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Medium |
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You are given an array of intervals, where intervals[i] = [starti, endi] and each starti is unique.
The right interval for an interval i is an interval j such that startj >= endi and startj is minimized. Note that i may equal j.
Return an array of right interval indices for each interval i. If no right interval exists for interval i, then put -1 at index i.
Example 1:
Input: intervals = [[1,2]] Output: [-1] Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: intervals = [[3,4],[2,3],[1,2]] Output: [-1,0,1] Explanation: There is no right interval for [3,4]. The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3. The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.
Example 3:
Input: intervals = [[1,4],[2,3],[3,4]] Output: [-1,2,-1] Explanation: There is no right interval for [1,4] and [3,4]. The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.
Constraints:
1 <= intervals.length <= 2 * 104intervals[i].length == 2-106 <= starti <= endi <= 106- The start point of each interval is unique.
We can store the start point and index of each interval into an array arr, and sort it by the start point. Then we iterate through the interval array, for each interval [_, ed], we can use binary search to find the first interval whose start point is greater than or equal to ed, which is its right-side interval. If found, we store its index into the answer array, otherwise, we store -1.
The time complexity is
class Solution:
def findRightInterval(self, intervals: List[List[int]]) -> List[int]:
n = len(intervals)
ans = [-1] * n
arr = sorted((st, i) for i, (st, _) in enumerate(intervals))
for i, (_, ed) in enumerate(intervals):
j = bisect_left(arr, (ed, -inf))
if j < n:
ans[i] = arr[j][1]
return ansclass Solution {
public int[] findRightInterval(int[][] intervals) {
int n = intervals.length;
int[][] arr = new int[n][0];
for (int i = 0; i < n; ++i) {
arr[i] = new int[] {intervals[i][0], i};
}
Arrays.sort(arr, (a, b) -> a[0] - b[0]);
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
int j = search(arr, intervals[i][1]);
ans[i] = j < n ? arr[j][1] : -1;
}
return ans;
}
private int search(int[][] arr, int x) {
int l = 0, r = arr.length;
while (l < r) {
int mid = (l + r) >> 1;
if (arr[mid][0] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}class Solution {
public:
vector<int> findRightInterval(vector<vector<int>>& intervals) {
int n = intervals.size();
vector<pair<int, int>> arr;
for (int i = 0; i < n; ++i) {
arr.emplace_back(intervals[i][0], i);
}
sort(arr.begin(), arr.end());
vector<int> ans;
for (auto& e : intervals) {
int j = lower_bound(arr.begin(), arr.end(), make_pair(e[1], -1)) - arr.begin();
ans.push_back(j == n ? -1 : arr[j].second);
}
return ans;
}
};func findRightInterval(intervals [][]int) (ans []int) {
arr := make([][2]int, len(intervals))
for i, v := range intervals {
arr[i] = [2]int{v[0], i}
}
sort.Slice(arr, func(i, j int) bool { return arr[i][0] < arr[j][0] })
for _, e := range intervals {
j := sort.Search(len(arr), func(i int) bool { return arr[i][0] >= e[1] })
if j < len(arr) {
ans = append(ans, arr[j][1])
} else {
ans = append(ans, -1)
}
}
return
}function findRightInterval(intervals: number[][]): number[] {
const n = intervals.length;
const arr: number[][] = Array.from({ length: n }, (_, i) => [intervals[i][0], i]);
arr.sort((a, b) => a[0] - b[0]);
return intervals.map(([_, ed]) => {
let [l, r] = [0, n];
while (l < r) {
const mid = (l + r) >> 1;
if (arr[mid][0] >= ed) {
r = mid;
} else {
l = mid + 1;
}
}
return l < n ? arr[l][1] : -1;
});
}