search-a-2d-matrix.py

# Time:  O(logm + logn)
# Space: O(1)
#
# Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
# 
# Integers in each row are sorted from left to right.
# The first integer of each row is greater than the last integer of the previous row.
# For example,
# 
# Consider the following matrix:
# 
# [
#   [1,   3,  5,  7],
#   [10, 11, 16, 20],
#   [23, 30, 34, 50]
# ]
# Given target = 3, return true.
#

class Solution:
    # @param matrix, a list of lists of integers
    # @param target, an integer
    # @return a boolean
    def searchMatrix(self, matrix, target):
        m = len(matrix)
        n = len(matrix[0])
        i, j = 0, m * n
        
        while i < j:
            mid = i + (j - i) / 2
            val = matrix[mid / n][mid % n]
            if val == target:
                return True
            elif val < target:
                i = mid + 1
            else:
                j = mid
        return False

if __name__ == "__main__":
    matrix = [[1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
    print Solution().searchMatrix(matrix, 3)