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Java-based LeetCode algorithm problem solutions, regularly updated
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package g1001_1100.s1096_brace_expansion_ii;
// #Hard #String #Breadth_First_Search #Stack #Backtracking
// #2022_02_18_Time_23_ms_(60.36%)_Space_50.3_MB_(28.57%)
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class Solution {
public List braceExpansionII(String expression) {
Set res = flatten(expression);
List sorted = new ArrayList<>(res);
Collections.sort(sorted);
return sorted;
}
private Set flatten(String expression) {
Set res = new HashSet<>();
// A temp set to store cartesian product results.
Set curSet = new HashSet<>();
int idx = 0;
while (idx < expression.length()) {
if (expression.charAt(idx) == '{') {
// end will be the index of matching "}"
int end = findClosingBrace(expression, idx);
Set set = flatten(expression.substring(idx + 1, end));
curSet = concatenateSet(curSet, set);
idx = end + 1;
} else if (Character.isLowerCase(expression.charAt(idx))) {
// Create set with single element
Set set =
new HashSet<>(Arrays.asList(Character.toString(expression.charAt(idx))));
curSet = concatenateSet(curSet, set);
idx++;
} else if (expression.charAt(idx) == ',') {
res.addAll(curSet);
curSet.clear();
idx++;
}
}
// Don't forget!
res.addAll(curSet);
return res;
}
private Set concatenateSet(Set set1, Set set2) {
if (set1.isEmpty() || set2.isEmpty()) {
return !set2.isEmpty() ? new HashSet<>(set2) : new HashSet<>(set1);
}
Set res = new HashSet<>();
for (String s1 : set1) {
for (String s2 : set2) {
res.add(s1 + s2);
}
}
return res;
}
private int findClosingBrace(String expression, int start) {
int count = 0;
int idx = start;
while (idx < expression.length()) {
if (expression.charAt(idx) == '{') {
count++;
} else if (expression.charAt(idx) == '}') {
count--;
}
if (count == 0) {
break;
}
idx++;
}
return idx;
}
}
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