Angle Between Hands of a Clock

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Steps

setup

Initialize Precomputed Array

Create a 12x60 2D array filled with zeros to store angles for each hour and minute.

💡 This array will hold all possible angles, enabling instant lookup later.

Line:

self.precomputed = [[0]*60 for _ in range(12)]

💡 Pre-allocating the array sets up the data structure needed for efficient precomputation.

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Start Outer Loop Over Hours

Set the hour pointer 'h' to 0 to begin computing angles for hour 0.

💡 We will compute angles for each minute at this hour before moving to the next hour.

Line:

for h in range(12):

💡 The outer loop controls which hour's angles are being computed.

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Start Inner Loop Over Minutes

Set the minute pointer 'm' to 0 to begin computing the angle for hour 0, minute 0.

💡 We compute the angle for each minute at the current hour before moving to the next minute.

Line:

for m in range(60):

💡 The inner loop controls which minute's angle is being computed for the current hour.

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Compute Hour Hand Angle

Calculate the hour hand angle for hour 0 and minute 0 using the formula: hour * 30 + minutes * 0.5.

💡 This formula accounts for the hour hand moving 30 degrees per hour plus 0.5 degrees per minute.

Line:

hour_angle = h * 30 + m * 0.5

💡 The hour hand angle changes gradually as minutes pass, not just in discrete 30-degree jumps.

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Compute Minute Hand Angle

Calculate the minute hand angle for minute 0 using the formula: minutes * 6.

💡 The minute hand moves 6 degrees per minute, completing a full circle in 60 minutes.

Line:

minute_angle = m * 6

💡 Minute hand angle is straightforward and depends only on the minute value.

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Calculate Absolute Difference Between Angles

Compute the absolute difference between the hour and minute hand angles.

💡 This difference represents the raw angle between the two hands before adjusting for the smaller angle.

Line:

diff = abs(hour_angle - minute_angle)

💡 The difference can be more than 180 degrees, so we must consider the smaller angle.

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Store Minimum Angle in Precomputed Array

Store the smaller angle between the two possible angles (diff and 360 - diff) in the precomputed array at position [h][m].

💡 The smaller angle is the actual angle between the clock hands, never more than 180 degrees.

Line:

self.precomputed[h][m] = min(diff, 360 - diff)

💡 This step ensures the angle stored is the smallest possible between the two hands.

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Advance Hour Pointer to 3

Move the hour pointer 'h' to 3 to compute angles for hour 3.

💡 We jump to hour 3 to focus on the canonical example time 3:30.

Line:

for h in range(12):  # h = 3

💡 Focusing on hour 3 prepares for computing the angle at 3:30.

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Set Minute Pointer to 30 for Hour 3

Set the minute pointer 'm' to 30 to compute the angle for hour 3, minute 30.

💡 We focus on minute 30 to illustrate the canonical example time 3:30.

Line:

for m in range(60):  # m = 30

💡 Minute 30 is a key time to understand fractional hour hand movement.

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Compute Hour and Minute Angles at 3:30

Calculate hour hand angle = 3 * 30 + 30 * 0.5 = 105 degrees and minute hand angle = 30 * 6 = 180 degrees.

💡 At 3:30, the hour hand is halfway between 3 and 4, and the minute hand points at 6.

Line:

hour_angle = h * 30 + m * 0.5
minute_angle = m * 6

💡 The hour hand moves continuously, not just on the hour marks.

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Calculate and Store Minimum Angle for 3:30

Calculate absolute difference = 75 degrees and store the smaller angle (75 degrees) in precomputed[3][30].

💡 The angle between the hands at 3:30 is 75 degrees, the expected output.

Line:

diff = abs(hour_angle - minute_angle)
self.precomputed[h][m] = min(diff, 360 - diff)

💡 This confirms the algorithm correctly computes the canonical example angle.

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Query Precomputed Angle for 3:30

Retrieve the precomputed angle for hour 3 and minute 30 from the array.

💡 Using the precomputed array allows instant retrieval without recalculating.

Line:

return self.precomputed[hour % 12][minutes]

💡 Precomputation enables O(1) query time for any hour and minute.

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Output Final Answer

Return the precomputed angle 75 as the final answer for the input time 3:30.

