LeetCode in Kotlin
3302. Find the Lexicographically Smallest Valid Sequence
Medium
You are given two strings word1 and word2.
A string x is called almost equal to y if you can change at most one character in x to make it identical to y.
A sequence of indices seq is called valid if:
- The indices are sorted in ascending order.
- Concatenating the characters at these indices in
word1in the same order results in a string that is almost equal toword2.
Return an array of size word2.length representing the lexicographically smallest valid sequence of indices. If no such sequence of indices exists, return an empty array.
Note that the answer must represent the lexicographically smallest array, not the corresponding string formed by those indices.
Example 1:
Input: word1 = “vbcca”, word2 = “abc”
Output: [0,1,2]
Explanation:
The lexicographically smallest valid sequence of indices is [0, 1, 2]:
- Change
word1[0]to'a'. word1[1]is already'b'.word1[2]is already'c'.
Example 2:
Input: word1 = “bacdc”, word2 = “abc”
Output: [1,2,4]
Explanation:
The lexicographically smallest valid sequence of indices is [1, 2, 4]:
word1[1]is already'a'.- Change
word1[2]to'b'. word1[4]is already'c'.
Example 3:
Input: word1 = “aaaaaa”, word2 = “aaabc”
Output: []
Explanation:
There is no valid sequence of indices.
Example 4:
Input: word1 = “abc”, word2 = “ab”
Output: [0,1]
Constraints:
1 <= word2.length < word1.length <= 3 * 105word1andword2consist only of lowercase English letters.
Solution
class Solution {
fun validSequence(word1: String, word2: String): IntArray? {
val c1 = word1.toCharArray()
val c2 = word2.toCharArray()
val dp = IntArray(c1.size + 1)
var j = c2.size - 1
for (i in c1.indices.reversed()) {
if (j >= 0 && c1[i] == c2[j]) {
dp[i] = dp[i + 1] + 1
j--
} else {
dp[i] = dp[i + 1]
}
}
val ans = IntArray(c2.size)
var i = 0
j = 0
while (i < c1.size && j < c2.size) {
if (c1[i] == c2[j]) {
ans[j] = i
j++
} else {
if (dp[i + 1] >= c2.size - 1 - j) {
ans[j] = i
j++
i++
break
}
}
i++
}
if (j < c2.size && i == c1.size) {
return IntArray(0)
}
while (j < c2.size && i < c1.size) {
if (c2[j] == c1[i]) {
ans[j] = i
j++
}
i++
}
return ans
}
}