Search in Rotated Array II (Duplicates)

🎯

Search in Rotated Array II (Duplicates)

mediumBINARY_SEARCHAmazonGoogle

Imagine you have a sorted list of product IDs that got rotated during a system update, and some IDs appear multiple times. You need to quickly check if a certain product ID exists in this rotated list.

💡 This problem is a variant of binary search on a rotated sorted array but complicated by duplicates. Beginners often struggle because duplicates can hide the pivot and make it unclear which half is sorted, breaking the usual binary search logic.

📋

Problem Statement

Given an integer array nums sorted in non-decreasing order, which is rotated at an unknown pivot, and may contain duplicates, write a function to determine if a given target value exists in nums. Return true if target is found, otherwise false.

1 ≤ nums.length ≤ 10^5-10^4 ≤ nums[i] ≤ 10^4-10^4 ≤ target ≤ 10^4

💡

Example

Input"nums = [2,5,6,0,0,1,2], target = 0"

Outputtrue

Target 0 is present in the rotated array.

Input"nums = [2,5,6,0,0,1,2], target = 3"

Outputfalse

Target 3 is not present in the array.

• Array with all duplicates, e.g. [1,1,1,1], target = 1 → true
• Array with all duplicates, target not present, e.g. [1,1,1,1], target = 2 → false
• Array with single element, target present, e.g. [1], target = 1 → true
• Array with single element, target absent, e.g. [1], target = 0 → false

⚠️

Common Mistakes

Not handling duplicates properly when nums[left] == nums[mid] == nums[right]

Binary search logic breaks, causing infinite loops or incorrect results

✅ Shrink search space by incrementing left and decrementing right when duplicates obscure sorted half

Using classic binary search conditions without considering rotation and duplicates

Incorrectly deciding which half is sorted, leading to wrong pointer moves

✅ Add conditions to check sorted halves carefully and handle duplicates explicitly

Not checking for target at mid before other conditions

Miss immediate success cases, causing unnecessary iterations

✅ Always check if nums[mid] == target first and return true immediately

Shrinking duplicates only from one side

May still get stuck if duplicates exist on both ends

✅ Shrink duplicates from both left and right ends when they are equal to mid

Using mid = (left + right) / 2 without floor or safe calculation

In some languages, integer division or overflow can cause errors

✅ Use floor division or mid = left + (right - left) / 2 to avoid overflow

🧠

Brute Force (Linear Search)

💡 This approach exists to establish a baseline and ensure correctness before optimizing. It helps beginners understand the problem fully by checking every element.

Intuition

Simply scan through the entire array and check if any element equals the target.

Algorithm

1. Iterate over each element in the array.
2. Compare the current element with the target.
3. If a match is found, return true immediately.
4. If the loop ends without a match, return false.

💡 This algorithm is straightforward but inefficient; it helps build intuition about the problem and correctness before optimization.

</>

Code

def search(nums, target):
    for num in nums:
        if num == target:
            return True
    return False

# Driver code
if __name__ == '__main__':
    print(search([2,5,6,0,0,1,2], 0))  # True
    print(search([2,5,6,0,0,1,2], 3))  # False

Line Notes

for num in nums:Iterate over every element to check for target presence

if num == target:Check if current element matches target

return TrueImmediately return True if target found to avoid unnecessary checks

return FalseReturn False if no element matched after full iteration

public class Solution {
    public boolean search(int[] nums, int target) {
        for (int num : nums) {
            if (num == target) {
                return true;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.search(new int[]{2,5,6,0,0,1,2}, 0)); // true
        System.out.println(sol.search(new int[]{2,5,6,0,0,1,2}, 3)); // false
    }
}

Line Notes

for (int num : nums)Loop through each element in the array

if (num == target)Check if current element equals target

return true;Return true immediately when target is found

return false;Return false if target not found after full loop

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

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        for (int num : nums) {
            if (num == target) {
                return true;
            }
        }
        return false;
    }
};

int main() {
    Solution sol;
    vector<int> nums1 = {2,5,6,0,0,1,2};
    cout << boolalpha << sol.search(nums1, 0) << endl; // true
    cout << boolalpha << sol.search(nums1, 3) << endl; // false
    return 0;
}

Line Notes

for (int num : nums)Iterate over each element in the vector

if (num == target)Check if current element matches target

return true;Return true immediately when target found

return false;Return false if target not found after full iteration

function search(nums, target) {
    for (const num of nums) {
        if (num === target) {
            return true;
        }
    }
    return false;
}

// Test cases
console.log(search([2,5,6,0,0,1,2], 0)); // true
console.log(search([2,5,6,0,0,1,2], 3)); // false

Line Notes

for (const num of nums)Loop through each element in the array

if (num === target)Check if current element equals target

return true;Return true immediately when target is found

return false;Return false if target not found after full iteration

⏱

Complexity

TimeO(n)

SpaceO(1)

We scan through the entire array once, so time is linear in array size. No extra space is used.

