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Coincidence Search in C++
Suppose we have a list of unique integers called nums. We have to find the number of integers that could still be successfully found using a standard binary search.
So, if the input is like [2,6,4,3,10], then the output will be 3, as if we use binary search to look for 4, we can find it at first iteration. We can also find 2 and 10 after two iterations.
To solve this, we will follow these steps −
Define a function help(), this will take target, an array & nums,
low := 0
high := size of nums - 1
while low <= high, do −
mid := low + (high - low) / 2
if nums[mid] is same as target, then −
return true
if nums[mid] < target, then −
low := mid + 1
Otherwise
high := mid - 1
return false
From the main method, do the following −
ret := 0
for each element i in nums
ret := ret + help(i, nums)
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h> using namespace std; class Solution { public: bool help(int target, vector<int> & nums) { int low = 0; int high = nums.size() - 1; while (low <= high) { int mid = low + (high - low) / 2; if (nums[mid] == target) return true; if (nums[mid] < target) { low = mid + 1; } else { high = mid - 1; } } return false; } int solve(vector<int> & nums) { int ret = 0; for (int i : nums) { ret += help(i, nums); } return ret; } }; main() { Solution ob; vector<int> v = {2,6,4,3,10}; cout << (ob.solve(v)); }
Input
{2,6,4,3,10}
Output
3