Time complexity analysis of search problem

February 3, 2022

I was introduced to an example interview question, where given a 2D array of characters and a word, find the word in the array, where the next character of the word is either to the right or below the previous character, and return a list of the co-ordinates of each character found.

I wrote the following code to solve this:

co_ords = []

def init_recursion(word, grid):
    co_ords.clear()
    return find_word(word, grid, 0, 0, True, 0)

def find_word(word, grid, x, y, explore, pos):
    word_len = len(word) - pos
    if word_len == 0:
        return True

    x_size = len(grid[0])
    y_size = len(grid)

    if word_len > (x_size - x) + (y_size - y):
        return False

    try:
        char_on = grid[y][x]

        if word[pos] == char_on:
            co_ords.append((y, x))
            if find_word(word, grid, x + 1, y, False, pos + 1):
                return True
            elif find_word(word, grid, x, y + 1, False, pos + 1):
                return True
            else:
                co_ords.pop()
        if word_len - 1 > (x_size - x) + (y_size - y):
            return False
        if explore and find_word(word, grid, x + 1, y, True, 0):
            return True
        else:
            return explore and find_word(word, grid, x, y + 1, True, 0)
    except IndexError:
        return False

# Valid grid and word combinations:
grid1 = [
    ['c', 'c', 'x', 't', 'i', 'b'],
    ['c', 'c', 'a', 't', 'n', 'i'],
    ['a', 'c', 'n', 'n', 't', 't'],
    ['t', 'c', 's', 'i', 'p', 't'],
    ['a', 'o', 'o', 'o', 'a', 'o'],
    ['o', 'a', 'a', 'a', 'o', 'o'],
    ['k', 'a', 'i', 'c', 'k', 'i']
]
word1 = 'catnip' >>> [(1, 1), (1, 2), (1, 3), (2, 3), (3, 3), (3, 4)]
word2 = 'cccc' >>> [(0, 0), (0, 1), (1, 1), (2, 1)] (one choice of many)
word3 = 's' >>> [(3, 2)]
word4 = 'bit' >>> [(0, 5), (1, 5), (2, 5)]

My complexity analysis for this is as follows:
Where:

R = number of rows
C = number of columns
W = characters in word

The find_word function will loop through the entire 2D grid RC – W² times, and for each exploration call, could recuse up to Wlog(W) times attempting to find a match. There would therefore be a total of Wlog(W)(RC – W²) recursive calls at the worst case.

Hence my questions:

One, have I got the complexity analysis of the above correct?
Two, in big O notation (worst case time complexity), how would this be represented?
I think that it could be one of the following:
1. O(RCW(log(W)) – W³)
2. O(RCW(log(W)))
  But I’m not sure which – normally the smaller term would be ignored in big O, however I’m not sure, as W is present in both the n³ and n³log(n) terms.

>Solution :

I disagree with your search depth being O(Wlog(W)). In the worst case, where every intermediate letter (except the last) matches the word, your recursion will have a branching factor of 2, and will explore 2^W paths. Take, as an example, an extremely large grid, filled entirely with As, and a sole B in the lower right corner, and a search word AAA...AB. You can see how that would explode exponentially (with respect to W).

You are correct too, in that you’d ignore the smaller term, so overall I think the time complexity should be O(RC*2^W).