Rectangle Overlap

🎯

Rectangle Overlap

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Imagine you have two windows on your computer screen, and you want to check if they overlap at all. This problem asks you to determine if two rectangles on a plane overlap.

💡 This problem is about understanding how to check if two shapes intersect by comparing their coordinates. Beginners often struggle because they try to visualize the problem without breaking it down into simple 1D overlap checks on both axes.

📋

Problem Statement

Given two axis-aligned rectangles represented as lists of four integers [x1, y1, x2, y2], where (x1, y1) are the coordinates of the bottom-left corner and (x2, y2) are the coordinates of the top-right corner, determine if the two rectangles overlap. Return true if they overlap, otherwise false.

Coordinates are integers within the range [-10^9, 10^9]Rectangles are axis-aligned (edges parallel to axes)Input rectangles may be degenerate (zero area)

💡

Example

Input"rec1 = [0,0,2,2], rec2 = [1,1,3,3]"

Outputtrue

The rectangles overlap in the area from (1,1) to (2,2).

Input"rec1 = [0,0,1,1], rec2 = [1,0,2,1]"

Outputfalse

The rectangles only touch at the edge x=1 but do not overlap.

• Rectangles touch at exactly one corner → false
• One rectangle completely inside another → true
• Rectangles share an edge but no area → false
• Rectangles with zero area (line or point) → false

⚠️

Common Mistakes

Considering touching edges as overlap

Returns true when rectangles only touch at edges or corners

✅ Use strict inequality (>) instead of >= when checking overlap

Mixing up coordinates (e.g., using bottom-left as top-right)

Incorrect overlap results or runtime errors

✅ Carefully use the given coordinate definitions and verify input format

Checking only one axis for overlap

Returns true even if rectangles do not overlap on the other axis

✅ Always check overlap on both x and y axes

Not handling zero-area rectangles

May incorrectly return true for line or point rectangles

✅ Ensure overlap length is strictly positive, not zero

Using >= instead of > in overlap conditions

False positives for overlap when rectangles only touch

✅ Use strict inequalities to ensure positive overlap area

🧠

Brute Force (Check Overlap by Comparing Coordinates)

💡 This approach exists to build intuition by directly checking the conditions for overlap without shortcuts. It helps beginners understand the problem from first principles.

Intuition

Two rectangles overlap if their projections on both the x-axis and y-axis overlap. We check if one rectangle is completely to the left, right, above, or below the other. If none of these is true, they overlap.

Algorithm

1. Check if one rectangle is to the left of the other by comparing x coordinates.
2. Check if one rectangle is above the other by comparing y coordinates.
3. If neither condition is true, the rectangles overlap.
4. Return true if overlap detected, else false.

💡 The challenge is to realize that checking non-overlap conditions is easier than checking overlap directly.

</>

Code

def isRectangleOverlap(rec1, rec2):
    # If one rectangle is to the left of the other
    if rec1[2] <= rec2[0] or rec2[2] <= rec1[0]:
        return False
    # If one rectangle is above the other
    if rec1[3] <= rec2[1] or rec2[3] <= rec1[1]:
        return False
    return True

# Driver code
if __name__ == '__main__':
    print(isRectangleOverlap([0,0,2,2], [1,1,3,3]))  # True
    print(isRectangleOverlap([0,0,1,1], [1,0,2,1]))  # False

Line Notes

if rec1[2] <= rec2[0] or rec2[2] <= rec1[0]:Checks if one rectangle is completely to the left of the other, so no horizontal overlap

if rec1[3] <= rec2[1] or rec2[3] <= rec1[1]:Checks if one rectangle is completely above the other, so no vertical overlap

return FalseIf either no horizontal or no vertical overlap, rectangles do not overlap

return TrueIf none of the no-overlap conditions hold, rectangles must overlap

public class RectangleOverlap {
    public static boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {
            return false;
        }
        if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {
            return false;
        }
        return true;
    }

    public static void main(String[] args) {
        System.out.println(isRectangleOverlap(new int[]{0,0,2,2}, new int[]{1,1,3,3})); // true
        System.out.println(isRectangleOverlap(new int[]{0,0,1,1}, new int[]{1,0,2,1})); // false
    }
}

