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Question 1: Minimum Cost to Adjust Pixel Intensities

Keywords: image processing, dynamic programming, matrix manipulation

Problem Statement:

A grid-like representation of an image is divided into n rows and m columns, with each cell containing pixel intensity values. The aim is to improve the visibility of specific objects in the image.

The goal is to adjust the intensity of each pixel to ensure that no pixel in the previous rows (within the same column) has a higher or equal intensity than the given pixel. Each increase in pixel intensity has a cost of one unit per unit of increase.

Solution Approach:

1. Greedy Approach: Traverse the matrix from top to bottom. For each row, ensure that the pixel intensity is strictly greater than the pixel in the row above (in the same column). If the condition is not met, increase the pixel intensity, and add the cost of this increase.
2. Cost Calculation: Keep track of the total cost as we adjust the intensities to meet the requirements.

Sample Code:

def getMinimumCost(pixelIntensity):
    n = len(pixelIntensity)
    m = len(pixelIntensity[0])
    total_cost = 0

    for col in range(m):
        for row in range(1, n):
            if pixelIntensity[row][col] <= pixelIntensity[row - 1][col]:
                increment = pixelIntensity[row - 1][col] - pixelIntensity[row][col] + 1
                total_cost += increment
                pixelIntensity[row][col] += increment

    return total_cost

# Example usage:
pixelIntensity = [
    [2, 4, 6],
    [4, 2, 7],
    [6, 4, 7]
]
print(getMinimumCost(pixelIntensity))  # Output: 6

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Question 2: SQL Grade Assignment

Keywords: SQL, query, conditional logic

Problem Statement:

You are given a table named students that contains the total marks of students in a class. The teacher wants to assign grades to the students based on the following criteria:

• Marks > 90 -> Grade A+
• Marks > 70 -> Grade A
• Marks > 50 -> Grade B
• Marks >= 40 -> Grade C
• Marks < 40 -> Fail

Write an SQL query to return the ID, name, marks, and grade for each student.

Sample SQL Query:

SELECT
    id,
    name,
    marks,
    CASE
        WHEN marks > 90 THEN 'A+'
        WHEN marks > 70 THEN 'A'
        WHEN marks > 50 THEN 'B'
        WHEN marks >= 40 THEN 'C'
        ELSE 'Fail'
    END AS grade
FROM students;

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