LeetCode in Kotlin
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
class Solution {
fun shortestCommonSupersequence(str1: String, str2: String): String {
val m = str1.length
val n = str2.length
val dp = Array(m + 1) { IntArray(n + 1) }
for (i in 0..m) {
for (j in 0..n) {
if (i == 0) {
dp[i][j] = j
} else if (j == 0) {
dp[i][j] = i
} else if (str1[i - 1] == str2[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
var l = dp[m][n]
val arr = CharArray(l)
var i = m
var 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[i - 1] == str2[j - 1]) {
arr[--l] = str1[i - 1]
i--
j--
} else if (dp[i - 1][j] < dp[i][j - 1]) {
arr[--l] = str1[i - 1]
i--
} else {
arr[--l] = str2[j - 1]
j--
}
}
while (i > 0) {
arr[--l] = str1[i - 1]
i--
}
while (j > 0) {
arr[--l] = str2[j - 1]
j--
}
return String(arr)
}
}