LeetCode #436 — MEDIUM

Find Right Interval

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of intervals, where intervals[i] = [start_i, end_i] and each start_i is unique.

The right interval for an interval i is an interval j such that start_j >= end_i and start_j 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 * 10⁴
intervals[i].length == 2
-10⁶ <= start_i <= end_i <= 10⁶
The start point of each interval is unique.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[[1,2]]

Example 2

[[3,4],[2,3],[1,2]]

Example 3

[[1,4],[2,3],[3,4]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #436: Find Right Interval
class 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;
    }
}

// Accepted solution for LeetCode #436: Find Right Interval
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
}

# Accepted solution for LeetCode #436: Find Right Interval
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 ans

// Accepted solution for LeetCode #436: Find Right Interval
struct Solution;
use std::collections::BTreeMap;

impl Solution {
    fn find_right_interval(intervals: Vec<Vec<i32>>) -> Vec<i32> {
        let n = intervals.len();
        let mut res: Vec<i32> = vec![-1; n];
        let mut btm: BTreeMap<i32, usize> = BTreeMap::new();
        for i in 0..n {
            let l = intervals[i][0];
            btm.insert(l, i);
        }
        for i in 0..n {
            let r = intervals[i][1];
            for (_, &j) in btm.range(r..).take(1) {
                res[i] = j as i32;
            }
        }
        res
    }
}

#[test]
fn test() {
    let intervals = vec_vec_i32![[1, 2]];
    let res = vec![-1];
    assert_eq!(Solution::find_right_interval(intervals), res);
    let intervals = vec_vec_i32![[3, 4], [2, 3], [1, 2]];
    let res = vec![-1, 0, 1];
    assert_eq!(Solution::find_right_interval(intervals), res);
    let intervals = vec_vec_i32![[1, 4], [2, 3], [3, 4]];
    let res = vec![-1, 2, -1];
    assert_eq!(Solution::find_right_interval(intervals), res);
}

// Accepted solution for LeetCode #436: Find Right Interval
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;
    });
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.