Binary Search

Binary Search can usually be used to find an optimal solution/value, if the results are monotonous (i.e. one-directional):

• Boolean: TTTTTTTTTTTTTTFFFFFFFFFFFFFFFF

• Numbers: [1, 2, 3, 5, 6, 88, 999]

The data needs to be sorted in a one-directional order, such that every time we are testing for a mid point, we can always eliminate half of the data, resulting in an extremely efficient solution.

Template

Note: The formula for mid , the while loop condition, and updating l and r can vary slightly depending on the problem context. Also, when the numbers go into negative domain, it can be different and treated with care!

int l = 0, r = n;
while(l < r){
    int mid = (l+r+1)/2;
    if(function(mid))
        l = mid;
    else
        r = mid-1;
}
return l;

Trials and Errors

This is a common way to employ Binary Search to efficiently find a desired optimal value (min/max) under a specific problem context. The method is to try possible values one by one and test if it satisfies the conditions

If we notice that:

1. The outcome of each trial is in monotonous order.

2. There are obvious easy/efficient ways to test a number in the problem. (e.g. O(N) time complexity)

Binary Search will likely be an ideal solution.