Description
Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.
Example 1:
Input:
s = "abpcplea", d = ["ale","apple","monkey","plea"]
Output:
"apple"
Example 2:
Input:
s = "abpcplea", d = ["a","b","c"]
Output:
"a"
Note:
- All the strings in the input will only contain lower-case letters.
- The size of the dictionary won't exceed 1,000.
- The length of all the strings in the input won't exceed 1,000.
Solution
Sort & Two-pointer, O(n logn + mn), S(1)
Sort dict to make sure longer word with smaller lexicographical order first. Then iterator dict and decide isSubsequence. Once a match is found, return directly.
class Solution {
public String findLongestWord(String s, List<String> d) {
Collections.sort(d, (a, b) -> a.length() != b.length()
? b.length() - a.length() : a.compareTo(b));
for (String t : d) {
if (isSubsequence(s, t)) {
return t;
}
}
return "";
}
private boolean isSubsequence(String s, String t) {
if (s.length() < t.length()) {
return false;
}
int j = 0; // point to t
for (int i = 0; i < s.length() && j < t.length(); ++i) {
if (s.charAt(i) == t.charAt(j)) {
++j;
}
}
return j == t.length();
}
}
Two-pointer, O(mn), S(1)
Don't need to sort dict. Keep current max subsequence match, and iterate dict and update max.
public class Solution {
public boolean isSubsequence(String x, String y) {
int j = 0;
for (int i = 0; i < y.length() && j < x.length(); i++)
if (x.charAt(j) == y.charAt(i))
j++;
return j == x.length();
}
public String findLongestWord(String s, List < String > d) {
String max_str = "";
for (String str: d) {
if (isSubsequence(str, s)) {
if (str.length() > max_str.length() || (str.length() == max_str.length() && str.compareTo(max_str) < 0))
max_str = str;
}
}
return max_str;
}
}