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_436.java
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_436.java
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package com.fishercoder.solutions;
import com.fishercoder.common.classes.Interval;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
/**436. Find Right Interval
*
* Given a set of intervals, for each of the interval i, check if there exists an interval j whose start
* point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.
*
* For any interval i, you need to store the minimum interval j's index,
* which means that the interval j has the minimum start point to build the "right" relationship for interval i.
* If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.
*
* Note:
*
* You may assume the interval's end point is always bigger than its start point.
* You may assume none of these intervals have the same start point.
*
* Example 1:
* Input: [ [1,2] ]
* Output: [-1]
* Explanation: There is only one interval in the collection, so it outputs -1.
*
* Example 2:
* Input: [ [3,4], [2,3], [1,2] ]
* Output: [-1, 0, 1]
* Explanation: There is no satisfied "right" interval for [3,4].
* For [2,3], the interval [3,4] has minimum-"right" start point;
* For [1,2], the interval [2,3] has minimum-"right" start point.
*
* Example 3:
* Input: [ [1,4], [2,3], [3,4] ]
* Output: [-1, 2, -1]
* Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
* For [2,3], the interval [3,4] has minimum-"right" start point.
* */
public class _436 {
public static class Solution1 {
public int[] findRightInterval(Interval[] intervals) {
if (intervals == null || intervals.length == 0) {
return new int[0];
}
int[] result = new int[intervals.length];
result[0] = -1;
Interval last = intervals[intervals.length - 1];
Interval first = intervals[0];
Map<Interval, Integer> map = new HashMap();
for (int i = 0; i < intervals.length; i++) {
map.put(intervals[i], i);
}
Collections.sort(Arrays.asList(intervals), (o1, o2) -> o1.start - o2.start);
for (int i = 1; i < intervals.length; i++) {
//TODO: use binary search for the minimum start interval for interval[i-1] instead of a while loop
int tmp = i - 1;
int tmpI = i;
while (tmpI < intervals.length && intervals[tmpI].start < intervals[tmp].end) {
tmpI++;
}
if (tmpI < intervals.length) {
result[map.get(intervals[tmp])] = map.get(intervals[tmpI]);
} else {
result[map.get(intervals[tmp])] = -1;
}
}
if (result[intervals.length - 1] == 0 && last.end > first.start) {
result[intervals.length - 1] = -1;
}
return result;
}
}
}