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How to Find a Specific Pair in a Matrix

October 20, 2024

Finding a specific pair in a matrix that meets a certain condition is a popular problem in coding interviews and competitive programming. The challenge typically involves identifying a pair of elements in the matrix that satisfies a specific criterion, such as having the maximum difference, or finding two elements that sum up to a given number. These problems test your ability to efficiently traverse and manipulate matrix data structures.

In this blog, we’ll explore an optimized approach to finding a specific pair in a matrix where we aim to find two elements such that the difference between the two elements is maximized. We’ll also discuss the brute-force method and an optimized approach for better performance.

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How to Find a Specific Pair in a Matrix

Table of Contents

Problem Statement

Given an n x n matrix, find a pair (a, b) such that a is positioned before b in the matrix (i.e., a appears before b when traversing the matrix row by row) and the difference b - a is maximized.

Example:

Input:
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]

Output: 8 (Difference between 9 and 1)

In this example, the matrix has 9 as the largest element and 1 as the smallest element, so the difference 9 - 1 equals 8.

Approaches to Solve the Problem

1. Brute Force Approach

The most straightforward way to solve this problem is to iterate over all possible pairs in the matrix, check the difference for each pair, and keep track of the maximum difference.

Steps:

Traverse every element in the matrix.
For each element, compare it with every element that appears after it.
Keep track of the maximum difference b - a where a is before b.

Code Example (Brute Force):

def findMaxDifference(matrix):
    n = len(matrix)
    max_diff = float('-inf')

    # Traverse through all pairs in the matrix
    for i in range(n):
        for j in range(n):
            for x in range(i, n):
                for y in range((j if i == x else 0), n):
                    if matrix[x][y] > matrix[i][j]:
                        max_diff = max(max_diff, matrix[x][y] - matrix[i][j])

    return max_diff

# Example usage
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]

print("Maximum difference is:", findMaxDifference(matrix))

Time Complexity: O(n^4), where n is the number of rows or columns in the matrix. This approach is highly inefficient for large matrices.

2. Optimized Approach

A more efficient way to solve this problem is by keeping track of the maximum elements as we traverse the matrix from the bottom-right corner to the top-left corner. This allows us to minimize comparisons and avoid redundant calculations.

Key Idea:

Start from the bottom-right corner and move to the top-left, keeping track of the maximum element seen so far.
For each element a at position (i, j), check the maximum element seen so far in the sub-matrix starting from (i+1, j+1) to the bottom-right corner.
The difference max_so_far - matrix[i][j] is calculated, and the maximum difference is updated.

Steps:

Initialize a variable max_so_far to store the maximum element in the sub-matrix starting from (i+1, j+1).
Traverse the matrix from bottom-right to top-left.
For each element matrix[i][j], calculate the difference max_so_far - matrix[i][j] and update the result if the difference is larger than the current result.
Update max_so_far with matrix[i][j] if matrix[i][j] is greater than max_so_far.

Code Example (Optimized Approach):

def findMaxDifference(matrix):
    n = len(matrix)
    max_diff = float('-inf')

    # Create a matrix to store the maximum elements from (i, j) to the bottom-right corner
    max_matrix = [[0 for _ in range(n)] for _ in range(n)]

    # Initialize the bottom-right element
    max_matrix[n-1][n-1] = matrix[n-1][n-1]

    # Fill the last row
    for j in range(n-2, -1, -1):
        max_matrix[n-1][j] = max(matrix[n-1][j], max_matrix[n-1][j+1])

    # Fill the last column
    for i in range(n-2, -1, -1):
        max_matrix[i][n-1] = max(matrix[i][n-1], max_matrix[i+1][n-1])

    # Fill the rest of the matrix
    for i in range(n-2, -1, -1):
        for j in range(n-2, -1, -1):
            max_matrix[i][j] = max(matrix[i][j], max(max_matrix[i+1][j], max_matrix[i][j+1]))

    # Traverse the matrix and find the maximum difference
    for i in range(n-1):
        for j in range(n-1):
            max_diff = max(max_diff, max_matrix[i+1][j+1] - matrix[i][j])

    return max_diff

# Example usage
matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]

print("Maximum difference is:", findMaxDifference(matrix))

Explanation:

We first create a matrix max_matrix where each element stores the maximum value from that position to the bottom-right corner.
Once the max_matrix is built, we calculate the maximum difference by checking every possible pair (a, b) where b lies in the sub-matrix starting at (i+1, j+1).

Time Complexity: O(n²), where n is the number of rows or columns in the matrix. This is significantly faster than the brute-force approach.

Space Complexity: O(n²) for the max_matrix storage.

Comparison of Approaches

Approach	Time Complexity	Space Complexity
Brute Force	O(n^4)	O(1)
Optimized Approach	O(n²)	O(n²)

The optimized approach is significantly more efficient than the brute-force method, making it ideal for larger matrices.

Conclusion

Finding a specific pair in a matrix with the maximum difference can be solved using both brute-force and optimized approaches. While the brute-force method is easier to understand, it is impractical for large matrices due to its high time complexity. The optimized approach that utilizes a max_matrix to store the maximum values in the sub-matrix drastically reduces the number of comparisons and is far more efficient.

Mastering this problem will improve your understanding of matrix traversal, dynamic programming, and optimization techniques, which are critical for solving more complex matrix-related problems in competitive programming and coding interviews.

Call-to-Action

Now that you’ve learned how to find a specific pair in a matrix with the maximum difference, try implementing this solution for other matrix-based problems on coding platforms like LeetCode, GeeksforGeeks, and HackerRank. For more tutorials and in-depth coding guides, subscribe to our newsletter!