp);
            res.add(rl); 
          }
          pp2.rptGroup = g;
          res.get(g).add(pp2);
        }
      }
    } else {
      // more involved - has multi-terms
      ArrayList<HashSet<PhrasePositions>> tmp = new ArrayList<>();
      ArrayList<FixedBitSet> bb = ppTermsBitSets(rpp, rptTerms);
      unionTermGroups(bb);
      HashMap<Term,Integer> tg = termGroups(rptTerms, bb);
      HashSet<Integer> distinctGroupIDs = new HashSet<>(tg.values());
      for (int i=0; i<distinctGroupIDs.size(); i++) {
        tmp.add(new HashSet<PhrasePositions>());
      }
      for (PhrasePositions pp : rpp) {
        for (Term t: pp.terms) {
          if (rptTerms.containsKey(t)) {
            int g = tg.get(t);
            tmp.get(g).add(pp);
            assert pp.rptGroup==-1 || pp.rptGroup==g;  
            pp.rptGroup = g;
          }
        }
      }
      for (HashSet<PhrasePositions> hs : tmp) {
        res.add(new ArrayList<>(hs));
      }
    }
    return res;
  }

  /** Actual position in doc of a PhrasePosition, relies on that position = tpPos - offset) */
  private final int tpPos(PhrasePositions pp) {
    return pp.position + pp.offset;
  }

  /** find repeating terms and assign them ordinal values */
  private LinkedHashMap<Term,Integer> repeatingTerms() {
    LinkedHashMap<Term,Integer> tord = new LinkedHashMap<>();
    HashMap<Term,Integer> tcnt = new HashMap<>();
    for (PhrasePositions pp : phrasePositions) {
      for (Term t : pp.terms) {
        Integer cnt = tcnt.compute(t, (key, old) -> old == null ? 1 : 1 + old);
        if (cnt==2) {
          tord.put(t,tord.size());
        }
      }
    }
    return tord;
  }

  /** find repeating pps, and for each, if has multi-terms, update this.hasMultiTermRpts */
  private PhrasePositions[] repeatingPPs(HashMap<Term,Integer> rptTerms) {
    ArrayList<PhrasePositions> rp = new ArrayList<>();
    for (PhrasePositions pp : phrasePositions) {
      for (Term t : pp.terms) {
        if (rptTerms.containsKey(t)) {
          rp.add(pp);
          hasMultiTermRpts |= (pp.terms.length > 1);
          break;
        }
      }
    }
    return rp.toArray(new PhrasePositions[0]);
  }
  
  /** bit-sets - for each repeating pp, for each of its repeating terms, the term ordinal values is set */
  private ArrayList<FixedBitSet> ppTermsBitSets(PhrasePositions[] rpp, HashMap<Term,Integer> tord) {
    ArrayList<FixedBitSet> bb = new ArrayList<>(rpp.length);
    for (PhrasePositions pp : rpp) {
      FixedBitSet b = new FixedBitSet(tord.size());
      Integer ord;
      for (Term t: pp.terms) {
        if ((ord=tord.get(t))!=null) {
          b.set(ord);
        }
      }
      bb.add(b);
    }
    return bb;
  }
  
  /** union (term group) bit-sets until they are disjoint (O(n^^2)), and each group have different terms */
  private void unionTermGroups(ArrayList<FixedBitSet> bb) {
    int incr;
    for (int i=0; i<bb.size()-1; i+=incr) {
      incr = 1;
      int j = i+1;
      while (j<bb.size()) {
        if (bb.get(i).intersects(bb.get(j))) {
          bb.get(i).or(bb.get(j));
          bb.remove(j);
          incr = 0;
        } else {
          ++j;
        }
      }
    }
  }
  
  /** map each term to the single group that contains it */ 
  private HashMap<Term,Integer> termGroups(LinkedHashMap<Term,Integer> tord, ArrayList<FixedBitSet> bb) throws IOException {
    HashMap<Term,Integer> tg = new HashMap<>();
    Term[] t = tord.keySet().toArray(new Term[0]);
    for (int i=0; i<bb.size(); i++) { // i is the group no.
      FixedBitSet bits = bb.get(i);
      for (int ord = bits.nextSetBit(0); ord != DocIdSetIterator.NO_MORE_DOCS; ord = ord + 1 >= bits.length() ? DocIdSetIterator.NO_MORE_DOCS : bits.nextSetBit(ord + 1)) {
        tg.put(t[ord],i);
      }
    }
    return tg;
  }

}
