Minimum Area Rectangle

▶

Steps

setup

Initialize point set and minimum area

Convert the list of points into a hash set for O(1) lookup. Initialize min_area to infinity to track the smallest rectangle area found.

💡 Using a hash set allows quick existence checks for points, which is crucial for efficient rectangle detection.

Line:

point_set = set(map(tuple, points))
min_area = float('inf')

💡 Hash set creation enables constant-time checks for the presence of points.

traverse

Start outer loop with i=0

Set pointer i to 0, selecting the first point [1,1] as the first corner candidate.

💡 We will compare this point with all subsequent points to find potential diagonal pairs.

Line:

for i in range(n):

💡 Outer loop selects the first point of the pair.

traverse

Start inner loop with j=1

Set pointer j to 1, selecting the second point [1,3] to pair with point i=0.

💡 We will check if points i and j can be diagonal corners of a rectangle.

Line:

for j in range(i+1, n):

💡 Inner loop selects the second point of the pair.

compare

Check if points i=0 and j=1 can be diagonal corners

Compare x and y coordinates of points [1,1] and [1,3]. Since x1 == x2, they cannot be diagonal corners.

💡 Diagonal corners must have different x and y coordinates to form a rectangle.

Line:

if x1 != x2 and y1 != y2:

💡 Points with the same x or y coordinate cannot be diagonal corners.

traverse

Move inner pointer j to 2

Increment j to 2, selecting point [3,1] to pair with i=0.

💡 We continue checking pairs with the first point fixed at i=0.

Line:

for j in range(i+1, n):

💡 Next candidate pair is points at indices 0 and 2.

compare

Check if points i=0 and j=2 can be diagonal corners

Compare points [1,1] and [3,1]. Since y1 == y2, they cannot be diagonal corners.

💡 Diagonal corners must differ in both x and y coordinates.

Line:

if x1 != x2 and y1 != y2:

💡 Points sharing the same y coordinate cannot be diagonal corners.

traverse

Move inner pointer j to 3

Increment j to 3, selecting point [3,3] to pair with i=0.

💡 We continue checking pairs with i fixed at 0.

Line:

for j in range(i+1, n):

💡 Next candidate pair is points at indices 0 and 3.

compare

Check if points i=0 and j=3 can be diagonal corners

Points [1,1] and [3,3] differ in both x and y, so they can be diagonal corners. Check if the other two corners exist in the point set.

💡 Finding the other two corners confirms if these points form a rectangle.

Line:

if (x1, y2) in point_set and (x2, y1) in point_set:

💡 Presence of the other two corners confirms a rectangle.

insert

Calculate area and update min_area

Calculate area = |1-3| * |1-3| = 2 * 2 = 4. Since 4 < inf, update min_area to 4.

💡 Updating min_area tracks the smallest rectangle found so far.

Line:

area = abs(x1 - x2) * abs(y1 - y2)
if area < min_area:
    min_area = area

💡 We found a valid rectangle with area 4, which is currently the smallest.

traverse

Move inner pointer j to 4

Increment j to 4, selecting point [2,2] to pair with i=0.

💡 Check if this pair can form a rectangle diagonal.

Line:

for j in range(i+1, n):

💡 Next candidate pair is points at indices 0 and 4.

compare

Check if points i=0 and j=4 can be diagonal corners

Points [1,1] and [2,2] differ in both x and y. Check if (1,2) and (2,1) exist in point_set.

💡 Both other corners must exist to form a rectangle.

Line:

if (x1, y2) in point_set and (x2, y1) in point_set:

💡 Missing corners mean no rectangle formed by this pair.

traverse

Move outer pointer i to 1

Increment i to 1, selecting point [1,3] as the first corner candidate.

💡 We now check pairs starting from the second point.

Line:

for i in range(n):

💡 Outer loop moves to next point.

traverse

Start inner loop with j=2

Set j to 2, selecting point [3,1] to pair with i=1.

💡 Check if these points can be diagonal corners.