💡 The algorithm outputs the smallest angle between the clock hands at the given time.

Line:

print(clock.angleClock(3, 30))  # Expected: 75

💡 Precomputation allows instant retrieval of the correct angle.

class ClockAngle:
    def __init__(self):
        self.precomputed = [[0]*60 for _ in range(12)]  # STEP 1: Initialize array
        for h in range(12):  # STEP 2: Outer loop over hours
            for m in range(60):  # STEP 3: Inner loop over minutes
                hour_angle = h * 30 + m * 0.5  # STEP 4: Compute hour hand angle
                minute_angle = m * 6  # STEP 5: Compute minute hand angle
                diff = abs(hour_angle - minute_angle)  # STEP 6: Calculate difference
                self.precomputed[h][m] = min(diff, 360 - diff)  # STEP 7: Store smaller angle

    def angleClock(self, hour: int, minutes: int) -> float:
        return self.precomputed[hour % 12][minutes]  # STEP 12: Query precomputed angle

if __name__ == '__main__':
    clock = ClockAngle()
    print(clock.angleClock(3, 30))  # STEP 13: Output final answer (75)
    print(clock.angleClock(12, 0))  # Additional test

📊

Angle Between Hands of a Clock - Watch the Algorithm Execute, Step by Step

Watching each step reveals how precomputation optimizes repeated angle calculations and clarifies the angle formulas.

setup

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Result: "empty 12x60 array initialized"

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Result: "array initialized, starting hour 0"

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Result: "starting minute 0 for hour 0"

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m

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Result: "hour_angle = 0 * 30 + 0 * 0.5 = 0"

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Result: "minute_angle = 0 * 6 = 0"

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m

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Result: "diff = abs(0 - 0) = 0"

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m

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Result: "precomputed[0][0] = 0"

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Result: "starting hour 3"

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Result: "starting minute 30 for hour 3"

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Result: "hour_angle=105, minute_angle=180"

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Result: "precomputed[3][30] = 75"

compare

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Result: 75

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Result: 75

Key Takeaways

✓ Precomputing all possible angles allows instant O(1) queries for any time.

This insight is hard to see from code alone because precomputation loops and query lookups are separated.

✓ The hour hand angle depends on both hour and minute, moving continuously rather than discretely.

Visualizing the gradual movement clarifies why the formula includes minutes * 0.5.

✓ The algorithm always chooses the smaller angle between the two hands, never exceeding 180 degrees.

Seeing the min(diff, 360 - diff) choice in action helps understand why this step is necessary.

Practice

1. You need to find the smallest positive number that is divisible by two given integers, and also find the greatest integer that divides both. Which approach guarantees the most efficient and reliable solution for this problem?

easy

A. Use a brute force search checking multiples and divisors until the answer is found.

B. Use a greedy algorithm that picks the largest divisor first and tries to build the LCM.

C. Apply the Euclidean algorithm iteratively to find GCD, then compute LCM using the GCD.

D. Use dynamic programming to store intermediate divisors and multiples for both numbers.

Solution

1. Step 1: Understand the problem requirements

   The problem requires finding the greatest common divisor (GCD) and least common multiple (LCM) of two integers efficiently.

2. Step 2: Evaluate approaches

   Brute force is inefficient for large inputs. Dynamic programming and greedy approaches do not naturally fit this problem. The Euclidean algorithm efficiently computes GCD in O(log(min(a,b))) time, and LCM can be derived from GCD.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Euclidean algorithm is the standard optimal method for GCD and LCM [OK]

Hint: Euclidean algorithm efficiently finds GCD and LCM [OK]

Common Mistakes:

• Thinking brute force is efficient enough
• Confusing greedy with Euclidean algorithm

2. Consider the following Python code that computes GCD and LCM of two numbers:

def gcd_iterative(a, b):
    while b != 0:
        a, b = b, a % b
    return a

def lcm_safe(a, b):
    gcd = gcd_iterative(a, b)
    return a // gcd * b

if __name__ == '__main__':
    a, b = 8, 12
    gcd = gcd_iterative(a, b)
    lcm = lcm_safe(a, b)
    print("GCD =", gcd)
    print("LCM =", lcm)

What is the output of this program?