💡 For n=100,000 elements, this means checking each element one by one, which can be slow in interviews.

Interview Verdict: Accepted but inefficient

This approach always works but is too slow for large inputs; it serves as a correctness baseline before optimization.

🧠

Modified Binary Search with Shrinking Both Ends

💡 This approach introduces binary search logic adapted to handle duplicates by shrinking both ends when duplicates obscure the sorted half.

Intuition

Use binary search to find the target, but when duplicates at start, mid, and end prevent deciding which half is sorted, shrink the search space from both ends.

Algorithm

1. Initialize two pointers, left and right, at the array ends.
2. While left <= right, calculate mid index.
3. If nums[mid] equals target, return true.
4. If nums[left], nums[mid], and nums[right] are equal, increment left and decrement right to shrink duplicates.
5. Else if left half is sorted (nums[left] <= nums[mid]), check if target lies in left half; adjust pointers accordingly.
6. Else, right half is sorted; check if target lies there; adjust pointers accordingly.
7. If loop ends without finding target, return false.

💡 The key challenge is handling duplicates that hide the sorted half; shrinking both ends helps restore binary search logic.

</>

Code

def search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return True
        if nums[left] == nums[mid] == nums[right]:
            left += 1
            right -= 1
        elif nums[left] <= nums[mid]:
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1
    return False

# Driver code
if __name__ == '__main__':
    print(search([2,5,6,0,0,1,2], 0))  # True
    print(search([2,5,6,0,0,1,2], 3))  # False

Line Notes

left, right = 0, len(nums) - 1Initialize pointers to cover entire array

mid = (left + right) // 2Calculate middle index to split search space

if nums[mid] == target:Check if mid element is target for immediate success

if nums[left] == nums[mid] == nums[right]:Duplicates at both ends and mid obscure sorted half, so shrink search space by moving both pointers inward

public class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return true;
            if (nums[left] == nums[mid] && nums[mid] == nums[right]) {
                left++;
                right--;
            } else if (nums[left] <= nums[mid]) {
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
        return false;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.search(new int[]{2,5,6,0,0,1,2}, 0)); // true
        System.out.println(sol.search(new int[]{2,5,6,0,0,1,2}, 3)); // false
    }
}

Line Notes

int left = 0, right = nums.length - 1;Initialize search boundaries

int mid = left + (right - left) / 2;Calculate mid index safely to avoid overflow

if (nums[mid] == target) return true;Return true immediately if target found at mid

if (nums[left] == nums[mid] && nums[mid] == nums[right])Duplicates at edges and mid require shrinking search space by moving both pointers inward

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

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return true;
            if (nums[left] == nums[mid] && nums[mid] == nums[right]) {
                left++;
                right--;
            } else if (nums[left] <= nums[mid]) {
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
        return false;
    }
};

int main() {
    Solution sol;
    vector<int> nums1 = {2,5,6,0,0,1,2};
    cout << boolalpha << sol.search(nums1, 0) << endl; // true
    cout << boolalpha << sol.search(nums1, 3) << endl; // false
    return 0;
}

Line Notes

int left = 0, right = nums.size() - 1;Set initial search boundaries

int mid = left + (right - left) / 2;Calculate mid index to avoid overflow

if (nums[mid] == target) return true;Return true if target found at mid

if (nums[left] == nums[mid] && nums[mid] == nums[right])Duplicates at edges and mid require shrinking search space by moving both pointers inward

function search(nums, target) {
    let left = 0, right = nums.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (nums[mid] === target) return true;
        if (nums[left] === nums[mid] && nums[mid] === nums[right]) {
            left++;
            right--;
        } else if (nums[left] <= nums[mid]) {
            if (nums[left] <= target && target < nums[mid]) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        } else {
            if (nums[mid] < target && target <= nums[right]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
    }
    return false;
}

// Test cases
console.log(search([2,5,6,0,0,1,2], 0)); // true
console.log(search([2,5,6,0,0,1,2], 3)); // false

Line Notes

let left = 0, right = nums.length - 1;Initialize pointers to array boundaries

let mid = Math.floor((left + right) / 2);Calculate middle index for binary search

if (nums[mid] === target) return true;Return true immediately if target found at mid

if (nums[left] === nums[mid] && nums[mid] === nums[right])Duplicates at edges and mid require shrinking search space by moving both pointers inward

⏱

Complexity

TimeO(n) in worst case due to duplicates, average O(log n)

SpaceO(1)

Duplicates can degrade binary search to linear time in worst case, but typically it runs in logarithmic time.

