LeetCode-in-Java
1092. Shortest Common Supersequence
Hard
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 <= 1000str1andstr2consist of lowercase English letters.
Solution
public 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 = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0) {
dp[i][j] = j;
} else if (j == 0) {
dp[i][j] = i;
} 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 - 1][j], dp[i][j - 1]);
}
}
}
// Length of the ShortestSuperSequence
int l = dp[m][n];
char[] arr = new char[l];
int i = m;
int j = n;
while (i > 0 && j > 0) {
/* If current character in str1 and str2 are same, then
current character is part of shortest supersequence */
if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
arr[--l] = str1.charAt(i - 1);
i--;
j--;
} else if (dp[i - 1][j] < dp[i][j - 1]) {
arr[--l] = str1.charAt(i - 1);
i--;
} else {
arr[--l] = str2.charAt(j - 1);
j--;
}
}
while (i > 0) {
arr[--l] = str1.charAt(i - 1);
i--;
}
while (j > 0) {
arr[--l] = str2.charAt(j - 1);
j--;
}
return new String(arr);
}
}