Binary Search - Medium Level - Question 1

Binary Search - Medium Level - Question 1

Leetcode 33 Search in Rotated Sorted Array

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Constraints:

1 <= nums.length <= 5000

-10^4 <= nums[i] <= 10^4

All values of nums are unique.

nums is guaranteed to be rotated at some pivot.

-10^4 <= target <= 10^4

Analysis:

This question is quite popular in interview.

Binary search should be the way to reach O(logn) time complexity, but this question is not easy to code, due to the boundary conditions.

One hidden condition is the value of the last element (or the first element). By comparing this value with the target (or the guessed value in binary search), we can determine which the half search range to be removed.

See the code below:

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = 0, right = nums.size(), n = nums.size();
        while(left < right) {
            if(nums[left] == target) return left;
            if(nums[right-1] == target) return right - 1;
            int mid = left + (right - left) / 2;
            if(nums[mid] == target) return mid;
            if(nums[right-1] > target) {
                if(nums[mid] > nums[right-1]) left = mid + 1;
                else {
                    if(nums[mid] > target) right = mid;
                    else left = mid + 1;
                }
            }
            else {
                if(nums[mid] < nums[right-1]) right = mid;
                else {
                    if(nums[mid] < target) left = mid + 1;
                    else right = mid;
                }
            }
        }
        return left >= 0 && left < n && nums[left] == target? left : -1;
    }
};

Question 2

Leetcode 153 Find Minimum in Rotated Sorted Array

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times.

[0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Constraints:

n == nums.length

1 <= n <= 5000

-5000 <= nums[i] <= 5000

All the integers of nums are unique.

nums is sorted and rotated between 1 and n times.

Analysis:

This question is very similar to the first one, which is a good practice to code the binary search.

One of the key observations is that: the one element is smaller than the one on its left, then the element is the minimum. Thus we can always to check this during binary search.

In addition, we need to use the hidden condition: the value of the first element or the last element, to help determine the search range.

See the code below:

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0, right = nums.size();
        while(left < right) {
            // whole range is sorted
            if(nums[left] <= nums[right - 1]) return nums[left];
            int mid = left + (right - left) / 2;
            // check the left neighbor
            if(mid>0 && nums[mid] < nums[mid-1]) return nums[mid];
            if(nums[mid] >= nums[0]) left = mid + 1;
            else right = mid;
        }
        return nums[left];
    } 
};

In addition,

1. Once the smallest element is found, the largest one is also found;

2. How about having elements with the same value?

Question 3

Leetcode 475 Heaters

Winter is coming! During the contest, your first job is to design a standard heater with a fixed warm radius to warm all the houses.

Every house can be warmed, as long as the house is within the heater's warm radius range. 

Given the positions of houses and heaters on a horizontal line, return the minimum radius standard of heaters so that those heaters could cover all houses.

Notice that all the heaters follow your radius standard, and the warm radius will the same.

Constraints:

1 <= houses.length, heaters.length <= 3 * 10^4

1 <= houses[i], heaters[i] <= 10^9

Analysis:

This question is not easy to think through.

One way to think about it is that: for each house, we need to figure out the shortest distance to all the heaters. Then the largest value of these shortest distances should be the final answer.

Then the question becomes: for each house, how to determine the shortest distance to all the heaters.

To increase the time complexity, we can sort the heaters by their locations. Then do a binary search with the location of the house, to find the shortest distance. Done. 

See the code below:

class Solution {
public:
    int findRadius(vector<int>& houses, vector<int>& heaters) {
        int res = 0;
        sort(heaters.begin(), heaters.end());
        for(auto a : houses){
            auto left = upper_bound(heaters.begin(), heaters.end(), a);
            if(left == heaters.begin()) res = max(res, *left - a);
            else if(left == heaters.end()) res = max(res, a - *(left-1));
            else res = max(res, min(*left - a, a - *(left - 1)));
        }
        return res;
    }
};

Upper Layer