**Difficulty:** Easy, **Asked-in:** Microsoft, Amazon, Adobe, Directi, Goldman Sachs, Paytm, SAP, Visa

**Key takeaway:** An excellent problem to learn problem-solving using the binary search.

You are given a row-wise and column-wise sorted 2D matrix and an integer k, write a program to find whether the integer k is present in the matrix or not. The 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.

**Examples**

```
Input: k = 6
mat[][] =
[
[1, 2, 6, 7],
[12, 13, 16, 21],
[23, 35, 36, 48]
]
Output: true
Explanation: value 6 is present in the matrix.
Input: k = 7
mat[][] =
[
[1, 2, 6],
[12, 13, 16],
[23, 35, 36]
]
Output: fasle
Explanation: value 7 is not present in the matrix.
```

- A brute force approach using nested loops
- An efficient approach using binary search

A simple approach would be to traverse the whole matrix using nested loops and check whether the k is present or not.

```
bool searchMatrix(int mat[][], int n, int m, int k)
{
for(int i = 0, i < n; i = i + 1)
{
for(j = 0; j < m; j = j + 1)
{
if(mat[i][j] == k)
return 1
}
}
return -1
}
```

We are using two nested loops, where the outer loop selects the row, and the inner loop keeps track of the column. So time complexity = O(n*m), Here n is the number of rows, and m is the number of columns. Space Complexity = O(1), we are not using any additional space.

Now the critical question is: How can we improve the time complexity? Can we use the sorted order property to solve it efficiently? Let's think!

Here each matrix row is sorted i.e. the first element of a row is greater than or equal to the last number of the preceding row. Therefore, the matrix can be viewed as a sorted one-dimensional array (both row-wise and column-wise sorted). Can we apply the idea of binary search? Let's think!

If we pick any ith row then all the values in the ith row would be:

- Sorted in increasing order.
- Greater than all the values in the (i - 1)th row.
- Less than all the values in the (i + 1)th row.

Here is a solution idea using the above observation:

- We first apply binary search on rows to find out the row in which k must lie, i.e., we need to find a row such that the first element of the row is greater than or equal to k and the last element of the row is less than or equal to k.
- Then, we apply binary search in that row to find out if k is present in the matrix or not.

```
bool searchMatrix(int mat[][], int n, int m, int k)
{
int l = 0, r = m - 1
while (l <= r)
{
int mid = l + (r - l) / 2
if (k >= mat[mid][0] && k <= mat[mid][n-1])
return searchRow(mat[mid], n, k)
if (k < mat[mid][0])
r = mid - 1
else
l = mid + 1
}
return false
}
```

**Binary search implementation**

```
bool searchRow(int M[], int n, int k)
{
int l = 0, r = n - 1
while (l <= r)
{
int mid = l + (r - l) / 2
if (M[mid] == k)
return true
if (k < M[mid])
r = mid - 1
else
l = mid + 1
}
return false
}
```

- Time complexity = Time complexity of searching row where k is present + Time complexity of searching value k = O(logn) + O(logm) = O(log n + log m).
- Space complexity = O(1). Think!

- Suppose our matrix is just sorted in-row and column-wise but not strictly sorted i.e the first integer of each row is not greater than the last integer of the previous row. Can we solve the problem using the above approach?
- How do we modify the above code, if we need to return the index (i, j) of the value k, if k is present in the matrix?
- Can we solve the above problem by searching the column and then searching elements in that column?
- How can we implement the above approach using a recursive binary search? What would be the space complexity?

- Using nested loops: Time = O(nm), Space = O(1)
- Using binary search: Time = O(logn + logm), Space = O(1)

- Search in a row-wise sorted 2D matrix
- Median of two sorted arrays
- Binary search algorithm
- Find the first and last position of an element in a sorted array
- Find the row with the maximum number of 1s
- Max in an array which is first increasing and then decreasing

Thanks to Navtosh Kumar for his contribution in creating the first version of the content. Enjoy learning, Enjoy coding, Enjoy algorithms!

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