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...