Delete Operation for Two Strings, by LeetCode: a variation of edit distance via DP

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Problem is this one: https://leetcode.com/problems/delete-operation-for-two-strings/:

Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.
Example 1:
Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

This is a variation of the classic String Edit Distance problem, whose solution is one of the iconic Dynamic Programming solutions. Knowing that this is a variation, all it is needed is a paper (or whiteboard) and try few variations to understand the logic: having a distance matrix where each element distance[i,j] represents the min distance when comparing word1[0..i] and word2[0..j]. The fun part is to figure out the logic of how to update distance[i,j] depending on whether word1[i] is equal or not to word2[j]. You kind of figure that our with few examples. Here is the raw whiteboard that I used, and on the side is my dog closely paying attention to my explanation. Code is down below, cheers, Marcelo.




public class Solution
{
public int MinDistance(string word1, string word2)
{
int[,] distance = new int[word1.Length + 1, word2.Length + 1];

 distance[0, 0] = 0;
for (int r = 1; r <= word1.Length; r++) distance[r, 0] = r;
for (int c = 1; c <= word2.Length; c++) distance[0, c] = c;
for (int r = 1; r <= word1.Length; r++)
{
for (int c = 1; c <= word2.Length; c++)
{
if (word1[r - 1] == word2[c - 1])
{
distance[r, c] = distance[r - 1, c - 1];
}
else
{
distance[r, c] = Math.Min(distance[r - 1, c], distance[r, c - 1]) + 1;
}
}
}

 return distance[word1.Length, word2.Length];
}
}

Comments

  1. TarasMarch 7, 2018 at 11:51 PM

    Very nice, Marcelo! It's possible to improve this solution by reducing the space complexity which is O(N^2) in your case. If you look at the access pattern you'll notice that you're actually using only 2 rows - current and previous, hence solution can be something like:

    class Solution {
    public:
    int minDistance(const string& word1, const string& word2) {
    vector prev(word2.size()+1);
    vector current(word2.size()+1);
    for (int j = 0; j <= word2.size(); ++j) prev[j] = j;
    for (int i = 1; i <= word1.size(); ++i) {
    current[0] = i;
    for (int j = 1; j <= word2.size(); ++j) {
    current[j] = word1[i-1] == word2[j-1] ? prev[j-1] : 1 + min(current[j-1], prev[j]);
    }
    swap(current, prev);
    }
    return prev[word2.size()];
    }
    };

    Cheers,
    Taras

    ReplyDelete
    Replies
    1. TarasMarch 7, 2018 at 11:53 PM

      and for those who don't like using anything but arrays:

      class Solution {
      public:
      int minDistance(const string& word1, const string& word2) {
      int d[2][word2.size()+1];
      int prev = 0;
      int current = 1;
      for (int j = 0; j <= word2.size(); ++j) d[prev][j] = j;
      for (int i = 1; i <= word1.size(); ++i) {
      d[current][0] = i;
      for (int j = 1; j <= word2.size(); ++j) {
      d[current][j] = word1[i-1] == word2[j-1] ? d[prev][j-1] : 1 + min(d[current][j-1], d[prev][j]);
      }
      swap(current, prev);
      }
      return d[prev][word2.size()];
      }
      };

      Delete
  2. Rahul ManghnaniJuly 22, 2020 at 2:22 PM

    A video explanation: https://youtu.be/-wxbv5l3Poc

    ReplyDelete

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