Squares of a Sorted Array

▶

Steps

setup

Initialize / Setup

Start with the original input array nums = [-4, -1, 0, 3, 10]. The pointer 'i' is set to 0 to begin squaring elements.

💡 Setting up the pointer 'i' at the start allows us to process each element sequentially.

Line:

for i in range(len(nums)):

💡 We will process each element one by one from left to right.

fill_cells

Square element at index 0

Square the element at index 0: (-4) * (-4) = 16. Update nums[0] to 16.

💡 Squaring negative numbers changes their sign and magnitude, which affects sorting later.

Line:

nums[i] = nums[i] * nums[i]

💡 The first element is now positive and larger than some others, changing the array's order.

fill_cells

Move pointer to index 1

Move pointer 'i' to index 1 to square the next element.

💡 Moving the pointer ensures each element is processed in order.

Line:

for i in range(len(nums)):

💡 The algorithm processes elements sequentially from left to right.

fill_cells

Square element at index 1

Square the element at index 1: (-1) * (-1) = 1. Update nums[1] to 1.

💡 Squaring negative numbers always results in positive values.

Line:

nums[i] = nums[i] * nums[i]

💡 The array is progressively transformed into squares.

fill_cells

Move pointer to index 2

Move pointer 'i' to index 2 to square the next element.

💡 Continuing the sequential processing of elements.

Line:

for i in range(len(nums)):

💡 Each element is handled one at a time.

fill_cells

Square element at index 2

Square the element at index 2: 0 * 0 = 0. Update nums[2] to 0.

💡 Zero squared remains zero, showing no change for this element.

Line:

nums[i] = nums[i] * nums[i]

💡 The array now contains squared values for the first three elements.

fill_cells

Move pointer to index 3

Move pointer 'i' to index 3 to square the next element.

💡 Continuing the iteration through the array.

Line:

for i in range(len(nums)):

💡 The algorithm processes all elements in order.

fill_cells

Square element at index 3

Square the element at index 3: 3 * 3 = 9. Update nums[3] to 9.

💡 Positive numbers squared remain positive and increase in magnitude.

Line:

nums[i] = nums[i] * nums[i]

💡 The array is now partially squared with updated values.

fill_cells

Move pointer to index 4

Move pointer 'i' to index 4 to square the last element.

💡 Final element to process in the squaring phase.

Line:

for i in range(len(nums)):

💡 All elements will be squared after this step.

fill_cells

Square element at index 4

Square the element at index 4: 10 * 10 = 100. Update nums[4] to 100.

💡 The last element is squared, completing the first phase.

Line:

nums[i] = nums[i] * nums[i]

💡 All elements are now squared but not sorted yet.

prune

Sort the array in place

Sort the squared array in ascending order: [16, 1, 0, 9, 100] → [0, 1, 9, 16, 100].

💡 Sorting rearranges the squared values to produce the final sorted squares.

Line:

nums.sort()

💡 Sorting is necessary because squaring disrupts the original order.

reconstruct

Return the sorted squares array

Return the fully squared and sorted array as the final output.

💡 The algorithm completes by returning the transformed array.

Line:

return nums

💡 The output matches the expected sorted squares.

def sortedSquares(nums):
    # STEP 1-11: Square each element
    for i in range(len(nums)):  # STEP 1
        nums[i] = nums[i] * nums[i]  # STEP 2,4,6,8,10
    nums.sort()  # STEP 11
    return nums  # STEP 12

if __name__ == '__main__':
    nums = [-4,-1,0,3,10]
    print(sortedSquares(nums))

📊

Squares of a Sorted Array - Watch the Algorithm Execute, Step by Step

Watching each element being squared and then the sorting process helps you understand how the algorithm transforms the input into the sorted squares output.

move_right

i

-4

0

-1

1

0

2

3

3

10

4

record

i

16

0

-1

1

0

2

3

3

10

4

move_right

16

0

i

-1

1

0

2

3

3

10

4

record

16

0

i

1

1

0

2

3

3

10

4

move_right

16

0

1

1

i

0

2

3

3

10

4

record

16

0

1

1

i

0

2

3

3

10

4

move_right

16

0

1

1

0

2

i

3

3

10

4

record

16

0

1

1

0

2

i

9

3

10

4

move_right

16

0

1

1

0

2

9

3

i

10

4

record

16

0

1

1

0

2

9

3

i

100

4

compare

0

0

1

1

9

2

16

3

100

4

record

0

0

1

1

9

2

16

3

100

4

Result: [0,1,9,16,100]

Key Takeaways

✓ Squaring each element in place transforms the array but disrupts the original order.

It's hard to see from code alone how squaring negative numbers affects the array's order.

✓ Sorting the squared array is necessary to restore ascending order after squaring.

Visualizing the array before and after sorting clarifies why sorting is required.

✓ The pointer 'i' moves sequentially through the array, ensuring each element is processed exactly once.