💡 For large arrays with many duplicates, this approach might scan many elements, but usually it is much faster than brute force.

Interview Verdict: Accepted and optimal for duplicates

This is the best practical approach for this problem, balancing correctness and efficiency.

🧠

Binary Search with Early Duplicate Skipping

💡 This approach optimizes the previous by aggressively skipping duplicates at the edges before binary search steps, reducing unnecessary comparisons.

Intuition

Before each binary search iteration, move left pointer forward while it equals the next element, and move right pointer backward similarly, to reduce duplicate clutter.

Algorithm

1. Initialize left and right pointers.
2. While left < right and nums[left] == nums[left + 1], increment left to skip duplicates.
3. While right > left and nums[right] == nums[right - 1], decrement right to skip duplicates.
4. Perform binary search logic similar to Approach 2.
5. Return true if target found, else false.

💡 Skipping duplicates early reduces the chance of worst-case linear scans, improving average performance.

</>

Code

def search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        while left < right and nums[left] == nums[left + 1]:
            left += 1
        while right > left and nums[right] == nums[right - 1]:
            right -= 1
        mid = (left + right) // 2
        if nums[mid] == target:
            return True
        if nums[left] <= nums[mid]:
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1
    return False

# Driver code
if __name__ == '__main__':
    print(search([2,5,6,0,0,1,2], 0))  # True
    print(search([2,5,6,0,0,1,2], 3))  # False

Line Notes

while left < right and nums[left] == nums[left + 1]:Skip duplicates at the left edge to reduce search space and avoid ambiguity

while right > left and nums[right] == nums[right - 1]:Skip duplicates at the right edge similarly to shrink search space

mid = (left + right) // 2Calculate mid index for binary search

if nums[mid] == target:Return true immediately if target found to avoid unnecessary further search

public class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            while (left < right && nums[left] == nums[left + 1]) left++;
            while (right > left && nums[right] == nums[right - 1]) right--;
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return true;
            if (nums[left] <= nums[mid]) {
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
        return false;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.search(new int[]{2,5,6,0,0,1,2}, 0)); // true
        System.out.println(sol.search(new int[]{2,5,6,0,0,1,2}, 3)); // false
    }
}

Line Notes

while (left < right && nums[left] == nums[left + 1]) left++;Skip duplicates at left boundary to reduce search space and clarify sorted half

while (right > left && nums[right] == nums[right - 1]) right--;Skip duplicates at right boundary similarly

int mid = left + (right - left) / 2;Calculate mid index safely

if (nums[mid] == target) return true;Return true if target found at mid to avoid unnecessary search

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

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            while (left < right && nums[left] == nums[left + 1]) left++;
            while (right > left && nums[right] == nums[right - 1]) right--;
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return true;
            if (nums[left] <= nums[mid]) {
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
        return false;
    }
};

int main() {
    Solution sol;
    vector<int> nums1 = {2,5,6,0,0,1,2};
    cout << boolalpha << sol.search(nums1, 0) << endl; // true
    cout << boolalpha << sol.search(nums1, 3) << endl; // false
    return 0;
}

Line Notes

while (left < right && nums[left] == nums[left + 1]) left++;Skip duplicates at left boundary to reduce search space and clarify sorted half

while (right > left && nums[right] == nums[right - 1]) right--;Skip duplicates at right boundary similarly

int mid = left + (right - left) / 2;Calculate mid index safely to avoid overflow

if (nums[mid] == target) return true;Return true immediately if target found

function search(nums, target) {
    let left = 0, right = nums.length - 1;
    while (left <= right) {
        while (left < right && nums[left] === nums[left + 1]) left++;
        while (right > left && nums[right] === nums[right - 1]) right--;
        let mid = Math.floor((left + right) / 2);
        if (nums[mid] === target) return true;
        if (nums[left] <= nums[mid]) {
            if (nums[left] <= target && target < nums[mid]) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        } else {
            if (nums[mid] < target && target <= nums[right]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
    }
    return false;
}

