Set Mismatch (One Duplicate, One Missing)

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Steps

setup

Initialize index i and array

We start with index i at 0 and the input array [1, 2, 2, 4]. This sets up the initial state for cyclic sorting.

💡 Initializing i at 0 is essential to start scanning the array from the beginning.

Line:

i = 0
n = len(nums)

💡 The algorithm begins scanning from the first element to place each number in its correct position.

compare

Check if nums[i] is at correct position

At i=0, nums[0] = 1. The correct position for 1 is index 0 (nums[i] - 1). Since nums[0] == nums[correct], no swap needed.

💡 If the element is already in the correct position, we move forward to the next index.

Line:

correct = nums[i] - 1
if nums[i] != nums[correct]:
    # swap
else:
    i += 1

💡 Elements already in place do not require swaps, so the pointer moves forward.

compare

Check if nums[i] is at correct position

At i=1, nums[1] = 2. The correct position for 2 is index 1. Since nums[1] == nums[correct], no swap needed.

💡 Again, the element is in its correct position, so we move i forward.

Line:

correct = nums[i] - 1
if nums[i] != nums[correct]:
    # swap
else:
    i += 1

💡 The algorithm confirms the element is correctly placed and advances.

compare

Check if nums[i] is at correct position

At i=2, nums[2] = 2. The correct position for 2 is index 1. Since nums[2] != nums[1], a swap is needed.

💡 When the current element is not in its correct position and differs from the element at the correct position, we swap.

Line:

correct = nums[i] - 1
if nums[i] != nums[correct]:
    nums[i], nums[correct] = nums[correct], nums[i]

💡 Swapping places the current element closer to its correct position.

compare

After swap, check again at i=2

After swapping, nums[2] is still 2 and nums[correct] is nums[1] = 2. Since they are equal, no swap occurs and i increments.

💡 If swapping does not change the element at i, it means a duplicate is detected and we move forward.

Line:

if nums[i] != nums[correct]:
    # swap
else:
    i += 1

💡 Duplicate elements cause the algorithm to move forward without infinite swapping.

compare

Check if nums[i] is at correct position

At i=3, nums[3] = 4. The correct position for 4 is index 3. Since nums[3] == nums[3], no swap needed.

💡 The last element is in the correct position, so we move i forward.

Line:

correct = nums[i] - 1
if nums[i] != nums[correct]:
    # swap
else:
    i += 1

💡 The algorithm has finished the cyclic sort loop as i reaches n.

traverse

Start scanning to find mismatch

Now we iterate over the array from i=0 to find the first index where nums[i] != i + 1.

💡 This scan identifies the duplicate and missing numbers by detecting the mismatch.

Line:

for i in range(n):
    if nums[i] != i + 1:
        return [nums[i], i + 1]

💡 Mismatch between value and index+1 reveals the duplicate and missing pair.

compare

Check if nums[0] == 1

At i=0, nums[0] = 1 which equals i + 1 = 1, so continue scanning.

💡 Matching value means no error at this index.

Line:

if nums[i] != i + 1:
    return [nums[i], i + 1]

💡 Correctly placed elements are skipped.

compare

Check if nums[1] == 2

At i=1, nums[1] = 2 which equals i + 1 = 2, so continue scanning.

💡 Still no mismatch, continue scanning.

Line:

if nums[i] != i + 1:
    return [nums[i], i + 1]

💡 Elements in correct position are confirmed.

compare

Detect mismatch at i=2

At i=2, nums[2] = 2 but i + 1 = 3, so mismatch found. Duplicate is 2, missing is 3.

💡 Mismatch reveals the duplicate and missing numbers directly.

Line:

if nums[i] != i + 1:
    return [nums[i], i + 1]

💡 Mismatch between value and expected index+1 identifies the error pair.

def findErrorNums(nums):
    i = 0  # STEP 1
    n = len(nums)
    while i < n:  # STEP 2-6
        correct = nums[i] - 1  # STEP 2
        if nums[i] != nums[correct]:  # STEP 4
            nums[i], nums[correct] = nums[correct], nums[i]  # STEP 4
        else:
            i += 1  # STEP 2,3,5,6
    for i in range(n):  # STEP 7-10
        if nums[i] != i + 1:  # STEP 8-10
            return [nums[i], i + 1]  # STEP 10
    return [-1, -1]

📊

Set Mismatch (One Duplicate, One Missing) - Watch the Algorithm Execute, Step by Step

Watching each swap and pointer move reveals how cyclic sort efficiently places numbers and how the mismatch reveals the duplicate and missing values.

compare

i

1

0

2

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2

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4

3

Result: "[]"

move_right

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Result: "[]"

move_right

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Result: "[]"

swap

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0

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1

i

2

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Result: "[]"

move_right

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i

4

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Result: "[]"

move_right

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Result: "[]"

compare

i

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1

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4

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Result: "[]"

move_right

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Result: "[]"

move_right

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i

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Result: "[]"

record

1

0

2

1

i

2

2

4

3

Result: "[2, 3]"

Key Takeaways

✓ Cyclic sort places each number at its correct index, making it easy to spot duplicates and missing numbers by scanning once.

This insight is hard to see from code alone because the swapping logic and pointer increments are subtle without visualization.

✓ When a duplicate is encountered, the algorithm avoids infinite swaps by moving the pointer forward, ensuring termination.

Understanding this prevents confusion about why the algorithm doesn't get stuck swapping duplicates endlessly.

