PROBLEM DESCRIPTION
Given an array of meeting time intervals intervals where intervals[i] = [starti, endi], return the minimum number of conference rooms required.
SOLUTION
BRUTE FORCE (TLE)
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class Solution {
public int[] minInterval(int[][] intervals, int[] queries) {
int[] ans = new int[queries.length];
int minSize = -1;
for(int i=0; i<queries.length; i++){
minSize = Integer.MAX_VALUE;
int current = queries[i];
for(int j=0; j<intervals.length; j++){
if(current >= intervals[j][0] && current <= intervals[j][1]){
minSize = Math.min(minSize, intervals[j][1] - intervals[j][0] + 1);
}
}
if(minSize == Integer.MAX_VALUE) minSize = -1;
ans[i] = minSize;
}
return ans;
}
}
OPTIMIZATION
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class Solution {
public int[] minInterval(int[][] intervals, int[] queries) {
int n = intervals.length;
//sort the interval based on start position
Arrays.sort(intervals, (o1, o2) -> o1[0] > o2[0]?1: (o1[0] < o2[0]?-1:0) );
//create a copy of queries array and sort it
int[] Q = queries.clone();
Arrays.sort(Q);
//sorted hashmap to store <Size, RightPosition>
TreeMap<Integer, Integer> pq = new TreeMap<>();
//store answer in a hashmap
Map<Integer, Integer> res = new HashMap<>();
//init
int i = 0;
//for each query
for(Integer q: Q){
//while there are more intervals to check and current position is >= start of current interval
//it may possibly form the answer, so we add it to the min heap (TreeMap)
while(i<n && q >= intervals[i][0]){
//calculate size of the interval
int size = intervals[i][1] - intervals[i][0] + 1;
//put the size of the interval and also the right position (to be used later)
pq.put(size, intervals[i][1]);
//go to the next interval
i++;
}
//we will pop intervals from min heap / TreeMap which in not in the required range
//if min heap is not empty and current position is after the end of interval, remove it from min heap because it cannot form the answer
while(pq.size() > 0 && pq.firstEntry().getValue() < q){
pq.pollFirstEntry();
}
//if all intervals were removed, store answer as -1
if(pq.size() == 0){
res.put(q, -1);
//else, store the answer as size on top of the min heap
}else{
res.put(q, pq.firstKey());
}
}
//the final answer array will be formed using the res hashmap
int[] finalAnswer = new int[queries.length];
for(int k=0; k<finalAnswer.length; k++){
finalAnswer[k] = res.get(queries[k]);
}
return finalAnswer;
}
}