Seeing the pointer move step-by-step helps understand the iteration process clearly.

Practice

1. You are given an integer array and a target value. The task is to count the number of triplets in the array such that the sum of the triplet is less than the target. Which approach guarantees an optimal solution in terms of time complexity?

easy

A. Use three nested loops to check all triplets and count those with sum less than target.

B. Sort the array, then for each element use two pointers to find pairs that form triplets with sum less than target.

C. Use dynamic programming to store counts of sums and build up to triplets.

D. Use a greedy approach by always picking the smallest elements first to form triplets.

Solution

1. Step 1: Understand problem constraints

   The problem requires counting triplets with sum less than a target, which suggests checking combinations efficiently.

2. Step 2: Identify approach that reduces complexity

   Sorting and using two pointers allows scanning pairs efficiently for each fixed element, reducing complexity from O(n^3) to O(n^2).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sorting + two pointers is a classic pattern for 3Sum variants [OK]

Hint: Sorting + two pointers reduces triple loops to quadratic [OK]

Common Mistakes:

• Assuming brute force is optimal
• Thinking DP applies directly here
• Believing greedy always works for sum constraints

2. Consider the following code snippet implementing the 3Sum problem. What is the final returned list when the input is [-1, 0, 1]?

easy

A. []

B. [[0, 1, -1]]

C. [[-1, 0, 1]]

D. [[-1, -1, 2]]

Solution

1. Step 1: Sort input and initialize

   Sorted input is [-1, 0, 1]. Start with i=0 (nums[i]=-1), seen = {}.

2. Step 2: Iterate j and check complements

   j=1 (nums[j]=0), complement=1. 1 not in seen, add 0 to seen. j=2 (nums[j]=1), complement=0 in seen -> triplet [-1,0,1] added.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Triplet [-1,0,1] found correctly [OK]

Hint: Sorted input and complement check find correct triplet [OK]

Common Mistakes:

• Confusing order of triplet
• Missing complement in seen
• Off-by-one in loops

3. You are given an unsorted array and a number k. The task is to find a subset of k elements such that the difference between the maximum and minimum elements in this subset is minimized. Which algorithmic approach guarantees an optimal solution efficiently?

easy

A. Use a greedy approach by always picking the smallest element and then the next closest elements without sorting.

B. Sort the array and use a sliding window of size k to find the minimum difference between the first and last elements in each window.

C. Use dynamic programming to store minimum differences for all subsets of size k.

D. Use brute force to generate all subsets of size k and compute their differences.

Solution

1. Step 1: Understand the problem constraints

   Finding a subset of k elements with minimum max-min difference requires contiguous elements after sorting.

2. Step 2: Identify the optimal approach

   Sorting the array places elements in ascending order, so the minimum difference subset must be a contiguous window of size k. Sliding window efficiently checks all such windows.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sorting + sliding window is classic for minimum range in subsets [OK]

Hint: Sort then slide window to find minimal range [OK]

Common Mistakes:

• Trying greedy without sorting misses optimal subsets
• Using DP unnecessarily complicates the problem
• Brute force is correct but impractical

4. The following code attempts to implement the Dutch National Flag algorithm but contains a subtle bug. Identify the line causing the bug.

medium

A. Line where mid is incremented after swapping 0

B. Line where low is incremented

C. Line where mid is incremented after encountering 1

D. Line where high is decremented after swapping 2

Solution

1. Step 1: Identify pointer updates after swapping 0

   After swapping nums[low] and nums[mid] when nums[mid]==0, both low and mid must be incremented to avoid infinite loops or incorrect ordering.

2. Step 2: Locate missing increment

   The code increments low but does not increment mid after swapping 0, causing mid to stay at the same index and potentially reprocessing the same element indefinitely.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Missing mid increment after swapping 0 causes infinite loop or wrong output [OK]

Hint: mid must increment after swapping 0 to progress [OK]

Common Mistakes:

• Forgetting mid increment after swap
• Confusing pointer roles
• Incrementing mid after swapping 2

5. What is the time complexity of the optimal two pointers approach for the trapping rain water problem, and why might some candidates mistakenly think it is higher?

medium

A. O(n^2) because each pointer moves inside a nested loop

B. O(n log n) due to sorting or priority queue operations

C. O(n) but with O(n) extra space for auxiliary arrays

D. O(n) because each element is visited at most once by either pointer

Solution

1. Step 1: Analyze pointer movements

   Each pointer (left and right) moves inward only, never revisiting elements, so total steps ≤ n.

2. Step 2: Understand common misconception

   Some think nested if-else inside while implies nested loops, leading to O(n^2) assumption, but loop is single while with two pointers.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Two pointers traverse array once -> O(n) time [OK]

Hint: Two pointers move inward once each -> O(n) time [OK]

Common Mistakes:

• Confusing nested if-else with nested loops
• Assuming extra arrays increase time complexity
• Thinking sorting is involved