✓ The mismatch between the value and its index+1 after sorting directly reveals the duplicate and missing numbers.

This concrete example shows how the final scan leverages the sorted array structure to find the answer efficiently.

Practice

1. You are given an array of integers where each integer is between 1 and n (inclusive), and some elements appear twice while others appear once. Which algorithmic approach guarantees finding all duplicate elements in O(n) time and O(1) extra space?

easy

A. Cyclic sort to place each number at its correct index, then scanning for mismatches

B. Using a hash set to track seen elements and identify duplicates

C. Sorting the array and then scanning for consecutive duplicates

D. Applying a dynamic programming approach to count occurrences

Solution

1. Step 1: Understand problem constraints

   The array contains numbers from 1 to n, allowing index-based placement.

2. Step 2: Identify approach that uses in-place swaps

   Cyclic sort places each number at index = number - 1, enabling O(n) time and O(1) space.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Cyclic sort exploits value-index mapping for optimal duplicate detection [OK]

Hint: Use cyclic sort for O(n) time and O(1) space [OK]

Common Mistakes:

• Assuming sorting is always O(n) time
• Believing hash sets use O(1) space
• Confusing DP applicability for duplicates

2. Given the following code snippet implementing the cyclic sort approach, what is the returned output when the input array is [2, 2, 3]?

easy

A. [2, 1]

B. [2, 1]

C. [2, 1]

D. [2, 1]

Solution

1. Step 1: Trace the cyclic sort swaps

   Initial array: [2, 2, 3]. i=0, nums[0]=2, correct_index=1. nums[0] == nums[1], so increment i to 1. i=1, nums[1]=2, correct_index=1, same index, increment i to 2. i=2, nums[2]=3, correct_index=2, correct place, increment i to 3.

2. Step 2: Identify mismatch in final array

   Final array remains [2, 2, 3]. Check indices: i=0 expects 1 but has 2 (duplicate), i=1 expects 2 and has 2 (correct), i=2 expects 3 and has 3 (correct). The mismatch at i=0 means duplicate=2, missing=1.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Output matches expected corrupt pair [2, 1] [OK]

Hint: Trace swaps carefully; duplicates cause no swap but increment index [OK]

Common Mistakes:

• Off-by-one in indexing
• Returning wrong pair order
• Missing increment after swap

3. Consider the following Python function that finds the first k missing positive numbers using cyclic sort and a boolean array. What is the output of find_k_missing_numbers([3, -1, 4, 5, 5], 3)?

easy

A. [2, 3, 6]

B. [1, 2, 3]

C. [1, 4, 6]

D. [1, 2, 6]

Solution

1. Step 1: Perform cyclic sort on input [3, -1, 4, 5, 5]

   After sorting, the array becomes [1, -1, 3, 4, 5] with 1 at index 0, -1 at index 1, 3 at index 2, 4 at index 3, and 5 at index 4. Negative numbers are ignored in placement.

2. Step 2: Identify missing numbers and extras

   Missing numbers are 2 (since index 1 != 2), and since k=3, the next missing number beyond array length is 6. So missing = [1, 2, 6].

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Output matches expected missing positives [OK]

Hint: Cyclic sort places numbers correctly; missing are indices not matching values [OK]

Common Mistakes:

• Confusing indices and values after cyclic sort
• Forgetting to add missing numbers beyond array length
• Miscounting duplicates as missing

4. What is the time complexity of the optimal cyclic sort approach to find all duplicates in an array of size n, and why might some candidates mistakenly think it is higher?

medium

A. O(n^2) because nested swaps can happen repeatedly

B. O(n log n) because of the sorting-like swaps

C. O(n) because each element is swapped at most once to its correct position

D. O(n) but with O(n) extra space for tracking duplicates

Solution

1. Step 1: Analyze the while loop behavior

   Each swap places at least one element in its correct position, so total swaps ≤ n.

2. Step 2: Understand why time is linear

   Since each element is moved at most once, total operations are O(n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Swapping each element once ensures linear time [OK]

Hint: Each element moves at most once -> O(n) time [OK]

Common Mistakes:

• Assuming swaps cause O(n^2) due to nested loops
• Confusing cyclic sort with comparison-based sorting
• Forgetting that extra space is O(1)

5. Suppose the problem is modified so that the input array can contain duplicates and the missing positive range must consider repeated numbers as well (e.g., if 2 appears twice, the missing range should still be correct). Which modification to the cyclic sort approach correctly handles duplicates without causing infinite loops or incorrect missing ranges?

hard

A. Skip swapping when nums[i] == nums[correct_pos] to avoid infinite loops caused by duplicates

B. Use a hash set to track visited numbers during cyclic sort swaps

C. Sort the array first to group duplicates, then apply cyclic sort

D. Increment i unconditionally after every swap regardless of duplicates

Solution

1. Step 1: Understand duplicate handling in cyclic sort

   Duplicates cause infinite swapping if we swap identical numbers repeatedly.

2. Step 2: Modify swap condition

   Adding a check to skip swapping when nums[i] == nums[correct_pos] prevents infinite loops and preserves correctness.

3. Step 3: Confirm other options fail

   Hash sets add extra space, sorting breaks O(n) time, unconditional increment after swap may skip needed swaps.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Skipping swaps on duplicates is standard fix for cyclic sort with duplicates [OK]

Hint: Avoid swapping identical elements to handle duplicates safely [OK]

Common Mistakes:

• Ignoring duplicates causing infinite loops
• Adding extra space unnecessarily