Line Notes

if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {Checks horizontal non-overlap by comparing right and left edges

if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {Checks vertical non-overlap by comparing top and bottom edges

return false;Return false immediately if no overlap on either axis

return true;If no non-overlap condition met, rectangles overlap

#include <iostream>
#include <vector>
using namespace std;

bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
    if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0])
        return false;
    if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1])
        return false;
    return true;
}

int main() {
    vector<int> rec1 = {0,0,2,2};
    vector<int> rec2 = {1,1,3,3};
    cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl; // true
    rec1 = {0,0,1,1};
    rec2 = {1,0,2,1};
    cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl; // false
    return 0;
}

Line Notes

if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0])Checks if one rectangle is completely left of the other horizontally

if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1])Checks if one rectangle is completely below the other vertically

return false;No overlap if either horizontal or vertical non-overlap detected

return true;Otherwise, rectangles overlap

function isRectangleOverlap(rec1, rec2) {
    if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {
        return false;
    }
    if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {
        return false;
    }
    return true;
}

// Test cases
console.log(isRectangleOverlap([0,0,2,2], [1,1,3,3])); // true
console.log(isRectangleOverlap([0,0,1,1], [1,0,2,1])); // false

Line Notes

if (rec1[2] <= rec2[0] || rec2[2] <= rec1[0]) {Checks horizontal non-overlap by comparing edges

if (rec1[3] <= rec2[1] || rec2[3] <= rec1[1]) {Checks vertical non-overlap by comparing edges

return false;Return false if rectangles do not overlap on either axis

return true;Return true if rectangles overlap on both axes

⏱

Complexity

TimeO(1)

SpaceO(1)

Only a fixed number of comparisons are done regardless of input size.

💡 This means the algorithm runs instantly even for very large coordinate values.

Interview Verdict: Accepted

This approach is optimal for this problem and will pass all tests efficiently.

🧠

Using 1D Interval Overlap Helper Function

💡 This approach introduces a helper function to check 1D overlap, improving code clarity and reusability, which is a good practice in interviews.

Intuition

Since rectangle overlap depends on overlap in both x and y intervals, we can write a function to check if two intervals overlap and reuse it for both axes.

Algorithm

1. Define a helper function that returns true if two 1D intervals overlap.
2. Use this helper to check if the x intervals of the rectangles overlap.
3. Use this helper to check if the y intervals of the rectangles overlap.
4. Return true only if both x and y intervals overlap.

💡 Breaking the problem into smaller reusable parts makes the solution easier to understand and maintain.

</>

Code

def intervals_overlap(a_start, a_end, b_start, b_end):
    return a_start < b_end and b_start < a_end

def isRectangleOverlap(rec1, rec2):
    return intervals_overlap(rec1[0], rec1[2], rec2[0], rec2[2]) and intervals_overlap(rec1[1], rec1[3], rec2[1], rec2[3])

# Driver code
if __name__ == '__main__':
    print(isRectangleOverlap([0,0,2,2], [1,1,3,3]))  # True
    print(isRectangleOverlap([0,0,1,1], [1,0,2,1]))  # False

Line Notes

def intervals_overlap(a_start, a_end, b_start, b_end):Defines a reusable function to check 1D interval overlap

return a_start < b_end and b_start < a_endChecks if intervals overlap by verifying start of one is before end of other

return intervals_overlap(rec1[0], rec1[2], rec2[0], rec2[2]) and intervals_overlap(rec1[1], rec1[3], rec2[1], rec2[3])Checks overlap on both x and y axes using helper

print(isRectangleOverlap([0,0,2,2], [1,1,3,3]))Tests the function with overlapping rectangles

