Minimum Light Radius - Google Top Interview Questions

Problem Statement :

You are given a list of integers nums representing coordinates of houses on a 1-dimensional line. 

You have 3 street lights that you can put anywhere on the coordinate line and a light at coordinate x lights up houses in [x - r, x + r], inclusive. 

Return the smallest r required such that we can place the 3 lights and all the houses are lit up.


n ≤ 100,000 where n is the length of nums

Example 1


nums = [3, 4, 5, 6]




If we place the lamps on 3.5, 4.5 and 5.5 then with r = 0.5 we can light up all 4 houses.

Solution :


                        Solution in C++ :

int need(vector<int>& v, double r) {
    int ret = 0;
    for (int i = 0; i < v.size();) {
        double upto = v[i] + 2 * r;
        int j = i + 1;
        while (j < v.size() && v[j] <= upto) j++;
        i = j;
    return ret;

double solve(vector<int>& nums) {
    if (nums.size() == 0) return 0;
    sort(nums.begin(), nums.end());
    double lhs = 0;
    double rhs = nums.back() - nums[0];
    for (int qq = 0; qq < 40; qq++) {
        double mid = (lhs + rhs) / 2;
        if (need(nums, mid) <= 3)
            rhs = mid;
            lhs = mid;
    return rhs;

                        Solution in Java :

import java.util.*;

class Solution {
    int[] nums;
    public double solve(int[] nums) {
        if (nums.length <= 3)
            return 0;
        this.nums = nums;
        int l = 0, r = nums[nums.length - 1];
        while (l < r) {
            int m = l + (r - l) / 2;
            if (good(m)) {
                r = m;
            } else {
                l = m + 1;
        return (l + 0.0) / 2;
    public boolean good(int range) {
        int end = range + nums[0], ct = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > end) {
                end = nums[i] + range;
        return ct <= 2;

                        Solution in Python : 
class Solution:
    def solve(self, nums):
        N = len(nums)
        if N <= 3:
            return 0
        LIGHTS = 3

        def can(diameter):
            start = nums[0]
            end = start + diameter
            for i in range(LIGHTS):
                idx = bisect_right(nums, end)
                if idx >= N:
                    return True
                start = nums[idx]
                end = start + diameter
            return False

        lo = -1
        hi = nums[-1] - nums[0]
        # Invariant: the answer lies in the interval (lo, hi]
        while lo < hi - 1:
            mid = (lo + hi) // 2
            if can(mid):
                hi = mid
                lo = mid
        return hi / 2

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