Solution Approach

The solution implements a binary search approach to find if a number is a perfect square. Here's how the implementation works:

Binary Search Setup: The code uses Python's bisect_left function to perform the binary search. The search range is set from 1 to num, represented as range(1, num + 1).

Key Function: The crucial part is the key=lambda x: x * x parameter. This transforms our search space - instead of searching through the numbers [1, 2, 3, ..., num], we're effectively searching through their squares [1, 4, 9, ..., num²].

Finding the Position: bisect_left finds the leftmost position where num would fit in the sequence of perfect squares. Since bisect_left returns a 0-based index and our range starts at 1, we add 1 to get the actual value: l = bisect_left(...) + 1.

Verification: After finding l, we check if l * l == num. If this condition is true, it means num is exactly the square of l, making it a perfect square.

Example Walkthrough:

• For num = 16: The binary search finds that 16 fits at position 3 (0-indexed) in the sequence [1, 4, 9, 16, 25, ...]. Adding 1 gives us l = 4. Since 4 * 4 = 16, the function returns true.
• For num = 14: The binary search finds that 14 would fit between 9 and 16. It returns position 3, giving us l = 4. Since 4 * 4 = 16 ≠ 14, the function returns false.

The time complexity is O(log num) due to the binary search, and the space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with num = 20 to see how the binary search approach works:

Step 1: Initialize Binary Search

• We're searching in the range [1, 2, 3, 4, 5, ..., 20]
• The key function transforms each value x to x * x, so we're effectively searching where 20 fits in the sequence [1, 4, 9, 16, 25, 36, 49, ...]

Step 2: Binary Search Process

• Start with search range [1, 20], middle = 10
  • Check: 10 * 10 = 100 > 20, so search left half [1, 9]
• New range [1, 9], middle = 5
  • Check: 5 * 5 = 25 > 20, so search left half [1, 4]
• New range [1, 4], middle = 2
  • Check: 2 * 2 = 4 < 20, so search right half [3, 4]
• New range [3, 4], middle = 3
  • Check: 3 * 3 = 9 < 20, so search right half [4, 4]
• Final range [4, 4], middle = 4
  • Check: 4 * 4 = 16 < 20, need to go higher

Step 3: Result from bisect_left

• bisect_left returns index 4 (0-based) as the position where 20 would be inserted
• After adding 1 to convert from 0-based index: l = 5

Step 4: Verification

• Check if l * l == num: 5 * 5 = 25 ≠ 20
• Since 25 ≠ 20, return false

Therefore, 20 is not a perfect square. The closest perfect squares are 16 (4²) and 25 (5²), with 20 falling between them.

Time and Space Complexity

The time complexity is O(log n), where n is the given number. The bisect_left function performs a binary search on the range from 1 to num, which contains n elements. Binary search divides the search space in half at each step, resulting in logarithmic time complexity.

The space complexity is O(1). Although range(1, num + 1) is used as an argument to bisect_left, in Python 3, range objects are lazy iterators that don't create the entire list in memory. The bisect_left function only evaluates the key function for specific indices during the binary search process, using constant extra space for variables like search boundaries and the mid-point calculation.

Common Pitfalls

Pitfall 1: Using Wrong Binary Search Template Variant

The Problem: Using while left < right with right = mid instead of the standard template with while left <= right and right = mid - 1.

Wrong Implementation:

while left < right:
    mid = (left + right) // 2
    if mid * mid >= num:
        right = mid  # WRONG - should be mid - 1
    else:
        left = mid + 1
return left * left == num  # WRONG - should track first_true_index

Solution: Use the standard template with first_true_index to track the answer:

first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if mid * mid >= num:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
return first_true_index != -1 and first_true_index * first_true_index == num

Pitfall 2: Integer Overflow Issue

The multiplication mid * mid can exceed the maximum integer value in languages with fixed integer sizes.

Problem Example: When num = 2147483647 (maximum 32-bit integer), and mid = 46341, calculating mid * mid = 2147488281 would overflow in a 32-bit system.

Solution: Use long/long long types for the multiplication:

if ((long) mid * mid >= num)  // Java

if (static_cast<long long>(mid) * mid >= num)  // C++

Pitfall 3: Not Handling Edge Cases Properly

The code might not handle num = 1 correctly if the search range is optimized incorrectly.

Problem Example: If you set right = num // 2, then for num = 1, right = 0, making the search range invalid.

Solution: Keep the search range as [1, num] which handles all cases correctly, including num = 1.

Pitfall 4: Using Wrong Middle Point Calculation

Using (left + right) // 2 instead of left + (right - left) // 2 can cause overflow in languages with fixed integer sizes when left + right exceeds the maximum integer value.