LeetCode #497 — MEDIUM

Random Point in Non-overlapping Rectangles

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

The Problem

Problem Statement

You are given an array of non-overlapping axis-aligned rectangles rects where rects[i] = [a_i, b_i, x_i, y_i] indicates that (a_i, b_i) is the bottom-left corner point of the i^th rectangle and (x_i, y_i) is the top-right corner point of the i^th rectangle. Design an algorithm to pick a random integer point inside the space covered by one of the given rectangles. A point on the perimeter of a rectangle is included in the space covered by the rectangle.

Any integer point inside the space covered by one of the given rectangles should be equally likely to be returned.

Note that an integer point is a point that has integer coordinates.

Implement the Solution class:

Solution(int[][] rects) Initializes the object with the given rectangles rects.
int[] pick() Returns a random integer point [u, v] inside the space covered by one of the given rectangles.

Example 1:

Input
["Solution", "pick", "pick", "pick", "pick", "pick"]
[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]
Output
[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]

Explanation
Solution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);
solution.pick(); // return [1, -2]
solution.pick(); // return [1, -1]
solution.pick(); // return [-1, -2]
solution.pick(); // return [-2, -2]
solution.pick(); // return [0, 0]

Constraints:

1 <= rects.length <= 100
rects[i].length == 4
-10⁹ <= a_i < x_i <= 10⁹
-10⁹ <= b_i < y_i <= 10⁹
x_i - a_i <= 2000
y_i - b_i <= 2000
All the rectangles do not overlap.
At most 10⁴ calls will be made to pick.

Step 01

Brute Force Baseline

Problem summary: You are given an array of non-overlapping axis-aligned rectangles rects where rects[i] = [ai, bi, xi, yi] indicates that (ai, bi) is the bottom-left corner point of the ith rectangle and (xi, yi) is the top-right corner point of the ith rectangle. Design an algorithm to pick a random integer point inside the space covered by one of the given rectangles. A point on the perimeter of a rectangle is included in the space covered by the rectangle. Any integer point inside the space covered by one of the given rectangles should be equally likely to be returned. Note that an integer point is a point that has integer coordinates. Implement the Solution class: Solution(int[][] rects) Initializes the object with the given rectangles rects. int[] pick() Returns a random integer point [u, v] inside the space covered by one of the given rectangles.

Baseline thinking

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

Pattern signal: Array · Math · Binary Search · Segment Tree

Example 1

["Solution","pick","pick","pick","pick","pick"]
[[[[-2,-2,1,1],[2,2,4,6]]],[],[],[],[],[]]

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 #497: Random Point in Non-overlapping Rectangles
class Solution {
    private int[] s;
    private int[][] rects;
    private Random random = new Random();

    public Solution(int[][] rects) {
        int n = rects.length;
        s = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + (rects[i][2] - rects[i][0] + 1) * (rects[i][3] - rects[i][1] + 1);
        }
        this.rects = rects;
    }

    public int[] pick() {
        int n = rects.length;
        int v = 1 + random.nextInt(s[n]);
        int left = 0, right = n;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (s[mid] >= v) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        int[] rect = rects[left - 1];
        return new int[] {rect[0] + random.nextInt(rect[2] - rect[0] + 1),
            rect[1] + random.nextInt(rect[3] - rect[1] + 1)};
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(rects);
 * int[] param_1 = obj.pick();
 */

// Accepted solution for LeetCode #497: Random Point in Non-overlapping Rectangles
type Solution struct {
	s     []int
	rects [][]int
}

func Constructor(rects [][]int) Solution {
	n := len(rects)
	s := make([]int, n+1)
	for i, v := range rects {
		s[i+1] = s[i] + (v[2]-v[0]+1)*(v[3]-v[1]+1)
	}
	return Solution{s, rects}
}

func (this *Solution) Pick() []int {
	n := len(this.rects)
	v := 1 + rand.Intn(this.s[len(this.s)-1])
	left, right := 0, n
	for left < right {
		mid := (left + right) >> 1
		if this.s[mid] >= v {
			right = mid
		} else {
			left = mid + 1
		}
	}
	rect := this.rects[left-1]
	x, y := rect[0]+rand.Intn(rect[2]-rect[0]+1), rect[1]+rand.Intn(rect[3]-rect[1]+1)
	return []int{x, y}
}

# Accepted solution for LeetCode #497: Random Point in Non-overlapping Rectangles
class Solution:
    def __init__(self, rects: List[List[int]]):
        self.rects = rects
        self.s = [0] * len(rects)
        for i, (x1, y1, x2, y2) in enumerate(rects):
            self.s[i] = self.s[i - 1] + (x2 - x1 + 1) * (y2 - y1 + 1)

    def pick(self) -> List[int]:
        v = random.randint(1, self.s[-1])
        idx = bisect_left(self.s, v)
        x1, y1, x2, y2 = self.rects[idx]
        return [random.randint(x1, x2), random.randint(y1, y2)]


# Your Solution object will be instantiated and called as such:
# obj = Solution(rects)
# param_1 = obj.pick()

// Accepted solution for LeetCode #497: Random Point in Non-overlapping Rectangles
use rand::distributions::WeightedIndex;
use rand::prelude::*;

struct Solution {
    rng: ThreadRng,
    rects: Vec<Vec<i32>>,
    size: usize,
    dist: WeightedIndex<i32>,
}

impl Solution {
    fn new(rects: Vec<Vec<i32>>) -> Self {
        let rng = thread_rng();
        let size = rects.len();
        let weights: Vec<i32> = rects
            .iter()
            .map(|v| (v[2] - v[0] + 1) * (v[3] - v[1] + 1))
            .collect();
        let dist = WeightedIndex::new(weights).unwrap();
        Solution {
            rng,
            rects,
            size,
            dist,
        }
    }

    fn pick(&mut self) -> Vec<i32> {
        let rect = &self.rects[self.rng.sample(&self.dist)];
        let x = self.rng.gen_range(rect[0], rect[2] + 1);
        let y = self.rng.gen_range(rect[1], rect[3] + 1);
        vec![x, y]
    }
}

// Accepted solution for LeetCode #497: Random Point in Non-overlapping Rectangles
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #497: Random Point in Non-overlapping Rectangles
// class Solution {
//     private int[] s;
//     private int[][] rects;
//     private Random random = new Random();
// 
//     public Solution(int[][] rects) {
//         int n = rects.length;
//         s = new int[n + 1];
//         for (int i = 0; i < n; ++i) {
//             s[i + 1] = s[i] + (rects[i][2] - rects[i][0] + 1) * (rects[i][3] - rects[i][1] + 1);
//         }
//         this.rects = rects;
//     }
// 
//     public int[] pick() {
//         int n = rects.length;
//         int v = 1 + random.nextInt(s[n]);
//         int left = 0, right = n;
//         while (left < right) {
//             int mid = (left + right) >> 1;
//             if (s[mid] >= v) {
//                 right = mid;
//             } else {
//                 left = mid + 1;
//             }
//         }
//         int[] rect = rects[left - 1];
//         return new int[] {rect[0] + random.nextInt(rect[2] - rect[0] + 1),
//             rect[1] + random.nextInt(rect[3] - rect[1] + 1)};
//     }
// }
// 
// /**
//  * Your Solution object will be instantiated and called as such:
//  * Solution obj = new Solution(rects);
//  * int[] param_1 = obj.pick();
//  */

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(log n)

Space

O(1)

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.