Line:

for j in range(i+1, n):

💡 Next candidate pair is points at indices 1 and 2.

compare

Check if points i=1 and j=2 can be diagonal corners

Points [1,3] and [3,1] differ in both x and y. Check if (1,1) and (3,3) exist in point_set.

💡 Verify the other two corners to confirm rectangle.

Line:

if (x1, y2) in point_set and (x2, y1) in point_set:

💡 This is the same rectangle found earlier, area 4.

compare

Calculate area and compare with min_area

Calculate area = |1-3| * |3-1| = 2 * 2 = 4. Since 4 is not less than current min_area 4, do not update.

💡 We only update min_area if a smaller rectangle is found.

Line:

area = abs(x1 - x2) * abs(y1 - y2)
if area < min_area:
    min_area = area

💡 Multiple pairs can form rectangles of the same area.

traverse

Move inner pointer j to 3

Increment j to 3, selecting point [3,3] to pair with i=1.

💡 Check next pair for rectangle possibility.

Line:

for j in range(i+1, n):

💡 Next candidate pair is points at indices 1 and 3.

compare

Check if points i=1 and j=3 can be diagonal corners

Points [1,3] and [3,3] share the same y coordinate, so they cannot be diagonal corners.

💡 Diagonal corners must differ in both x and y.

Line:

if x1 != x2 and y1 != y2:

💡 Points with same y coordinate cannot be diagonal corners.

traverse

Move inner pointer j to 4

Increment j to 4, selecting point [2,2] to pair with i=1.

💡 Check next pair for rectangle possibility.

Line:

for j in range(i+1, n):

💡 Next candidate pair is points at indices 1 and 4.

compare

Check if points i=1 and j=4 can be diagonal corners

Points [1,3] and [2,2] differ in both x and y. Check if (1,2) and (2,3) exist in point_set.

💡 Both other corners must exist to form a rectangle.

Line:

if (x1, y2) in point_set and (x2, y1) in point_set:

💡 Missing corners mean no rectangle formed by this pair.

reconstruct

End loops and return result

All pairs checked. Since min_area is 4 (not infinity), return 4 as the minimum rectangle area.

💡 Returning the smallest area found or 0 if none found.

Line:

return 0 if min_area == float('inf') else min_area

💡 The algorithm successfully found the minimum rectangle area.

def minAreaRect(points):
    point_set = set(map(tuple, points))  # STEP 1
    min_area = float('inf')  # STEP 1
    n = len(points)
    for i in range(n):  # STEP 2, 12
        for j in range(i+1, n):  # STEP 3,5,7,10,13,16,18
            x1, y1 = points[i]
            x2, y2 = points[j]
            if x1 != x2 and y1 != y2:  # STEP 4,6,8,11,14,17,19
                if (x1, y2) in point_set and (x2, y1) in point_set:  # STEP 8,11,14,19
                    area = abs(x1 - x2) * abs(y1 - y2)  # STEP 9,15
                    if area < min_area:  # STEP 9
                        min_area = area  # STEP 9
    return 0 if min_area == float('inf') else min_area  # STEP 20

points = [[1,1],[1,3],[3,1],[3,3],[2,2]]
print(minAreaRect(points))

📊

Minimum Area Rectangle - Watch the Algorithm Execute, Step by Step

Watching each pair comparison and the hash set lookups helps you understand how the algorithm efficiently finds rectangles without checking every quadruple of points.

setup

[1,1]

0

[1,3]

1

[3,1]

2

[3,3]

3

[2,2]

4

Result: "inf"

move_i

i

[1,1]

0

[1,3]

1

[3,1]

2

[3,3]

3

[2,2]

4

Result: "inf"

move_j

i

[1,1]

0

j

[1,3]

1

[3,1]

2

[3,3]

3

[2,2]

4

Result: "inf"

compare

i

[1,1]

0

j

[1,3]

1

[3,1]

2

[3,3]

3

[2,2]

4

Result: "inf"

move_j

i

[1,1]

0

[1,3]

