Given a set S of n points and a query point p, find the point in S closest to p.
For uniformly distributed points, a good strategy is to represent the space as a grid and maintain a list of inner points for each cell. For a given query point, we can check the cell the point falls into and its neighbouring cells. For a sqrt(n) * sqrt(n) grid we’ll have one point per cell, on average. So, on average, finding the point in S closest to p, requires traversing a constant number of cells.

Longest common subsequence solution

 
h2. Longest common subsequence solution

Given two strings A (n characters) and B (m characters), find their longest common subsequence. (eg. The longest common sub sequence for abcabc and abcbcca is abcbc.)

!<{margin-right: 20px; auto;display:block;}blog/square-root-trick?image02.png 90%!
We can start from the last saved row to n to compute the solution path from the n - 1 line to [n/k] * k line. And then go lower and lower to compute the part of the solution between the ik (i+1)k. Computing part of the path between row ik and row (i+1)k takes k * m space and k * m time. Computing the whole path takes n/k * (k * m) = nm time and km space, keeping every kth row in memory takes [n/k]m. again we minimize memory by using k = sqrt n

Caveats

h2. Caveats

The above problems have better solutions using interval trees or some other clever tricks. The sqrt trick is nice, since it improves the naive solution a lot without much effort.

Additional problems

h2. Additional problems

1. (Josephus problem) n people numbered from 1 to n sit in a circle and play a game. Starting from the first person and every kth person is eliminated. Write an algorithm that prints out the order in which people are eliminated.
2. (Level ancestor) You are given an tree of size n. ancestor(node, levelsUp) finds the node’s ancestor that is levelsUp steps up. For example, ancestor(node, 1) returns the father and ancestor(node, 2) returns the grandfather. Implement ancestor(node, levelsUp) efficiently. (O(sqrt(n)) per query)