1092. Shortest Common Supersequence

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.

# @lc code=start
using LeetCode

function shortest_common_supersequence(str1::String, str2::String)
    m, n = length(str1), length(str2)
    dp = OffsetArray(fill(0, m + 1, n + 1), -1, -1)
    for i in 1:m, j in 1:n
        dp[i, j] = max(
            dp[i - 1, j], dp[i, j - 1], dp[i - 1, j - 1] + Int(str1[i] == str2[j])
        )
    end
    res = ""
    coord = CartesianIndex(m, n)
    while coord != CartesianIndex(0, 0)
        c1 = coord - CartesianIndex(1, 0)
        c2 = coord - CartesianIndex(0, 1)
        if c1 ∈ CartesianIndices(dp) && dp[coord] == dp[c1]
            res *= str1[coord[1]]
            coord -= CartesianIndex(1, 0)
        elseif c2 ∈ CartesianIndices(dp) && dp[coord] == dp[c2]
            res *= str2[coord[2]]
            coord -= CartesianIndex(0, 1)
        else
            res *= str1[coord[1]]
            coord -= CartesianIndex(1, 1)
        end
    end
    return reverse(res)
end
# @lc code=end

shortest_common_supersequence (generic function with 1 method)

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