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// Copyright 2011 Google Inc. All Rights Reserved.
//
// Licensed 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.
#include "edit_distance.h"
#include <algorithm>
#include <vector>
int EditDistance(const StringPiece& s1,
const StringPiece& s2,
bool allow_replacements,
int max_edit_distance) {
// The algorithm implemented below is the "classic"
// dynamic-programming algorithm for computing the Levenshtein
// distance, which is described here:
//
// http://en.wikipedia.org/wiki/Levenshtein_distance
//
// Although the algorithm is typically described using an m x n
// array, only one row plus one element are used at a time, so this
// implementation just keeps one vector for the row. To update one entry,
// only the entries to the left, top, and top-left are needed. The left
// entry is in row[x-1], the top entry is what's in row[x] from the last
// iteration, and the top-left entry is stored in previous.
int m = s1.len_;
int n = s2.len_;
vector<int> row(n + 1);
for (int i = 1; i <= n; ++i)
row[i] = i;
for (int y = 1; y <= m; ++y) {
row[0] = y;
int best_this_row = row[0];
int previous = y - 1;
for (int x = 1; x <= n; ++x) {
int old_row = row[x];
if (allow_replacements) {
row[x] = min(previous + (s1.str_[y - 1] == s2.str_[x - 1] ? 0 : 1),
min(row[x - 1], row[x]) + 1);
}
else {
if (s1.str_[y - 1] == s2.str_[x - 1])
row[x] = previous;
else
row[x] = min(row[x - 1], row[x]) + 1;
}
previous = old_row;
best_this_row = min(best_this_row, row[x]);
}
if (max_edit_distance && best_this_row > max_edit_distance)
return max_edit_distance + 1;
}
return row[n];
}
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