Common complexity classes (O(1), O(n), O(log n), O(n²)) in Data Structures Theory - Step-by-Step Execution

Concept Flow - Common complexity classes (O(1), O(n), O(log n), O(n²))

Start: Input size n

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Choose algorithm

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Measure steps as function of n

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Classify complexity

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O(1): Constant steps

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O(log n): Steps grow slowly

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O(n): Steps grow linearly

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O(n²): Steps grow quadratically

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Predict performance for large n

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End

This flow shows how we start with input size, pick an algorithm, count steps, classify complexity, and predict performance.

Execution Sample

Data Structures Theory

def example(n):
    for i in range(n):
        for j in range(n):
            print(i, j)

This code prints pairs (i, j) for all i and j from 0 to n-1, showing O(n²) complexity.

Analysis Table

Step	i	j	Condition i<n	Condition j<n	Action	Output
1	0	0	0<3 True	0<3 True	Print (0,0)	(0,0)
2	0	1	0<3 True	1<3 True	Print (0,1)	(0,1)
3	0	2	0<3 True	2<3 True	Print (0,2)	(0,2)
4	0	3	0<3 True	3<3 False	Inner loop ends
5	1	0	1<3 True	0<3 True	Print (1,0)	(1,0)
6	1	1	1<3 True	1<3 True	Print (1,1)	(1,1)
7	1	2	1<3 True	2<3 True	Print (1,2)	(1,2)
8	1	3	1<3 True	3<3 False	Inner loop ends
9	2	0	2<3 True	0<3 True	Print (2,0)	(2,0)
10	2	1	2<3 True	1<3 True	Print (2,1)	(2,1)
11	2	2	2<3 True	2<3 True	Print (2,2)	(2,2)
12	2	3	2<3 True	3<3 False	Inner loop ends
13	3	-	3<3 False	-	Outer loop ends

💡 Outer loop i=3 fails condition i<n (3<3 is False), so loops end.

State Tracker

Variable	Start	After Step 1	After Step 4	After Step 8	After Step 12	Final
i	-	0	0	1	2	3
j	-	0	3	3	3	-

Key Insights - 3 Insights

Why does the inner loop reset j to 0 after it reaches n?

Because after j reaches n and the inner loop ends (see Step 4, 8, 12), the outer loop increments i and the inner loop starts again with j=0 (Step 5, 9). This is why j resets each time.

Why is this algorithm O(n²) and not O(n)?

Because for each of the n iterations of i, the inner loop runs n times, making total steps about n * n = n² (see execution_table rows 1-12).

What would happen if we removed the inner loop?

Then the steps would be just n (one loop), making complexity O(n). The execution_table would have fewer rows and j would not change.

Visual Quiz - 3 Questions

Look at the execution_table at Step 7. What are the values of i and j?

A i=0, j=2

B i=1, j=2

C i=2, j=1

D i=1, j=1

💡 Hint

Check the 'i' and 'j' columns in execution_table row for Step 7.

At which step does the outer loop condition become false?

A Step 12

B Step 4

C Step 13

D Step 1

💡 Hint

Look at the 'Condition i

If n was increased to 4, how many total print actions would occur?

A 16

B 8

C 12

D 4

💡 Hint

Recall that total prints = n * n, see execution_table steps 1-12 for n=3.

Concept Snapshot

Common complexity classes:
O(1): Constant steps, no matter input size.
O(n): Steps grow linearly with input size.
O(log n): Steps grow slowly, halving input each time.
O(n²): Steps grow quadratically, nested loops.
Used to predict algorithm speed as input grows.

Full Transcript

This visual execution shows how algorithms behave with different input sizes n. We start with input size n, pick an algorithm, and count how many steps it takes. For example, nested loops cause steps to grow as n squared, shown by printing pairs (i, j) for i and j from 0 to n-1. The execution table traces each step, showing variable values and conditions. The variable tracker shows how i and j change over time. Key moments clarify why loops reset and why complexity is quadratic. The quiz tests understanding of variable values and loop endings. The snapshot summarizes common complexity classes and their meaning.