public class RectangleOverlap {
    private static boolean intervalsOverlap(int aStart, int aEnd, int bStart, int bEnd) {
        return aStart < bEnd && bStart < aEnd;
    }

    public static boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        return intervalsOverlap(rec1[0], rec1[2], rec2[0], rec2[2]) &&
               intervalsOverlap(rec1[1], rec1[3], rec2[1], rec2[3]);
    }

    public static void main(String[] args) {
        System.out.println(isRectangleOverlap(new int[]{0,0,2,2}, new int[]{1,1,3,3})); // true
        System.out.println(isRectangleOverlap(new int[]{0,0,1,1}, new int[]{1,0,2,1})); // false
    }
}

Line Notes

private static boolean intervalsOverlap(int aStart, int aEnd, int bStart, int bEnd) {Helper method to check if two intervals overlap

return aStart < bEnd && bStart < aEnd;Overlap exists if intervals intersect in any positive length

return intervalsOverlap(rec1[0], rec1[2], rec2[0], rec2[2]) && intervalsOverlap(rec1[1], rec1[3], rec2[1], rec2[3]);Checks overlap on both axes

System.out.println(isRectangleOverlap(...));Runs test cases to verify correctness

#include <iostream>
#include <vector>
using namespace std;

bool intervalsOverlap(int aStart, int aEnd, int bStart, int bEnd) {
    return aStart < bEnd && bStart < aEnd;
}

bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
    return intervalsOverlap(rec1[0], rec1[2], rec2[0], rec2[2]) &&
           intervalsOverlap(rec1[1], rec1[3], rec2[1], rec2[3]);
}

int main() {
    vector<int> rec1 = {0,0,2,2};
    vector<int> rec2 = {1,1,3,3};
    cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl; // true
    rec1 = {0,0,1,1};
    rec2 = {1,0,2,1};
    cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl; // false
    return 0;
}

Line Notes

bool intervalsOverlap(int aStart, int aEnd, int bStart, int bEnd) {Helper function to check 1D interval overlap

return aStart < bEnd && bStart < aEnd;Returns true if intervals intersect with positive length

return intervalsOverlap(rec1[0], rec1[2], rec2[0], rec2[2]) && intervalsOverlap(rec1[1], rec1[3], rec2[1], rec2[3]);Checks overlap on both x and y axes

cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl;Prints boolean result for test cases

function intervalsOverlap(aStart, aEnd, bStart, bEnd) {
    return aStart < bEnd && bStart < aEnd;
}

function isRectangleOverlap(rec1, rec2) {
    return intervalsOverlap(rec1[0], rec1[2], rec2[0], rec2[2]) &&
           intervalsOverlap(rec1[1], rec1[3], rec2[1], rec2[3]);
}

// Test cases
console.log(isRectangleOverlap([0,0,2,2], [1,1,3,3])); // true
console.log(isRectangleOverlap([0,0,1,1], [1,0,2,1])); // false

Line Notes

function intervalsOverlap(aStart, aEnd, bStart, bEnd) {Defines helper to check 1D interval overlap

return aStart < bEnd && bStart < aEnd;Returns true if intervals overlap positively

return intervalsOverlap(rec1[0], rec1[2], rec2[0], rec2[2]) && intervalsOverlap(rec1[1], rec1[3], rec2[1], rec2[3]);Checks overlap on both axes

console.log(isRectangleOverlap(...));Runs test cases to verify correctness

⏱

Complexity

TimeO(1)

SpaceO(1)

Only constant time checks and function calls are made.

💡 This approach is just as fast as brute force but better organized.

Interview Verdict: Accepted

This approach is clean and efficient, suitable for interviews.

🧠

Mathematical Overlap Check Using Max and Min

💡 This approach uses a mathematical perspective by comparing max of left edges and min of right edges to check overlap length, which is a common pattern in geometry problems.

Intuition

If the maximum of the left edges is less than the minimum of the right edges on both axes, the rectangles overlap. Otherwise, they don't.

