583. Delete Operation for Two Strings - Solved in Java/C#/Python/C++/JavaScript/Go/Ruby

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LeetCode link: 583. Delete Operation for Two Strings, difficulty: Medium.

Given two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.

In one step, you can delete exactly one character in either string.

Example 1:

Input: word1 = "sea", word2 = "eat"

Output: 2

Explanation:

You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

Example 2:

Input: word1 = "leetcode", word2 = "etco"

Output: 4

Constraints:

1 <= word1.length, word2.length <= 500
word1 and word2 consist of only lowercase English letters.

Intuition
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It is a question of comparing two strings which is about dealing with "two swappable arrays".
After doing similar questions many times, we will form the intuition of using two-dimensional arrays for dynamic programming.

Pattern of "Dynamic Programming"

"Dynamic Programming" requires the use of the dp array to store the results. The value of dp[i][j] can be converted from its previous (or multiple) values through a formula. Therefore, the value of dp[i][j] is derived step by step, and it is related to the previous dp record value.

"Dynamic programming" is divided into five steps

Determine the meaning of each value of the array dp.
Initialize the value of the array dp.
Fill in the dp grid data in order according to an example.
Based on the dp grid data, derive the recursive formula.
Write a program and print the dp array. If it is not as expected, adjust it.

Detailed description of these five steps

Determine the meaning of each value of the array dp.
- First determine whether dp is a one-dimensional array or a two-dimensional array. A one-dimensional rolling array means that the values of the array are overwritten at each iteration. Most of the time, using one-dimensional rolling array instead of two-dimensional array can simplify the code; but for some problems, such as operating "two swappable arrays", for the sake of ease of understanding, it is better to use two-dimensional array.
- Try to use the meaning of the return value required by the problem as the meaning of dp[i] (one-dimensional) or dp[i][j] (two-dimensional). It works about 60% of the time. If it doesn't work, try other meanings.
- Try to save more information in the design. Repeated information only needs to be saved once in a dp[i].
- Use simplified meanings. If the problem can be solved with boolean value, don't use numeric value.
Initialize the value of the array dp. The value of dp involves two levels:
1. The length of dp. Usually: condition array length plus 1 or condition array length.
2. The value of dp[i] or dp[i][j]. dp[0] or dp[0][0] sometimes requires special treatment.
Fill in the dp grid data in order according to an example.
- The "recursive formula" is the core of the "dynamic programming" algorithm. But the "recursive formula" is obscure. If you want to get it, you need to make a table and use data to inspire yourself.
- If the original example is not good enough, you need to redesign one yourself.
- According to the example, fill in the dp grid data "in order", which is very important because it determines the traversal order of the code.
- Most of the time, from left to right, from top to bottom. But sometimes it is necessary to traverse from right to left, from bottom to top, from the middle to the right (or left), such as the "palindrome" problems. Sometimes, it is necessary to traverse a line twice, first forward and then backward.
- When the order is determined correctly, the starting point is determined. Starting from the starting point, fill in the dp grid data "in order". This order is also the order in which the program processes.
- In this process, you will get inspiration to write a "recursive formula". If you can already derive the formula, you do not need to complete the grid.
Based on the dp grid data, derive the recursive formula.
- There are three special positions to pay attention to: dp[i - 1][j - 1], dp[i - 1][j] and dp[i][j - 1], the current dp[i][j] often depends on them.
- When operating "two swappable arrays", due to symmetry, we may need to use dp[i - 1][j] and dp[i][j - 1] at the same time.
Write a program and print the dp array. If it is not as expected, adjust it.
- Focus on analyzing those values that are not as expected.

After reading the above, do you feel that "dynamic programming" is not that difficult? Try to solve this problem. 🤗

Step by Step Solutions

Determine the meaning of the dp[i][j].
- dp[i][j] represents the minimum number of steps required to make word1's first i letters and word2's first j letters the same.
- dp[i][j] is an integer.
Determine the dp array's initial value.
- Use an example:
```
After initialization, the 'dp' array would be:  
#     e a t
#   0 1 2 3 # dp[0]
# s 1 0 0 0
# e 2 0 0 0
# a 3 0 0 0
```
- dp[0][j] = j, because dp[0] represents the empty string, and the number of steps is just the number of chars to be deleted.
- dp[i][0] = i, the reason is the same as the previous line, just viewed in vertical direction.

Fill in the dp grid data "in order" according to an example.

