Arrange number 1..n so that each pair of consecutive numbers sum up to a square number

Problem Statement

Write a program to arrange numbers from 1 to n such that the sum of two consecutive numbers is a perfect square.

An interesting way of looking at this problem

Make a graph so that each node is one of the numbers and there is an edge between two nodes if these two numbers can add up to a square number. Find a path starting from one node and end up in another so that each node is visited once and all nodes are visited. Such path may not exist.

There should be an easier way to do this by seeking for nodes with only one edge in/out...

Given a dictionary of words and a string with all spaces removed, determine whether the string is composed of valid words

Problem Statement

Given a dictionary of words and a string with all spaces removed, determine whether the string is composed of valid words

Examples

helloworld-> hello world (valid)

isitniceinhere-> is it nice in here (valid)

zxyy-> invalid

Time Complexity Requirement

O(n^2)

Algorithm

True/False array T of size equal to the input. T[i] = true => the substring starting from i is composed of valid words.

Assume API method IsValidWord() is O(1) since we are given a dictionary.

for each element in the array:

T[i] = (IsValidWord(s.substring(i, 1)) && T[i+1])

|| (IsValidWord(s.substring(i, 2)) && T[i+2])

...

|| (IsValidWord(s.substring(i, n-i-1)) && T[n-1])

The result is T[0].

Complexity analysis

for each i, it takes O(n) to compute. and we start the computation from i = s.length - 1 and decrease i, and there are n element in the array to compute. So we get O(n^2) as the total time complexity.

Given two strings of a and b. Find the max window in a and b where the substrings in the window are anagrams.

Problem description:

1. Given two strings a and b, find the max value of "length" where for some i and j, a.substring(i, i+length) and b.substring(j, j+length) are anagrams.

2. Given two strings a and b, find the max window of matching patterns between a and b where the pattern matching doesn't have to be an exact match but also can be a permutation of each other. (which is essentially anagrams...)

Algorithm:

let m = a.length;
let n = b.length;
let matrix = int[m][n]

if a[i] == b[j] then matrix[i][j] = 1
else matrix[i][j] = 0

With the matrix populated by the routines above, find the largest square that meets the following two conditions:

• no column contains all 0s. For each column, at least one cell is 1
  
• no rows contains all 0s

The time complexity is O(n^2). I don't believe it can be done in O(nlogn). But I could be wrong.

Someone was asked the following question during interview: if m = n, would that help speeding up the complexity?

Well, I have no idea...