1

j

[3,1]

2

[3,3]

3

[2,2]

4

Result: "inf"

compare

i

[1,1]

0

[1,3]

1

j

[3,1]

2

[3,3]

3

[2,2]

4

Result: "inf"

move_j

i

[1,1]

0

[1,3]

1

[3,1]

2

j

[3,3]

3

[2,2]

4

Result: "inf"

compare

i

[1,1]

0

[1,3]

1

[3,1]

2

j

[3,3]

3

[2,2]

4

Result: "inf"

record

i

[1,1]

0

[1,3]

1

[3,1]

2

j

[3,3]

3

[2,2]

4

Result: 4

move_j

i

[1,1]

0

[1,3]

1

[3,1]

2

[3,3]

3

j

[2,2]

4

Result: 4

compare

i

[1,1]

0

[1,3]

1

[3,1]

2

[3,3]

3

j

[2,2]

4

Result: 4

move_i

[1,1]

0

i

[1,3]

1

[3,1]

2

[3,3]

3

[2,2]

4

Result: 4

move_j

[1,1]

0

i

[1,3]

1

j

[3,1]

2

[3,3]

3

[2,2]

4

Result: 4

compare

[1,1]

0

i

[1,3]

1

j

[3,1]

2

[3,3]

3

[2,2]

4

Result: 4

compare

[1,1]

0

i

[1,3]

1

j

[3,1]

2

[3,3]

3

[2,2]

4

Result: 4

move_j

[1,1]

0

i

[1,3]

1

[3,1]

2

j

[3,3]

3

[2,2]

4

Result: 4

compare

[1,1]

0

i

[1,3]

1

[3,1]

2

j

[3,3]

3

[2,2]

4

Result: 4

move_j

[1,1]

0

i

[1,3]

1

[3,1]

2

[3,3]

3

j

[2,2]

4

Result: 4

compare

[1,1]

0

i

[1,3]

1

[3,1]

2

[3,3]

3

j

[2,2]

4

Result: 4

return

[1,1]

0

[1,3]

1

[3,1]

2

[3,3]

3

[2,2]

4

Result: 4

Key Takeaways

✓ Using a hash set for points enables constant-time existence checks, drastically improving efficiency over brute force.

Without the hash set, checking if the other two corners exist would require nested loops, making the algorithm much slower.

✓ Only pairs of points with different x and y coordinates can be diagonal corners of a rectangle.

This insight reduces the number of pairs to check and is visually clear when watching pointer comparisons.

✓ The algorithm updates the minimum area only when a smaller rectangle is found, ensuring the final answer is the smallest possible.

Seeing the area comparisons step-by-step clarifies why some rectangles are ignored even if they exist.

Practice

1. Consider the following Python function that checks if two rectangles overlap. What is the return value when calling isRectangleOverlap([0,0,2,2], [1,1,3,3])?

easy

A. null

B. false

C. true

D. Raises an error

Solution

1. Step 1: Calculate x_overlap

   min(2,3) - max(0,1) = 2 - 1 = 1 > 0

2. Step 2: Calculate y_overlap

   min(2,3) - max(0,1) = 2 - 1 = 1 > 0

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Both overlaps positive means rectangles overlap [OK]

Hint: Check if min > max on both axes for overlap [OK]

Common Mistakes:

• Confusing > 0 with >= 0 causing false positives on touching edges
• Mixing up coordinates leading to negative overlap
• Assuming overlap if only one axis overlaps

2. Examine the following buggy Python code intended to compute the LCM safely without overflow:

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 * b // gcd

if __name__ == '__main__':
    a, b = 1000000, 1000000
    print("LCM =", lcm_safe(a, b))

What is the subtle bug in this code?

medium

A. Line 7: Multiplying a and b before dividing by gcd can cause integer overflow.

B. Line 3: The Euclidean algorithm lacks a base case for recursion.

C. Line 9: The input values are not validated for being positive integers.

D. Line 2: The while loop condition should be 'while a != 0' instead of 'while b != 0'.

Solution

1. Step 1: Analyze lcm_safe function

   The function computes LCM as a * b // gcd. Multiplying a and b first can overflow for large inputs before division.

