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. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [3, 4, 5, 6] Output 0.5 Explanation 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();) {
ret++;
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;
else
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;
Arrays.sort(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) {
ct++;
end = nums[i] + range;
}
}
return ct <= 2;
}
}
Solution in Python :
class Solution:
def solve(self, nums):
nums.sort()
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
else:
lo = mid
return hi / 2
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