Algorithm

1. Calculate the maximum of the left x-coordinates and the minimum of the right x-coordinates.
2. Calculate the maximum of the bottom y-coordinates and the minimum of the top y-coordinates.
3. If both x and y overlap lengths are positive, rectangles overlap.
4. Return true if overlap exists, else false.

💡 This approach directly computes the overlap length, making it easy to extend to find the overlap area if needed.

</>

Code

def isRectangleOverlap(rec1, rec2):
    x_overlap = min(rec1[2], rec2[2]) - max(rec1[0], rec2[0])
    y_overlap = min(rec1[3], rec2[3]) - max(rec1[1], rec2[1])
    return x_overlap > 0 and y_overlap > 0

# Driver code
if __name__ == '__main__':
    print(isRectangleOverlap([0,0,2,2], [1,1,3,3]))  # True
    print(isRectangleOverlap([0,0,1,1], [1,0,2,1]))  # False

Line Notes

x_overlap = min(rec1[2], rec2[2]) - max(rec1[0], rec2[0])Computes horizontal overlap length by comparing edges

y_overlap = min(rec1[3], rec2[3]) - max(rec1[1], rec2[1])Computes vertical overlap length similarly

return x_overlap > 0 and y_overlap > 0Overlap exists only if both lengths are positive

print(isRectangleOverlap([0,0,2,2], [1,1,3,3]))Tests the function with overlapping rectangles

public class RectangleOverlap {
    public static boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        int xOverlap = Math.min(rec1[2], rec2[2]) - Math.max(rec1[0], rec2[0]);
        int yOverlap = Math.min(rec1[3], rec2[3]) - Math.max(rec1[1], rec2[1]);
        return xOverlap > 0 && yOverlap > 0;
    }

    public static void main(String[] args) {
        System.out.println(isRectangleOverlap(new int[]{0,0,2,2}, new int[]{1,1,3,3})); // true
        System.out.println(isRectangleOverlap(new int[]{0,0,1,1}, new int[]{1,0,2,1})); // false
    }
}

Line Notes

int xOverlap = Math.min(rec1[2], rec2[2]) - Math.max(rec1[0], rec2[0]);Calculates horizontal overlap length

int yOverlap = Math.min(rec1[3], rec2[3]) - Math.max(rec1[1], rec2[1]);Calculates vertical overlap length

return xOverlap > 0 && yOverlap > 0;Returns true if both overlaps are positive

System.out.println(isRectangleOverlap(...));Runs test cases to verify correctness

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
    int xOverlap = min(rec1[2], rec2[2]) - max(rec1[0], rec2[0]);
    int yOverlap = min(rec1[3], rec2[3]) - max(rec1[1], rec2[1]);
    return xOverlap > 0 && yOverlap > 0;
}

int main() {
    vector<int> rec1 = {0,0,2,2};
    vector<int> rec2 = {1,1,3,3};
    cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl; // true
    rec1 = {0,0,1,1};
    rec2 = {1,0,2,1};
    cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl; // false
    return 0;
}

Line Notes

int xOverlap = min(rec1[2], rec2[2]) - max(rec1[0], rec2[0]);Computes horizontal overlap length

int yOverlap = min(rec1[3], rec2[3]) - max(rec1[1], rec2[1]);Computes vertical overlap length

return xOverlap > 0 && yOverlap > 0;Returns true if both overlaps are positive

cout << boolalpha << isRectangleOverlap(rec1, rec2) << endl;Prints boolean result for test cases

function isRectangleOverlap(rec1, rec2) {
    const xOverlap = Math.min(rec1[2], rec2[2]) - Math.max(rec1[0], rec2[0]);
    const yOverlap = Math.min(rec1[3], rec2[3]) - Math.max(rec1[1], rec2[1]);
    return xOverlap > 0 && yOverlap > 0;
}

// Test cases
console.log(isRectangleOverlap([0,0,2,2], [1,1,3,3])); // true
console.log(isRectangleOverlap([0,0,1,1], [1,0,2,1])); // false

Line Notes

const xOverlap = Math.min(rec1[2], rec2[2]) - Math.max(rec1[0], rec2[0]);Calculates horizontal overlap length

const yOverlap = Math.min(rec1[3], rec2[3]) - Math.max(rec1[1], rec2[1]);Calculates vertical overlap length

return xOverlap > 0 && yOverlap > 0;Returns true if both overlaps are positive

console.log(isRectangleOverlap(...));Runs test cases to verify correctness

⏱

Complexity

TimeO(1)

SpaceO(1)

Only constant time arithmetic and comparisons are performed.

