1092. Shortest Common Supersequence

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1092. Shortest Common Supersequence

Description

Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences . If there are multiple valid strings, return any of them.

A string s is a subsequence of string t if deleting some number of characters from t (possibly 0) results in the string s.

Example 1:

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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.

Example 2:

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Input: str1 = "aaaaaaaa", str2 = "aaaaaaaa"
Output: "aaaaaaaa"

Constraints:

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

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<vector<int>> dp;

    string shortestCommonSupersequence(string str1, string str2) {
        int m = str1.size(), n = str2.size();
        dp.resize(m, vector<int>(n, -1));
        int minLen = dfs(str1, str2, 0, 0);
        string res;
        int idx1 = 0, idx2 = 0;
        while (minLen > 0) {
            if (idx1 == str1.size()) {
                res += str2[idx2++];
                minLen--;
                continue;
            }
            if (idx2 == str2.size()) {
                res += str1[idx1++];
                minLen--;
                continue;
            }
            if (str1[idx1] == str2[idx2]) {
                res += str1[idx1];
                idx1++; idx2++;
                minLen--;
                continue;
            }
            if (idx1 + 1 < str1.size() && dp[idx1 + 1][idx2] == minLen - 1) {
                res += str1[idx1++];
                minLen--;
            } else {
                res += str2[idx2++];
                minLen--;
            }
        }
        return res;
    }

    int dfs(string& str1, string& str2, int idx1, int idx2) {
        if (idx1 == str1.size()) {
            return str2.size() - idx2;
        }
        if (idx2 == str2.size()) {
            return str1.size() - idx1;
        }
        if (dp[idx1][idx2] != -1) {
            return dp[idx1][idx2];
        }
        int res = INT_MAX;
        if (str1[idx1] == str2[idx2]) {
            res = 1 + dfs(str1, str2, idx1 + 1, idx2 + 1);
        } else {
            res = 1 + min(dfs(str1, str2, idx1 + 1, idx2), dfs(str1, str2, idx1, idx2 + 1));
        }
        dp[idx1][idx2] = res;
        return res;
    }
};