LeetCode 1092: Shortest Common Supersequence Solution

Master LeetCode problem 1092 (Shortest Common Supersequence ), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1092. Shortest Common Supersequence

Hard

String Dynamic Programming

Problem Explanation

Explanation

To find the shortest common supersequence of two strings, we can use dynamic programming. Let dp[i][j] represent the length of the shortest common supersequence of str1[0...i-1] and str2[0...j-1]. We can build this table by considering the following cases:

If the characters at the current positions are equal, then the current character is part of the supersequence. So, dp[i][j] = dp[i-1][j-1] + 1.
If the characters are different, we have two options: either include str1[i-1] in the supersequence or str2[j-1]. So, dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1.

Finally, to reconstruct the shortest common supersequence, we can start from dp[m][n] and backtrack by considering the characters and transitions made during the dp table construction.

Solution Code

class Solution {
    public String shortestCommonSupersequence(String str1, String str2) {
        int m = str1.length();
        int n = str2.length();
        int[][] dp = new int[m+1][n+1];

        for (int i = 1; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= n; j++) {
            dp[0][j] = j;
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (str1.charAt(i-1) == str2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                } else {
                    dp[i][j] = Math.min(dp[i-1][j], dp[i][j-1]) + 1;
                }
            }
        }

        StringBuilder sb = new StringBuilder();
        int i = m, j = n;
        while (i > 0 && j > 0) {
            if (str1.charAt(i-1) == str2.charAt(j-1)) {
                sb.insert(0, str1.charAt(i-1));
                i--;
                j--;
            } else if (dp[i-1][j] < dp[i][j-1]) {
                sb.insert(0, str1.charAt(i-1));
                i--;
            } else {
                sb.insert(0, str2.charAt(j-1));
                j--;
            }
        }

        while (i > 0) {
            sb.insert(0, str1.charAt(i-1));
            i--;
        }
        while (j > 0) {
            sb.insert(0, str2.charAt(j-1));
            j--;
        }

        return sb.toString();
    }
}

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