// Test cases
console.log(search([2,5,6,0,0,1,2], 0)); // true
console.log(search([2,5,6,0,0,1,2], 3)); // false

Line Notes

while (left < right && nums[left] === nums[left + 1]) left++;Skip duplicates at left boundary to reduce search space and clarify sorted half

while (right > left && nums[right] === nums[right - 1]) right--;Skip duplicates at right boundary similarly

let mid = Math.floor((left + right) / 2);Calculate mid index for binary search

if (nums[mid] === target) return true;Return true immediately if target found

⏱

Complexity

TimeO(n) worst case, average better than Approach 2

SpaceO(1)

Early skipping of duplicates reduces the search space faster, improving average runtime though worst case remains linear.

💡 This optimization helps avoid many redundant checks, especially when duplicates cluster at edges.

Interview Verdict: Accepted and slightly faster in practice

This is a practical optimization that can be mentioned or implemented if time permits in interviews.

📊

All Approaches - One-Glance Tradeoffs

💡 In interviews, always code the modified binary search with shrinking both ends (Approach 2) as it balances correctness and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n)	O(1)	No	N/A	Mention only - never code due to inefficiency
2. Modified Binary Search with Shrinking Both Ends	O(n) worst, O(log n) average	O(1)	No	N/A	Code this approach for best balance of speed and correctness
3. Binary Search with Early Duplicate Skipping	O(n) worst, better average than Approach 2	O(1)	No	N/A	Mention as optimization if time permits

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice all approaches, and prepare to explain tradeoffs clearly during interviews.

How to Present

Step 1: Clarify the problem and constraints, especially about duplicates and rotation.Step 2: State the brute force approach to show understanding and correctness.Step 3: Explain why classic binary search fails due to duplicates.Step 4: Introduce modified binary search with shrinking both ends.Step 5: Optionally mention early duplicate skipping optimization.Step 6: Code the modified binary search approach carefully.Step 7: Test with edge cases and explain complexity.

Time Allocation

Clarify: 2min → Approach: 5min → Code: 10min → Test: 3min. Total ~20min

What the Interviewer Tests

Interviewer tests your understanding of binary search modifications, handling of duplicates, and ability to optimize while maintaining correctness.

Common Follow-ups

• What if duplicates were not allowed? → Classic binary search on rotated array suffices.
• How to find the minimum element in such an array? → Similar binary search with duplicates handling.

💡 These follow-ups test your grasp of the pattern and ability to adapt binary search to related problems.

🔍

Pattern Recognition

When to Use

1) Array is sorted but rotated; 2) Contains duplicates; 3) Need to search for target efficiently; 4) Classic binary search fails due to duplicates

Signature Phrases

'rotated sorted array''may contain duplicates''search for target''return true if found, false otherwise'

NOT This Pattern When

Standard binary search on sorted arrays without rotation or duplicates

Similar Problems

Search in Rotated Sorted Array I - same problem without duplicates, simpler binary searchFind Minimum in Rotated Sorted Array II - similar duplicate handling with binary searchSearch in Rotated Sorted Array without Duplicates - classic binary search on rotated array

Practice

1. Consider the following Python code implementing the optimal approach to find k closest elements to x in a sorted array. What is the returned output when calling findClosestElements([1, 3, 5], 2, 4)?

easy

A. [1, 3]

B. [3, 4]

C. [1, 5]

D. [3, 5]

Solution

1. Step 1: Initialize variables

   Array length is 3, k=2, so low=0, high=1.

2. Step 2: First iteration of binary search

   Calculate mid = (0+1)//2 = 0. Compare x - arr[mid] = 4 - 1 = 3 and arr[mid + k] - x = arr[2] - 4 = 5 - 4 = 1. Since 3 > 1, set low = mid + 1 = 1.

3. Step 3: Loop ends as low == high == 1

   Return arr[1:3] = [3, 5].

