HackerRank 'The Grid Search' Solution

Martin Kysel · April 2, 2015

Short Problem Definition:

Given a 2D array of digits, try to find the location of a given 2D pattern of digits

The Grid Search


time complexity is O(n^2 \* m^2)

space complexity is O(1)


There are many sophisticated 2d pattern matching algorithms out there. Just think of computer vision, robotics, gaming… The issue with most of them is, that they are rather heuristics than complete matches. I first started solving this challenge by using brute force and well.. It passed the time constraints. I did not expect that. If anyone has a good way of reducing the number of subarray checks, please post it in the comments!

SubArray-Check reduction techniques:

  • keep the sum of the sub-array in a separate grid, check only if sum matches the pattern
  • Rabin-Karp string searching algorithm on each line to fast forward

def matchSubArray(arr, pat, x, y, pat_y, pat_x):
    for running_y in xrange(pat_y):
        for running_x in xrange(pat_x):
            if arr[running_y+y][running_x+x] != pat[running_y][running_x]:
                return False
    return True
def solveBruteForce(arr, pat, arr_y, arr_x, pat_y, pat_x):
    for y in xrange(arr_y-pat_y+1):
        for x in xrange(arr_x-pat_x+1):
            if matchSubArray(arr, pat, x, y, pat_y, pat_x):
                return True
    return False
if __name__ == '__main__':
    t = int(raw_input())
    for _ in xrange(t):
        arr_y, arr_x = map(int, raw_input().split())
        arr = [0] * arr_y
        for y in xrange(arr_y):
            arr[y] = list(raw_input())
        pat_y, pat_x = map(int, raw_input().split())
        pat = [0] * pat_y
        for y in xrange(pat_y):
            pat[y] = list(raw_input())
        if solveBruteForce(arr, pat, arr_y, arr_x, pat_y, pat_x):
            print "YES"
            print "NO"

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