💡 This approach is mathematically elegant and runs instantly for any input.

Interview Verdict: Accepted

This is a clean and optimal solution, often preferred in interviews for its clarity.

📊

All Approaches - One-Glance Tradeoffs

💡 The mathematical overlap check (Approach 3) is the best to code in interviews due to clarity and optimality.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force (Direct coordinate comparison)	O(1)	O(1)	No	N/A	Good to mention for intuition, but code the mathematical approach
2. Using 1D Interval Overlap Helper	O(1)	O(1)	No	N/A	Good for clean code, acceptable to code in interview
3. Mathematical Overlap Check Using Max and Min	O(1)	O(1)	No	N/A	Best approach to code in interviews for clarity and optimality

💼

Interview Strategy

💡 Use this guide to understand the problem deeply and practice explaining your thought process clearly before coding.

How to Present

Step 1: Clarify the problem and confirm input/output format.Step 2: Explain the brute force approach by checking non-overlap conditions.Step 3: Introduce a helper function for 1D interval overlap to improve clarity.Step 4: Present the mathematical approach using max and min to check overlap lengths.Step 5: Code the chosen approach and test with edge cases.

Time Allocation

Clarify: 2min → Approach: 3min → Code: 5min → Test: 5min. Total ~15min

What the Interviewer Tests

The interviewer tests your understanding of coordinate geometry, edge cases (like touching edges), and ability to write clean, correct code.

Common Follow-ups

• What if rectangles are rotated? → Requires more complex geometry checks.
• How to find the overlapping rectangle area? → Use overlap lengths to compute area.

💡 These follow-ups test your ability to extend the solution beyond the basic problem.

🔍

Pattern Recognition

When to Use

Use this pattern when you need to check if two axis-aligned rectangles or intervals overlap by comparing their coordinate ranges on each axis.

Signature Phrases

axis-aligned rectanglesoverlapbottom-left and top-right coordinatesintersecting area

NOT This Pattern When

Problems involving rotated rectangles or polygons require more advanced geometry, not simple interval overlap.

Similar Problems

Interval List Intersections - similar 1D interval overlap logicMeeting Rooms - checking if time intervals overlapMaximum Number of Events That Can Be Attended - interval scheduling overlaps

Practice

1. Consider the following Python code snippet implementing the optimal approach for Max Points on a Line:

from collections import defaultdict
import math

def maxPoints(points):
    n = len(points)
    if n <= 2:
        return n
    max_points = 1
    points.sort()
    for i in range(n):
        if max_points >= n - i or max_points > n // 2:
            break
        slopes = defaultdict(int)
        duplicates = 0
        for j in range(i + 1, n):
            dx = points[j][0] - points[i][0]
            dy = points[j][1] - points[i][1]
            if dx == 0 and dy == 0:
                duplicates += 1
                continue
            gcd = math.gcd(dy, dx)
            if gcd != 0:
                dy //= gcd
                dx //= gcd
            if dx < 0:
                dx = -dx
                dy = -dy
            slopes[(dy, dx)] += 1
        curr_max = max(slopes.values(), default=0) + duplicates + 1
        max_points = max(max_points, curr_max)
    return max_points

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

What is the output of this code?

easy

A. 2

B. 3

C. 1

D. 0

Solution

1. Step 1: Trace outer loop with i=0

   Points sorted: [(1,1), (2,2), (3,3)]. For i=0, slopes computed with points (2,2) and (3,3) both have slope (1,1) after normalization.