2. Step 2: Identify the bug

   To avoid overflow, division by gcd should happen before multiplication: (a // gcd) * b.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Multiplying large numbers before division risks overflow [OK]

Hint: Divide before multiply to avoid overflow in LCM calculation [OK]

Common Mistakes:

• Ignoring overflow risk in LCM calculation
• Misunderstanding Euclidean algorithm base case

3. The following code attempts to determine if a number is happy but contains a subtle bug. Identify the line with the bug.

def isHappy(n: int) -> bool:
    def digit_square_sum(num: int) -> int:
        total = 0
        while num > 0:
            digit = num % 10
            total += digit * digit
            num //= 10
        return total

    while n != 1:
        n = digit_square_sum(n)
    return True

medium

A. Line with 'while n != 1:' loop condition

B. Missing cycle detection causing infinite loop

C. Line with 'return True' after the loop

D. Line with 'n = digit_square_sum(n)' inside the loop

Solution

1. Step 1: Analyze loop termination condition

   The loop only terminates when n == 1, but if n falls into an unhappy cycle, it never reaches 1.

2. Step 2: Identify missing cycle detection

   Without tracking seen numbers or using a known cycle set, the code loops infinitely for unhappy numbers.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Infinite loop occurs due to missing cycle detection [OK]

Hint: No cycle detection -> infinite loop on unhappy numbers [OK]

Common Mistakes:

• Assuming loop condition suffices
• Ignoring cycle detection necessity
• Misreading return placement

4. Suppose you want to count primes less than n but now the input n can be up to 10^12, too large for a standard sieve. Which approach is best suited to handle this scenario efficiently?

hard

A. Use a segmented sieve that processes ranges of size sqrt(n) to reduce memory usage.

B. Use the standard sieve of Eratosthenes with a larger memory allocation.

C. Use brute force primality checks for each number up to n.

D. Use a dynamic programming approach to store prime counts up to n.

Solution

1. Step 1: Understand memory constraints

   Standard sieve requires O(n) space, which is infeasible for n=10^12.

2. Step 2: Identify suitable algorithm

   Segmented sieve divides the range into manageable blocks, using O(sqrt(n)) memory and still marking primes efficiently.

3. Step 3: Compare alternatives

   Brute force is too slow, DP is not applicable, and standard sieve is too large.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Segmented sieve is standard for large n prime counting [OK]

Hint: Segmented sieve handles large n with limited memory [OK]

Common Mistakes:

• Trying standard sieve for huge n
• Using brute force for large ranges
• Misapplying DP to prime counting

5. Suppose the problem is extended: now the input integer can be arbitrarily large (beyond 32-bit), and you must return the reversed integer modulo 10^9+7. Which modification to the recursive digit extraction approach correctly handles this variant?

hard

A. Remove overflow checks and rely on Python's big integer support to reverse directly

B. Use string conversion and reverse, then convert back modulo 10^9+7 without overflow checks

C. Iteratively extract digits and build reversed number modulo 10^9+7, avoiding recursion

D. Keep the same recursion but add modulo operation after each digit append to rev

Solution

1. Step 1: Understand new constraints

   Input can be arbitrarily large, so 32-bit overflow checks are irrelevant; modulo 10^9+7 must be applied.

2. Step 2: Identify correct approach

   Iterative digit extraction with modulo after each append avoids recursion stack overhead and handles large inputs efficiently.

3. Step 3: Why other options fail

   Recursive approach risks stack overflow; string conversion lacks modulo during reversal; removing overflow checks ignores modulo requirement.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Iterative modulo approach scales and respects new constraints [OK]

Hint: Modulo after each digit append avoids overflow [OK]

Common Mistakes:

• Using recursion without modulo causes stack overflow
• String reversal misses modulo during conversion
• Ignoring modulo leads to incorrect large results