easy

A. GCD = 4
LCM = 24

B. GCD = 2
LCM = 48

C. GCD = 4
LCM = 96

D. GCD = 8
LCM = 12

Solution

1. Step 1: Trace gcd_iterative(8, 12)

   Initial: a=8, b=12 Iteration 1: a=12, b=8%12=8 Iteration 2: a=8, b=12%8=4 Iteration 3: a=4, b=8%4=0 Return gcd=4

2. Step 2: Compute lcm_safe(8, 12)

   lcm = (8 // 4) * 12 = 2 * 12 = 24

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   GCD(8,12)=4 and LCM(8,12)=24 [OK]

Hint: GCD divides both numbers; LCM is product divided by GCD [OK]

Common Mistakes:

• Confusing order of modulo in Euclidean algorithm
• Calculating LCM as a*b without dividing by GCD

3. You are given a positive integer and repeatedly replace it by the sum of the squares of its digits. If this process eventually reaches 1, the number is considered "happy"; otherwise, it falls into a cycle that never reaches 1. Which approach guarantees detecting whether a number is happy in optimal time and space?

easy

A. Use a brute force approach storing all intermediate numbers in a hash set to detect cycles.

B. Use dynamic programming to store results for all numbers up to n to avoid recomputation.

C. Use a greedy approach that always picks the largest digit to speed up convergence.

D. Use Floyd's Cycle Detection algorithm (Tortoise and Hare) to detect cycles without extra space.

Solution

1. Step 1: Understand the problem pattern

   The problem involves detecting cycles in a sequence generated by repeatedly summing the squares of digits. Detecting cycles efficiently is key.

2. Step 2: Identify the optimal cycle detection method

   Floyd's Cycle Detection algorithm uses two pointers moving at different speeds to detect cycles in O(1) space, which is optimal compared to storing all seen numbers.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Floyd's algorithm detects cycles without extra memory [OK]

Hint: Cycle detection with two pointers uses constant space [OK]

Common Mistakes:

• Assuming greedy digit selection speeds up convergence
• Using DP unnecessarily increases space
• Brute force uses extra memory

4. You are given a set of points on a 2D plane and need to find the maximum number of points that lie on the same straight line. Which approach guarantees an optimal solution for this problem?

easy

A. Use a greedy approach by sorting points and connecting consecutive points only.

B. Use a divide-and-conquer approach by splitting points into halves and merging results.

C. Use dynamic programming to build up solutions based on subproblems of smaller point sets.

D. Use a nested loop to consider every pair of points, compute slopes as reduced fractions, and count points sharing the same slope from each base point.

Solution

1. Step 1: Understand problem requirements

   The problem requires finding the maximum number of points on a single line, which depends on slopes between points.

2. Step 2: Identify approach that handles all pairs and slope normalization

   Only the nested loop approach with slope normalization (reduced fractions) correctly groups points on the same line and handles duplicates and vertical lines.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Nested loops with slope hashing is the standard optimal approach [OK]

Hint: Check if approach handles all pairs and slope normalization [OK]

Common Mistakes:

• Assuming greedy or DP solves this geometry problem optimally

5. You are given two rectangles on a 2D plane, each defined by their bottom-left and top-right coordinates. You need to determine if they overlap (i.e., share any area). Which approach guarantees an optimal and correct solution for this problem?

easy

A. Use a greedy algorithm to iteratively check if one rectangle can be shifted to overlap the other.

B. Check if the projections of the rectangles on both the x-axis and y-axis overlap using min and max coordinate comparisons.

C. Apply dynamic programming to store intermediate overlap states between sub-rectangles.

D. Use a brute force grid scan to check every point inside the bounding box of both rectangles.

Solution

1. Step 1: Understand the problem requires checking overlap on both axes

   Rectangles overlap if and only if their projections on the x-axis and y-axis overlap simultaneously.

2. Step 2: Recognize that comparing min and max coordinates on each axis is optimal

   Using min and max to find the intersection length on each axis is a constant-time, direct mathematical check.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Checking intervals on both axes ensures correct overlap detection [OK]

Hint: Check overlap on both x and y axes using min/max [OK]

Common Mistakes:

• Assuming overlap if only one axis overlaps
• Trying to solve with DP or greedy unnecessarily
• Using brute force scanning which is inefficient