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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:

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:

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

 

Constraints:

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

Solutions

Solution 1

Thinking

The SCS length is \(|s_1|+|s_2|-\mathrm{LCS}\). \(m,n\le 1000\): build the LCS table, then walk the transitions backward, inserting the characters that are not shared.

\(f[i][j]\) is the LCS of the two prefixes. From \((m,n)\), a move from above or the left emits the discarded character; an equal pair emits the common character and steps diagonally.

The walk collects the reverse of the SCS; leftover characters on one side are appended when the other index hits zero.

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class Solution:
    def shortestCommonSupersequence(self, str1: str, str2: str) -> str:
        m, n = len(str1), len(str2)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if str1[i - 1] == str2[j - 1]:
                    f[i][j] = f[i - 1][j - 1] + 1
                else:
                    f[i][j] = max(f[i - 1][j], f[i][j - 1])
        ans = []
        i, j = m, n
        while i or j:
            if i == 0:
                j -= 1
                ans.append(str2[j])
            elif j == 0:
                i -= 1
                ans.append(str1[i])
            else:
                if f[i][j] == f[i - 1][j]:
                    i -= 1
                    ans.append(str1[i])
                elif f[i][j] == f[i][j - 1]:
                    j -= 1
                    ans.append(str2[j])
                else:
                    i, j = i - 1, j - 1
                    ans.append(str1[i])
        return ''.join(ans[::-1])

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class Solution {
    public String shortestCommonSupersequence(String str1, String str2) {
        int m = str1.length(), n = str2.length();
        int[][] f = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
                    f[i][j] = f[i - 1][j - 1] + 1;
                } else {
                    f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
                }
            }
        }
        int i = m, j = n;
        StringBuilder ans = new StringBuilder();
        while (i > 0 || j > 0) {
            if (i == 0) {
                ans.append(str2.charAt(--j));
            } else if (j == 0) {
                ans.append(str1.charAt(--i));
            } else {
                if (f[i][j] == f[i - 1][j]) {
                    ans.append(str1.charAt(--i));
                } else if (f[i][j] == f[i][j - 1]) {
                    ans.append(str2.charAt(--j));
                } else {
                    ans.append(str1.charAt(--i));
                    --j;
                }
            }
        }
        return ans.reverse().toString();
    }
}

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class Solution {
public:
    string shortestCommonSupersequence(string str1, string str2) {
        int m = str1.size(), n = str2.size();
        vector<vector<int>> f(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (str1[i - 1] == str2[j - 1])
                    f[i][j] = f[i - 1][j - 1] + 1;
                else
                    f[i][j] = max(f[i - 1][j], f[i][j - 1]);
            }
        }
        int i = m, j = n;
        string ans;
        while (i || j) {
            if (i == 0)
                ans += str2[--j];
            else if (j == 0)
                ans += str1[--i];
            else {
                if (f[i][j] == f[i - 1][j])
                    ans += str1[--i];
                else if (f[i][j] == f[i][j - 1])
                    ans += str2[--j];
                else
                    ans += str1[--i], --j;
            }
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

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func shortestCommonSupersequence(str1 string, str2 string) string {
    m, n := len(str1), len(str2)
    f := make([][]int, m+1)
    for i := range f {
        f[i] = make([]int, n+1)
    }
    for i := 1; i <= m; i++ {
        for j := 1; j <= n; j++ {
            if str1[i-1] == str2[j-1] {
                f[i][j] = f[i-1][j-1] + 1
            } else {
                f[i][j] = max(f[i-1][j], f[i][j-1])
            }
        }
    }
    ans := []byte{}
    i, j := m, n
    for i > 0 || j > 0 {
        if i == 0 {
            j--
            ans = append(ans, str2[j])
        } else if j == 0 {
            i--
            ans = append(ans, str1[i])
        } else {
            if f[i][j] == f[i-1][j] {
                i--
                ans = append(ans, str1[i])
            } else if f[i][j] == f[i][j-1] {
                j--
                ans = append(ans, str2[j])
            } else {
                i, j = i-1, j-1
                ans = append(ans, str1[i])
            }
        }
    }
    for i, j = 0, len(ans)-1; i < j; i, j = i+1, j-1 {
        ans[i], ans[j] = ans[j], ans[i]
    }
    return string(ans)
}

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function shortestCommonSupersequence(str1: string, str2: string): string {
    const m = str1.length;
    const n = str2.length;
    const f = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            if (str1[i - 1] == str2[j - 1]) {
                f[i][j] = f[i - 1][j - 1] + 1;
            } else {
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
            }
        }
    }
    let ans: string[] = [];
    let i = m;
    let j = n;
    while (i > 0 || j > 0) {
        if (i === 0) {
            ans.push(str2[--j]);
        } else if (j === 0) {
            ans.push(str1[--i]);
        } else {
            if (f[i][j] === f[i - 1][j]) {
                ans.push(str1[--i]);
            } else if (f[i][j] === f[i][j - 1]) {
                ans.push(str2[--j]);
            } else {
                ans.push(str1[--i]);
                --j;
            }
        }
    }
    return ans.reverse().join('');
}

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