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1092. Shortest Common Supersequence (Hard)

Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences.  If multiple answers exist, you may return any of them.

(A string S is a subsequence of string T if deleting some number of characters from T (possibly 0, and the characters are chosen anywhere from T) results in the string S.)

 

Example 1:

Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation: 
str1 = "abac" is a subsequence of "cabac" because we can delete the first "c".
str2 = "cab" is a subsequence of "cabac" because we can delete the last "ac".
The answer provided is the shortest such string that satisfies these properties.

 

Note:

  1. 1 <= str1.length, str2.length <= 1000
  2. str1 and str2 consist of lowercase English letters.

Related Topics:
Dynamic Programming

Similar Questions:

Solution 1. DP

Let dp[i][j] be the length of the shortest common supersequence of s[0..(i-1)] and t[0..(j-1)].

dp[i][j] = 1 + dp[i-1][j-1]                   if s[i-1] == t[j-1]
           1 + min(dp[i-1][j], dp[i][j-1])    if s[i-1] != t[j-1]
dp[0][i] = dp[i][0] = i

dp[M][N] is the length of the shortest common supersequence of s and t.

With this dp array, we can construct the shortest common supersequence.

// OJ: https://leetcode.com/problems/shortest-common-supersequence/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    string shortestCommonSupersequence(string s, string t) {
        int M = s.size(), N = t.size();
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 1; i <= N; ++i) dp[0][i] = i;
        for (int i = 1; i <= M; ++i) dp[i][0] = i;
        for (int i = 1; i <= M; ++i) {
            for (int j = 1; j <= N; ++j) {
                if (s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
                else dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        string ans(dp[M][N], ' ');
        for (int i = M, j = N, k = ans.size() - 1; k >= 0; --k) {
            if (i && j && s[i - 1] == t[j - 1]) ans[k] = s[--i], --j;
            else if (i && dp[i][j] == dp[i - 1][j] + 1) ans[k] = s[--i];
            else ans[k] = t[--j];
        }
        return ans;
    }
};

Solution 2. LCS

Let dp[i][j] be the length of longest common subsequence of s[0..(i-1)] and t[0..(j-1)].

dp[i][j] = 1 + dp[i-1][j-1]                   if s[i-1] == t[j-1]
           max(dp[i-1][j], dp[i][j-1])        if s[i-1] != t[j-1]
dp[0][i] = dp[i][0] = 0

dp[M][N] is the length of the LCS of s and t, and M + N - dp[M][N] is the length of the shortest common supersequence.

We can use dp array to construct the shortest common supersequence as well.

// OJ: https://leetcode.com/problems/shortest-common-supersequence/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
// Ref: https://leetcode.com/problems/shortest-common-supersequence/discuss/312757/JavaPython-3-O(mn)-clean-DP-code-w-picture-comments-and-analysis.
class Solution {
public:
    string shortestCommonSupersequence(string s, string t) {
        int M = s.size(), N = t.size();
        vector<vector<int>> dp(M + 1, vector<int>(N + 1));
        for (int i = 1; i <= M; ++i) {
            for (int j = 1; j <= N; ++j) {
                if (s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
                else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        string ans(M + N - dp[M][N], ' ');
        for (int i = M, j = N, k = ans.size() - 1; k >= 0; --k) {
            if (i && j && s[i - 1] == t[j - 1]) ans[k] = s[--i], --j;
            else if (i && dp[i][j] == dp[i - 1][j]) ans[k] = s[--i];
            else ans[k] = t[--j];
        }
        return ans;
    }
};