Time complexity describes how the running time of an algorithm grows as the input size grows. It does not measure exact seconds on a particular computer. Instead, it estimates how many basic operations the algorithm performs.
This helps us compare algorithms in a machine-independent way. A solution that looks fast for small input may become very slow for large input, so time complexity is one of the most important tools in algorithm analysis.
Two algorithms may solve the same problem correctly, but their performance can be very different when the input becomes large.
Time complexity usually measures the number of basic operations performed by an algorithm, such as:
We usually express time complexity using Big O notation, which focuses on the dominant growth term and ignores minor constants.
The following rules are commonly used while analyzing code:
| Code Pattern | Typical Complexity | Reason |
|---|---|---|
| Simple statement | O(1) | Takes constant time |
| Single loop running n times | O(n) | One pass over input |
| Two nested loops of size n | O(n^2) | n multiplied by n |
| Loop that halves the problem each step | O(log n) | Input size shrinks quickly |
| Sequential independent blocks | Add them, then keep dominant term | O(n) + O(n^2) becomes O(n^2) |
| Recursive algorithm | Use recurrence relation | Analyze how work repeats |
The running time does not depend on the size of the input. The algorithm performs a fixed number of operations.
def get_first(arr):
return arr[0]
def is_even(n):
return n % 2 == 0
def swap(a, b):
return b, aThe running time grows slowly because the problem size is reduced by a constant factor in each step. Binary Search is the classic example.
def binary_search(arr, target):
left, right = 0, len(arr) - 1
while left <= right:
mid = (left + right) // 2
if arr[mid] == target:
return mid
if arr[mid] < target:
left = mid + 1
else:
right = mid - 1
return -1
# Each step cuts the search space in half.The algorithm processes each element once, so running time grows directly with input size.
def find_max(arr):
max_val = arr[0]
for num in arr:
if num > max_val:
max_val = num
return max_valThis often appears in efficient divide-and-conquer algorithms such as Merge Sort and Heap Sort. The problem is divided in logarithmic levels, and each level does linear work.
def merge_sort(arr):
if len(arr) <= 1:
return arr
mid = len(arr) // 2
left = merge_sort(arr[:mid])
right = merge_sort(arr[mid:])
return merge(left, right)
# Recurrence: T(n) = 2T(n/2) + O(n)
# Result: O(n log n)Quadratic time usually appears when one loop is nested inside another loop over the same input.
def print_pairs(arr):
for i in range(len(arr)):
for j in range(len(arr)):
print(arr[i], arr[j])
# Number of pair checks grows roughly as n^2.Exponential time appears in algorithms where each step creates multiple recursive branches. These algorithms become impractical very quickly.
def fibonacci(n):
if n <= 1:
return n
return fibonacci(n - 1) + fibonacci(n - 2)
# Creates many repeated recursive calls.
# Time complexity is exponential.An algorithm may not behave the same way for every input. That is why time complexity is often discussed in three forms:
| Case | Meaning | Example |
|---|---|---|
| Best case | Minimum time taken for some input | Linear search finds the item at the first position |
| Average case | Expected time for typical input | Linear search usually checks several elements |
| Worst case | Maximum time taken for input of size n | Linear search checks the whole array |
| Algorithm | Best Case | Average Case | Worst Case |
|---|---|---|---|
| Linear Search | O(1) | O(n) | O(n) |
| Binary Search | O(1) | O(log n) | O(log n) |
| Bubble Sort | O(n) | O(n^2) | O(n^2) |
| Merge Sort | O(n log n) | O(n log n) | O(n log n) |
| Quick Sort | O(n log n) | O(n log n) | O(n^2) |
Loops are one of the easiest places to estimate complexity.
# O(n)
for i in range(n):
print(i)
# O(n^2)
for i in range(n):
for j in range(n):
print(i, j)
# O(log n)
i = 1
while i < n:
print(i)
i = i * 2
# O(n)
for i in range(n):
print(i)
for j in range(n):
print(j)Important idea:
Recursive algorithms are often analyzed using a recurrence relation. A recurrence shows how the work for size n depends on smaller inputs.
| Recurrence | Example | Result |
|---|---|---|
| T(n) = T(n - 1) + O(1) | Factorial | O(n) |
| T(n) = T(n / 2) + O(1) | Binary Search | O(log n) |
| T(n) = 2T(n / 2) + O(n) | Merge Sort | O(n log n) |
| T(n) = T(n - 1) + O(n) | Selection Sort style sum | O(n^2) |
| T(n) = 2T(n - 1) + O(1) | Naive Fibonacci | Exponential |
def factorial(n):
if n <= 1:
return 1
return n * factorial(n - 1)
# T(n) = T(n - 1) + O(1) -> O(n)
def binary_search_recursive(arr, target, left, right):
if left > right:
return -1
mid = (left + right) // 2
if arr[mid] == target:
return mid
if arr[mid] < target:
return binary_search_recursive(arr, target, mid + 1, right)
return binary_search_recursive(arr, target, left, mid - 1)
# T(n) = T(n / 2) + O(1) -> O(log n)Amortized analysis studies the average cost of a sequence of operations when most operations are cheap and a few are expensive.
The Master Theorem is a shortcut for solving recurrences of the form T(n) = aT(n / b) + f(n).
| Case | Condition | Result |
|---|---|---|
| Case 1 | f(n) is smaller than n^(log_b a) | T(n) = Theta(n^(log_b a)) |
| Case 2 | f(n) has the same order as n^(log_b a) | T(n) = Theta(n^(log_b a) log n) |
| Case 3 | f(n) is larger than n^(log_b a) | T(n) = Theta(f(n)) |
Example: Merge Sort has T(n) = 2T(n / 2) + O(n). Here, a = 2 and b = 2. Since n^(log_b a) = n, this falls under Case 2, so the result is Theta(n log n).
| Complexity | n = 10 | n = 1,000 | n = 100,000 | Practical Meaning |
|---|---|---|---|---|
| O(1) | 1 step | 1 step | 1 step | Very fast |
| O(log n) | About 3 | About 10 | About 17 | Excellent scaling |
| O(n) | 10 | 1,000 | 100,000 | Usually acceptable |
| O(n log n) | About 30 | About 10,000 | About 1.7 million | Efficient for sorting and divide-and-conquer |
| O(n^2) | 100 | 1 million | 10 billion | Too slow for large input |
| O(2^n) | 1,024 | Infeasible | Infeasible | Useful only for tiny inputs |
# O(n^2)
def two_sum_slow(arr, target):
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
if arr[i] + arr[j] == target:
return [i, j]
return None
# O(n)
def two_sum_fast(arr, target):
seen = {}
for i, num in enumerate(arr):
complement = target - num
if complement in seen:
return [seen[complement], i]
seen[num] = i
return NoneExplore 500+ free tutorials across 20+ languages and frameworks.