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by Ganesh Kumar MAugust 11th, 2020

We have to search for a value *x*** **in a sorted matrix ** M**. If

Let us consider the above matrix as an example. In this example, we are going to search for the value **12.** Since 12 is present in the matrix, the algorithm should return its coordinates **(2, 1**)

A simple approach is to traverse all the values in the matrix and check if it is equal to 12.

**Time Complexity**The worst case time complexity of the above algorithm will be

The above algorithm behaves worse for large values of n and m. Let us look into the efficient algorithm now.

- Start from Top Right position (
in the matrix*0, m - 1)**M* - If the value is equal to
return*x**(0, m - 1)* - Move one row down if the current value is less than
*x* - Move one column left if the current value is greater that
*x*

Let us apply this algorithm into our matrix ** M.** We are going to search for the value

1. Start from the Top Right value **5 **at *M***[0][4]. 5 **is less than 12, so 12 should be somewhere in the bottom of the matrix since all row and column values are sorted in ascending order. So we move one row down.

2. The value **10 **at *M***[1][4]** is also less than 12, so we move one row down

3. The value **15 **at *M***[2][4]** is greater than 12, so 12 should be somewhere in the left of the matrix, so we move one column left.

4. The value **14 **at *M***[2][3]** is also greater than 12, so we move one column left

5. The value **13 **at *M***[2][2]** is greater than 12, so we move one column left

6. The value at *M***[2][1]** is equal to 12, so we return its index **(2, 1)**

**Time Complexity**

The worst case time complexity of the above algorithm will be**O(n + m) = O(2n) = O(n)** when **n = m, ***because we will be iterating all rows and all columns once if ***x ***is at the bottom left position* **(n - 1, 0)**

You can find the Java solution for this algorithm in the Github repo.

https://github.com/ganeshkumarm1/DSAlgo/tree/master/src/com/dsalgo/Algorithms/BinarySearch2D

Thank you 🙏 👋🤘

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