2. Step 2: Calculate max points for i=0

   Slopes[(1,1)] = 2, duplicates=0, so curr_max = 2 + 0 + 1 = 3. max_points updated to 3. Early termination triggers next iterations skipped.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   All points lie on the same line y=x, so output is 3 [OK]

Hint: All points on line -> max points equals total points [OK]

Common Mistakes:

• Off-by-one in counting duplicates or base point
• Ignoring early termination condition

2. What is the time complexity of the optimal happy number detection algorithm that uses a fixed set of known unhappy cycle numbers and repeatedly computes the sum of squares of digits until reaching 1 or a cycle?

medium

A. O(n^2) due to repeated digit extraction and sum computations

B. O(n) where n is the input number

C. O(k * log n) where k is the number of iterations until termination and log n is digits count

D. O(1) constant time because the unhappy cycle set is fixed

Solution

1. Step 1: Identify operations per iteration

   Each iteration computes sum of squares of digits, which takes O(log n) time since number of digits is proportional to log n.

2. Step 2: Count number of iterations

   Number of iterations k until reaching 1 or cycle is variable but bounded; total time is O(k * log n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Time depends on iterations and digit count, not linear or constant [OK]

Hint: Sum of digit squares costs O(log n) per iteration [OK]

Common Mistakes:

• Assuming linear O(n) time
• Confusing fixed cycle set with constant time
• Mistaking digit extraction cost

3. Consider the following buggy code for finding the nth ugly number. Which line contains the subtle bug that causes incorrect results for some inputs?

import heapq

def nthUglyNumber(n):
    heap = [1]
    seen = set()
    ugly = 1
    for _ in range(n):
        ugly = heapq.heappop(heap)
        for factor in [2, 3, 5]:
            new_ugly = ugly * factor
            if new_ugly not in seen:
                seen.add(new_ugly)
                heapq.heappush(heap, new_ugly)
    return ugly

medium

A. Line 3: Initializing seen as an empty set instead of containing 1

B. Line 6: Popping from heap before checking if heap is empty

C. Line 9: Multiplying ugly by factor without checking for overflow

D. Line 11: Adding new_ugly to seen before pushing to heap

Solution

1. Step 1: Check initialization of seen set

   The seen set is empty initially, but 1 is already in the heap. This causes duplicates because 1 is not marked as seen.

2. Step 2: Understand impact of missing 1 in seen

   Without 1 in seen, the algorithm may push 1 multiple times, causing duplicates and incorrect nth ugly number.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Adding 1 to seen fixes duplicates and ensures correctness [OK]

Hint: Always mark initial heap elements as seen to avoid duplicates [OK]

Common Mistakes:

• Forgetting to add initial element to seen set
• Assuming heap uniqueness without seen tracking
• Ignoring edge cases with initial values

4. Suppose the problem is modified so that points can be reused multiple times (i.e., infinite copies of each point are available). Which of the following changes to the algorithm is necessary to correctly find the maximum number of points on a line under this new constraint?

hard

A. No change needed; the original algorithm already accounts for duplicates and infinite reuse.

B. Change the approach to use a mathematical formula for infinite geometric progressions instead of slope counting.

C. Modify the algorithm to consider infinite multiplicity by counting points with the same coordinates once and multiplying by the number of allowed uses.

D. The problem reduces to checking if any line passes through at least one point, so return the total number of points.

Solution

1. Step 1: Understand infinite reuse impact

   Infinite copies mean duplicates can be counted multiple times, so counting duplicates once is insufficient.

2. Step 2: Modify counting to multiply duplicates by allowed reuse

   The algorithm must track multiplicity and multiply counts accordingly to reflect infinite reuse.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Original algorithm counts duplicates once; infinite reuse requires multiplicative counting [OK]

Hint: Infinite reuse requires multiplicative counting of duplicates [OK]

Common Mistakes:

• Assuming original algorithm handles infinite reuse
• Ignoring multiplicity effects

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