Search a 2D Matrix II in Python

   Suppose we have one m x n matrix. We have to write an efficient algorithm that searches for a value in that matrix. This matrix has the following properties −

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

So if the matrix is like −

If target is 5, then return true, if target is 20, then return false

To solve this, we will follow these steps −

• len := number of columns, c1 := 0, c2 := len – 1
• while true
  • if matrix[c1, c2] = target, then return true
  • else if matrix[c1, c2] > target, then c2 := c2 – 1, continue
  • c1 := c1 + 1
  • if c1 >= row count or c2 < 0, then return false
  • return false

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def searchMatrix(self, matrix, target):
      try:
         length = len(matrix[0])
         counter1, counter2 = 0, length-1
         while True:
            if matrix[counter1][counter2] == target:
               return True
            elif matrix[counter1][counter2]>target:
               counter2-=1
               continue
            counter1 = counter1 + 1
            if counter1 >= len(matrix) or counter2<0:
               return False
         except:
            return False
ob1 = Solution()
print(ob1.searchMatrix([[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], 5))

Input

[[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]]
5

Output

True