Sorting before Two Pointer

The total time complexity is derived by analyzing each step of the algorithm and summing their individual complexities. Here's a detailed breakdown:

Steps of the Algorithm

1. Sorting the Array (arr.sort()):

   • The sort() function sorts the array in ascending order.
   • Sorting an array of size n takes O(n log n) using efficient sorting algorithms like Timsort (used in Python).

2. Two-Pointer Traversal (the while loop):

   • After sorting, the two-pointer technique is applied.
   • The two pointers (left and right) traverse the array at most once. In each iteration:
     • Either left is incremented, or right is decremented.
     • This ensures that the loop runs for at most O(n) iterations.

Combining the Steps

• Sorting takes O(n log n).
• The two-pointer traversal takes O(n).
• Since these steps are performed sequentially (not nested), their complexities are added: $$\text{Total Time Complexity}=O(n\log n)+O(n)$$

Simplifying the Complexity

• In Big O notation, only the dominant term matters as n grows large.
  • O(n log n) dominates O(n) because logarithmic growth adds a significant factor to linear growth.
• Therefore, the total complexity simplifies to: $$O(n\log n)$$

Why Approximation?

The symbol ≈ in O(n log n) + O(n) ≈ O(n log n) indicates that:

• The O(n) term is negligible compared to O(n log n) for large n.
• So the effective time complexity is considered O(n log n).

Practical Example

Let’s assume n = 1,000,000:

• Sorting: nlog⁡n=1,000,000⋅log⁡21,000,000≈20,000,000n \log n = 1,000,000 \cdot \log_2 1,000,000 \approx 20,000,000nlogn=1,000,000⋅log2​1,000,000≈20,000,000 operations.
• Traversal: $O(n)=1,000,000$ operations.
• Clearly, the sorting step dominates.