Coincidence Search - Google Top Interview Questions


Problem Statement :


You are given a list of unique integers nums. Return the number of integers that could still be successfully found using a standard binary search.

Binary search in pseudocode:

lo = 0
hi = nums.size - 1

while lo <= hi
    mid = floor((lo + hi) / 2)
    if nums[mid] == target
        return mid
    elif nums[mid] < target
        lo = mid + 1
    else
        hi = mid - 1

Constraints

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

Example 1

Input

nums = [1, 5, 3, 2, 9]

Output

3

Explanation

Since if we used binary search to look for 3, we would still find it in the first iteration. We would also 
happen to find 1 and 9 after two iterations.


Solution :



title-img 




                        Solution in C++ :

int go(vector<int>& nums, int l, int r, int a, int b) {
    if (l > r) return 0;
    auto m = (l + r) / 2;
    auto mid = nums[m];
    return int(a <= mid && mid <= b) + go(nums, l, m - 1, a, min(mid - 1, b)) +
           go(nums, m + 1, r, max(a, mid + 1), b);
}

int solve(vector<int>& nums) {
    return go(nums, 0, (int)nums.size() - 1, INT_MIN, INT_MAX);
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        if (nums.length < 2) {
            return nums.length;
        }

        int ans = 0;
        for (int i = 0; i < nums.length; i++) {
            int j = Arrays.binarySearch(nums, nums[i]);
            if (i == j) {
                ans++;
            }
        }

        return ans;
    }
}
                    


                        Solution in Python : 
                            
def remove_unfindable(nums, candidates):
    if nums:
        mid = (len(nums) - 1) // 2
        for num in nums[:mid]:
            if num > nums[mid]:
                candidates.discard(num)
        for num in nums[mid + 1 :]:
            if num < nums[mid]:
                candidates.discard(num)
        if any(num in candidates for num in nums[:mid]):
            remove_unfindable(nums[:mid], candidates)
        if any(num in candidates for num in nums[mid + 1 :]):
            remove_unfindable(nums[mid + 1 :], candidates)


class Solution:
    def solve(self, nums):
        candidates = set(nums)
        remove_unfindable(nums, candidates)
        return len(candidates)


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