1. word1 = "s", word2 = "eat"
#     e a t
#   0 1 2 3
# s 1 2 3 4 # dp[1]

2. word1 = "se", word2 = "eat"
#     e a t
#   0 1 2 3
# s 1 2 3 4
# e 2 1 2 3

3. word1 = "sea", word2 = "eat"
#     e a t
#   0 1 2 3
# s 1 2 3 4
# e 2 1 2 3
# a 3 2 1 2

Based on the dp grid data, derive the "recursive formula".

if word1[i - 1] == word2[j - 1]
    dp[i][j] = dp[i - 1][j - 1]
else
    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1

Write a program and print the dp array. If it is not as expected, adjust it.

Complexity

Time complexity

O(N * M)

Space complexity

O(N * M)

Java #

class Solution {
    public int minDistance(String word1, String word2) {
        var dp = new int[word1.length() + 1][word2.length() + 1];
        for (var i = 0; i < dp.length; i++) {
            dp[i][0] = i;
        }
        for (var j = 0; j < dp[0].length; j++) {
            dp[0][j] = j;
        }

        for (var i = 1; i < dp.length; i++) {
            for (var j = 1; j < dp[0].length; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;
                }
            }
        }

        return dp[dp.length - 1][dp[0].length - 1];
    }
}

C# #

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

        for (var i = 0; i < dp.GetLength(0); i++)
            dp[i, 0] = i;

        for (var j = 0; j < dp.GetLength(1); j++)
            dp[0, j] = j;

        for (var i = 1; i < dp.GetLength(0); i++)
        {
            for (var j = 1; j < dp.GetLength(1); j++)
            {
                if (word1[i - 1] == word2[j - 1])
                {
                    dp[i, j] = dp[i - 1, j - 1];
                }
                else
                {
                    dp[i, j] = Math.Min(dp[i - 1, j], dp[i, j - 1]) + 1;
                }
            }
        }

        return dp[dp.GetUpperBound(0), dp.GetUpperBound(1)];
    }
}

Python #

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        dp = [[0] * (len(word2) + 1) for _ in range(len(word1) + 1)]
        for i in range(len(dp)):
            dp[i][0] = i
        for j in range(len(dp[0])):
            dp[0][j] = j

        for i in range(1, len(dp)):
            for j in range(1, len(dp[0])):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1

        return dp[-1][-1]

C++ #

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
        for (auto i = 0; i < dp.size(); i++) {
            dp[i][0] = i;
        }
        for (auto j = 0; j < dp[0].size(); j++) {
            dp[0][j] = j;
        }

        for (auto i = 1; i < dp.size(); i++) {
            for (auto j = 1; j < dp[0].size(); j++) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;
                }
            }
        }

        return dp[dp.size() - 1][dp[0].size() - 1];
    }
};

JavaScript #

var minDistance = function (word1, word2) {
  const dp = Array(word1.length + 1).fill().map(
    () => Array(word2.length + 1).fill(0)
  )
  dp.forEach((_, i) => { dp[i][0] = i })
  dp[0].forEach((_, j) => { dp[0][j] = j })

  for (let i = 1; i < dp.length; i++) {
    for (let j = 1; j < dp[0].length; j++) {
      if (word1[i - 1] == word2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1]
      } else {
        dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1
      }
    }
  }

  return dp.at(-1).at(-1)
};

Go #

func minDistance(word1 string, word2 string) int {
    dp := make([][]int, len(word1) + 1)
    for i := range dp {
        dp[i] = make([]int, len(word2) + 1)
        dp[i][0] = i
    }
    for j := range dp[0] {
        dp[0][j] = j
    }

    for i := 1; i < len(dp); i++ {
        for j := 1; j < len(dp[0]); j++ {
            if word1[i - 1] == word2[j - 1] {
                dp[i][j] = dp[i - 1][j - 1]
            } else {
                dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1
            }
        }
    }

    return dp[len(dp) - 1][len(dp[0]) - 1]
}

Ruby #

def min_distance(word1, word2)
  dp = Array.new(word1.size + 1) do
    Array.new(word2.size + 1, 0)
  end
  dp.each_with_index do |_, i|
    dp[i][0] = i
  end
  dp[0].each_with_index do |_, j|
    dp[0][j] = j
  end

  (1...dp.size).each do |i|
    (1...dp[0].size).each do |j|
      dp[i][j] =
        if word1[i - 1] == word2[j - 1]
          dp[i - 1][j - 1]
        else
          [ dp[i - 1][j], dp[i][j - 1] ].min + 1
        end
    end
  end

  dp[-1][-1]
end

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