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1.
Introduction
1.1.
Problem Statement
1.2.
Sample Examples
2.
Brute Force Approach
2.1.
Algorithm
2.2.
Implementation
2.2.1.
Time Complexity
2.2.2.
Space Complexity
3.
Optimized Approach
3.1.
Algorithm
3.2.
Implementation
3.2.1.
Time Complexity
3.2.2.
Space Complexity
4.
4.1.
Why is the time complexity of the optimized approach is O(n)?
4.2.
What do we do in the brute-force approach?
4.3.
Can a matrix be allocated using vectors?
5.
Conclusion
Last Updated: Mar 27, 2024

# Search in a Row wise and Column wise Sorted Matrix

Roadmap to SDE career at Amazon
Speaker
Anubhav Sinha
SDE-2 @
25 Jun, 2024 @ 01:30 PM

## Introduction

This blog covers a matrix problem. Matrices are among the most important and often asked data structures in programming contests and technical interviews. There are various standard matrix problems and techniques. This blog tackles a coding task that involves

searching for an element in an m*n matrix and printing its position.

Recommended Topic, Array Implementation of Queue and  Rabin Karp Algorithm

### Problem Statement

Ninja has given you a n x n matrix and a number x, determine the position of x in the matrix if it exists. If not, print "Not Found."  Every row and column in the specified matrix is sorted in ascending order. The designed algorithm should have a linear time complexity.

### Sample Examples

Example 1

Input

x=36

Output

(2,0)

Explanation

The Element at (2,0) in the matrix is 36.

Example 2

Input

x=21

Output

Explanation

The Element is not present in the matrix.

## Brute Force Approach

The approach is to run two nested loops, one iterates through rows and the other through the columns of the matrix. Check each element of the matrix for the enquired element and if found return the index of the element. If at the end of the search, the element is not found, print element is not found.

### Algorithm

1. Run a nested loop, one for iterating every row and one for iterating every column.

2. Check each index for x and output the index of the element if it is found.

3. If the element cannot be found, print "Element not found".

### Implementation

``````// Program to search an element in row-wise
// and column-wise sorted matrix by brute-force approach
#include <bits/stdc++.h>

using namespace std;

int bruteSearch(int matrix[3][3], int p, int x)
{
if (p == 0)
return -1;

// start traversing through the entire matrix
for (int i = 0; i < p; i++)
{
for (int j = 0; j < p; j++)
// if element is found return the index
if (matrix[i][j] == x)
{
cout << "(" << i << ", "
<< j << ")"
<< "\n";
return 1;
}
}

return 0;
}

// Driver code
int main()
{
int matrix[3][3] = {{10, 14, 18},
{22, 26, 30},
{34, 38, 42}};
bruteSearch(matrix, 3, 30);
return 0;
}``````

Output

``(1,2)``

#### Time Complexity

The time complexity of the above approach is O(n2) as the solution iterates through every row and column of the matrix.

#### Space Complexity

The program's auxiliary space is O(1) because no additional data structure is involved.

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## Optimized Approach

To locate the element, a better way is to use Divide and Conquer. In each comparison, we begin by eliminating a row or column in each comparison until an element is discovered. We should begin our search from the top-right corner of the matrix.

### Algorithm

1. Say the provided element is x, then define two variables i = 0 and j = n-1 as row and column indexes.

2. Run a loop until we get i=n.

3. If the current element is more than x, reduce the count of j and Remove the current column.

4. If the current element is smaller than x, increment the count of i and remove the current row.

5. If the element is equal, print the position of the element.

### Implementation

``````// Program to search an element in a
// sorted matrix by divide and conquer method
#include <bits/stdc++.h>

using namespace std;

int optimizedSearch(int matrix[3][3], int n, int x)
{
if (n == 0)
return -1;

int s = matrix[0][0], l = matrix[n - 1][n - 1];
if (x < s || x > l)
return -1;

// set index for top right element in the matrix
int i = 0, j = n - 1;
while (i < n && j >= 0)
{
if (matrix[i][j] == x)
{
cout << "("
<< i << "," << j << ")";
return 1;
}
if (matrix[i][j] > x)
j--;

// Check if matrix[i][j] < x
else
i++;
}

return 0;
}

// Driver code
int main()
{
int matrix[3][3] = {{10, 14, 18},
{22, 26, 30},
{34, 38, 42}};

optimizedSearch(matrix, 3, 30);
return 0;
}``````

Output

``(1,2)``

#### Time Complexity

The time complexity of the optimized approach will be O(n). Since only one traversal is required, i can go from 0 to n and j can go from n - 1 to 0 in at most 2*n steps, where n x n is the size of the matrix.

#### Space Complexity

The program's auxiliary space is O(1) because no additional space is required.

## Frequently Asked Questions

### Why is the time complexity of the optimized approach is O(n)?

Since only one traversal is required, i can go from 0 to n and j can go from n-1 to 0 in at most 2*n steps,  hence time complexity is O(n).

### What do we do in the brute-force approach?

The approach is to run two nested loops, one iterates through rows and the other through the columns of the matrix. Check each element of the matrix for the enquired element and if found return the index of the element

### Can a matrix be allocated using vectors?

Yes, a 2-d matrix can be allocated using a vector of vectors.

## Conclusion

This article extensively discussed the problem of searching for an element in a row-wise and column-wise sorted 2-D matrix and returning its position and its time and space complexities.

We hope this blog has helped you enhance your knowledge regarding matrices. After reading about the matrices, are you not feeling excited to read/explore more articles on this topic? Don't worry, Coding Ninjas has you covered. To learn, Reset Matrix ProblemSorted MatrixBook Allocation Problem and Unique Matrix.

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