Obsidian Notes

Shortest Common Supersequence

3 min read

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.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.

Approach - Bottom Up

  • Follow the same logic but could be reverse
  • O(m*n), O(m*n)
class Solution {
    public String shortestCommonSupersequence(String str1, String str2) {
        int m = str1.length(), n = str2.length();
        int[][] dp = new int[m+1][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 (str1.charAt(i-1) == str2.charAt(j-1))
                    dp[i][j] = 1 + dp[i-1][j-1];
                else
                    dp[i][j] = 1 + Math.min(dp[i][j-1], dp[i-1][j]);
            }
        }   
 
        StringBuilder ans = new StringBuilder();
        int i = m, j = n;
        while (i > 0 && j > 0) {
            if (str1.charAt(i-1) == str2.charAt(j-1)) {
                ans.append(str1.charAt(i-1));
                i--;
                j--;
            } else if (dp[i-1][j] < dp[i][j-1]) {
                ans.append(str1.charAt(i-1));
                i--;
            } else {
                ans.append(str2.charAt(j-1));
                j--;
            }
        }
 
        while (i > 0) {
            ans.append(str1.charAt(i-1));
            i--;
        }
 
        while (j > 0) {
            ans.append(str2.charAt(j-1));
            j--;
        }
 
        return ans.reverse().toString();
    }
}

Approach - Recursion

  • This code is for to just find the length
class Solution {
    public int shortestCommonSupersequence(String str1, String str2) {
        return dfs(str1, str2, 0, 0);
    }
 
    private int dfs(String s1, String s2, int i, int j) {
        if (i == s1.length() || j == s2.length())
            return i + j;
 
        if (s1.charAt(i) == s2.charAt(j))
            return 1 + dfs(s1, s2, i+1, j+1);
        else
            return 1 + Math.min(dfs(s1,s2,i+1,j),dfs(s1,s2,i,j+1));
    }
}

Backlinks