583. Delete Operation for Two Strings

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583. Delete Operation for Two Strings

Description

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:

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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:

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Input: word1 = "leetcode", word2 = "etco"
Output: 4

Constraints:

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

Hints/Notes

  • 2024/07/17
  • dp
  • No solution from 0x3F

Solution

Language: C++

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class Solution {
public:
    string word1_, word2_;
    vector<vector<int>> dp;

    int minDistance(string word1, string word2) {
        word1_ = word1;
        word2_ = word2;
        dp.resize(word1.size(), vector<int>(word2.size(), -1));
        int l = traverse(0, 0);
        return word1.size() - l + word2.size() - l;
    }

    int traverse(int i, int j) {
        if (i == word1_.size() || j == word2_.size()) {
            return 0;
        }
        if (dp[i][j] != -1) {
            return dp[i][j];
        }
        int res = 0;
        if (word1_[i] == word2_[j]) {
            res = 1 + traverse(i + 1, j + 1);
        } else {
            res = max(traverse(i, j + 1), traverse(i + 1, j));
        }
        dp[i][j] = res;
        return res;
    }
};

The other dp approach:

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class Solution {
public:
    string word1_, word2_;
    vector<vector<int>> dp;

    int minDistance(string word1, string word2) {
        word1_ = word1;
        word2_ = word2;
        dp.resize(word1.size(), vector<int>(word2.size(), -1));
        int l = traverse(0, 0);
        return l;
    }

    int traverse(int i, int j) {
        if (i == word1_.size()) {
            return word2_.size() - j;
        }
        if (j == word2_.size()) {
            return word1_.size() - i;
        }
        if (dp[i][j] != -1) {
            return dp[i][j];
        }
        int res = 0;
        if (word1_[i] == word2_[j]) {
            res = traverse(i + 1, j + 1);
        } else {
            res = min(1 + traverse(i, j + 1), 1 + traverse(i + 1, j));
        }
        dp[i][j] = res;
        return res;
    }
};