583

583. Delete Operation for Two Strings (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.

Related Topics:
String, Dynamic Programming

Similar Questions:

• Edit Distance (Hard)
• Minimum ASCII Delete Sum for Two Strings (Medium)
• Longest Common Subsequence (Medium)

Solution 1. Longest Common Subsequence

A variation of LCS problem. The result should be M + N - 2 * len(LCS).

// OJ: https://leetcode.com/problems/delete-operation-for-two-strings
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int minDistance(string s, string t) {
        int M = s.size(), N = t.size();
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (s[i] == t[j]) dp[i + 1][j + 1] = 1 + dp[i][j];
                else dp[i + 1][j + 1] = max(dp[i + 1][j], dp[i][j + 1]);
            }
        }
        return M + N - 2 * dp[M][N];
    }
};

Solution 2. DP

Let dp[i + 1][j + 1] be the minimum number of steps required for A[0..i] and B[0..j].

for 0 <= i < M, 0 <= j < N:
dp[i+1][j+1] = dp[i][j]                         if s[i] == t[j]
         = 1 + min(dp[i+1][j], dp[i][j+1])      if s[i] != t[j]

Trivial cases:

dp[i][0] = i  for 0 <= i <= M
dp[0][j] = j  for 0 <= j <= N

These trivial cases can be simplified to

dp[i][j] = i + j   if i == 0 || j == 0,  for 0 <= i <= M and 0 <= j <= N

// OJ: https://leetcode.com/problems/delete-operation-for-two-strings/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int minDistance(string s, string t) {
        int M = s.size(), N = t.size();
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 0; i <= M; ++i) dp[i][0] = i;
        for (int j = 1; j <= N; ++j) dp[0][j] = j;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (s[i] == t[j]) dp[i + 1][j + 1] = dp[i][j];
                else dp[i + 1][j + 1] = 1 + min(dp[i + 1][j], dp[i][j + 1]);
            }
        }
        return dp[M][N];
    }
};

Or

// OJ: https://leetcode.com/problems/delete-operation-for-two-strings/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int minDistance(string A, string B) {
        int M = A.size(), N = B.size();
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 0; i <= M; ++i) {
            for (int j = 0; j <= N; ++j) {
                if (i == 0 || j == 0) dp[i][j] = i + j;
                else if (A[i - 1] == B[j - 1]) dp[i][j] = dp[i - 1][j - 1];
                else dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[M][N];
    }
};

Solution 3. DP with Space Optimization

Since dp[i+1][j+1] is only dependent on dp[i][j], dp[i+1][j] and dp[i][j+1], we can reduce the size of the dp array from M * N to N, with an additional variable saving dp[i][j].

// OJ: https://leetcode.com/problems/delete-operation-for-two-strings/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(min(M,N))
class Solution {
public:
    int minDistance(string s, string t) {
        int M = s.size(), N = t.size();
        if (M < N) swap(M, N), swap(s, t);
        vector<int> dp(N + 1);
        for (int i = 0; i <= M; ++i) {
            int prev = 0;
            for (int j = 0; j <= N; ++j) {
                int cur = dp[j];
                if (i == 0 || j == 0) dp[j] = i + j;
                else if (s[i - 1] == t[j - 1]) dp[j] = prev;
                else dp[j] = 1 + min(dp[j], dp[j - 1]);
                prev = cur;
            }
        }
        return dp[N];
    }
};