Leetcode•Mar 29, 2026

Find Right Interval

Hazrat Ali

Leetcode

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.

solution
function findRightInterval(intervals) {
    const n = intervals.length;

    let starts = [];
    for (let i = 0; i < n; i++) {
        starts.push([intervals[i][0], i]);
    }
    starts.sort((a, b) => a[0] - b[0]);
    function lowerBound(arr, target) {
        let left = 0, right = arr.length;
        while (left < right) {
            let mid = Math.floor((left + right) / 2);
            if (arr[mid][0] < target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }

    let ans = new Array(n);

    for (let i = 0; i < n; i++) {
        let end = intervals[i][1];

        let idx = lowerBound(starts, end);

        if (idx === starts.length) {
            ans[i] = -1;
        } else {
            ans[i] = starts[idx][1];
        }
    }

    return ans;
}


let intervals = [[3,4],[2,3],[1,2]];
console.log(findRightInterval(intervals));



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