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Binary search correctly chooses window starting at index 1 [OK]

Hint: Check mid and window edges distances carefully [OK]

Common Mistakes:

• Off-by-one error in slicing window
• Incorrect comparison direction in if condition
• Returning wrong slice boundaries

2. You are given an array of positive integers and a threshold value. You want to find the smallest integer divisor such that when each array element is divided by this divisor and rounded up, the sum of these results does not exceed the threshold. Which algorithmic approach guarantees finding this smallest divisor efficiently?

easy

A. Greedy approach that picks divisors based on local minimum sums without backtracking

B. Dynamic programming to explore all divisor combinations and sums

C. Binary search on the divisor value combined with a feasibility check function

D. Sorting the array and using two pointers to find the divisor

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding the smallest divisor that satisfies a sum condition involving ceiling division, which is monotonic with respect to the divisor.

2. Step 2: Identify the suitable algorithmic pattern

   Binary search on the answer space (divisor values) combined with a helper feasibility function efficiently narrows down the smallest divisor by checking if a candidate divisor meets the threshold condition.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Binary search exploits monotonicity of feasibility -> [OK]

Hint: Monotonic feasibility suggests binary search on answer space [OK]

Common Mistakes:

• Assuming greedy local choices yield global minimum divisor
• Trying DP which is inefficient and unnecessary here
• Using sorting and two pointers which don't apply to divisor search

3. The following code attempts to find the kth positive missing number from a sorted array arr. Which line contains a subtle bug that causes incorrect results for some inputs?

def findKthPositive(arr, k):
    n = len(arr)
    total_missing = arr[-1] - n
    if k > total_missing:
        return arr[-1] + (k - total_missing)
    left, right = 0, n - 1
    while left <= right:
        mid = (left + right) // 2
        missing = arr[mid] - mid  # Bug here
        if missing < k:
            left = mid + 1
        else:
            right = mid - 1
    return arr[right] + (k - (arr[right] - (right + 1)))

medium

A. Line returning arr[right] + (k - (arr[right] - (right + 1)))

B. Line calculating total_missing = arr[-1] - n

C. Line updating left = mid + 1

D. Line calculating missing = arr[mid] - mid

Solution

1. Step 1: Identify the missing count formula

   The correct missing count at index mid is arr[mid] - (mid + 1), since indices are zero-based but numbers start at 1.

2. Step 2: Analyze the bug impact

   Using arr[mid] - mid undercounts missing numbers by 1, causing incorrect binary search decisions and wrong final result.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Correct missing count formula is arr[mid] - (mid + 1) [OK]

Hint: Missing count must subtract (mid + 1), not mid [OK]

Common Mistakes:

• Off-by-one in missing count calculation
• Incorrect binary search boundary updates
• Ignoring missing numbers before first element

4. The following code attempts to find the insert position of a target in a sorted array using recursive binary search. Identify the line containing the subtle bug that can cause incorrect results or infinite recursion.

medium

A. Line 3: if left > right: return right

B. Line 4: mid = (left + right) // 2

C. Line 7: if nums[mid] == target: return mid

D. Line 9: return binary_search(mid + 1, right)

Solution

1. Step 1: Analyze base case return value

   When left > right, the correct insertion index is left, not right.

2. Step 2: Consequence of returning right

   Returning right can cause incorrect insertion index or infinite recursion if right < left.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Correct base case returns left to indicate insertion point [OK]

Hint: Base case must return left, not right [OK]

Common Mistakes:

• Returning right instead of left at base case
• Incorrect mid calculation causing infinite loop
• Misplaced recursive calls

5. Suppose the problem is modified so that the multiplication table allows repeated use of the same row and column indices (i.e., infinite repetitions of each product). Which of the following changes to the algorithm is correct to find the k-th smallest number under this new condition?

hard

A. Use the same binary search with the counting function unchanged, since repetition doesn't affect counts.

B. Switch to generating all products up to k and sorting them, since binary search no longer applies.

C. Use a min-heap to generate products in ascending order with duplicates allowed, extracting k elements.

D. Modify the counting function to remove the cap by n, counting all multiples less than or equal to x for each i.

Solution

1. Step 1: Understand the new problem condition

   Allowing infinite repetitions means the multiplication table is conceptually infinite, so counting by rows capped at n no longer applies.

2. Step 2: Analyze algorithmic implications

   Binary search with counting function fails because counts become infinite or undefined; generating all values is impossible.

3. Step 3: Use a min-heap approach

   A min-heap can generate products in ascending order with duplicates allowed, extracting the k-th smallest efficiently.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Min-heap approach handles infinite repetitions correctly [OK]

Hint: Infinite repetitions break counting; use min-heap [OK]

Common Mistakes:

• Assuming counting function unchanged
• Trying to generate all values