DSA - Important Time Complexities

Asymptotic Notations in Time Complexity

1. Big-O Notation O(f(n))

Worst-case time complexity.
Upper bound: algorithm takes at most f(n) time.
Used most commonly in contests/interviews.
Example: Binary Search → O(log n)

2. Omega Notation Ω(f(n))

Best-case time complexity.
Lower bound: algorithm takes at least f(n) time.
Not very useful alone, because best case may be rare.
Example: Linear Search → Ω(1) (if element is at first position)

3. Theta Notation Θ(f(n))

Tight bound (best and worst both are f(n)).
Algorithm always takes exactly f(n) time asymptotically.
Example: Merge Sort → Θ(n log n)

Complete Recurrence Solutions

Part A: Decremental Recurrences $T(n) = a \cdot T(n-b) + f(n)$

General Form

$\boxed{T(n) = a \cdot T(n - b) + f(n)}$

where:
- $a \geq 1$ (branching factor)
- $b > 0$ (constant decrement)
- $f(n)$ is the non-recursive cost

Solution by Unrolling

Expand the recurrence $k$ times where $k \approx n/b$ :

$T(n) = a^k \cdot T(n - k \cdot b) + \sum_{i=0}^{k-1} a^i \cdot f(n - i \cdot b)$

At base case: $k = n/b$ , so $T(0)$ or $T(1) = c$ (constant)

Case 1: $a < 1$ (Shrinking Recursion)

Behavior: Exponential term $a^k \to 0$ as $k \to \infty$
Dominant term: The summation
Result: $\boxed{T(n) \approx \sum_{i=0}^{n/b} a^i \cdot f(n - i \cdot b)}$
If $f(n)$ is non-decreasing, the largest term is $f(n)$ :

$T(n) = O\left(\frac{f(n)}{1-a}\right)$

This only work if $f(n)$ is non-decreasing

Case 2: $a = 1$ (Linear Accumulation)

Behavior: No exponential growth/decay
Recurrence becomes: $T(n) = T(n-b) + f(n)$
Solution: Direct summation of all levels

$\boxed{T(n) = \sum_{i=0}^{n/b - 1} f(n - i \cdot b)}$

Common subcases:

$f(n)$	Result	Can Be Understand as `T(n-b) * F(n)` ⭐
$c$ (constant)	$T(n) = \Theta(n)$	`n * c` => `O(n)`
$n$	$T(n) = \Theta(n^2)$	`n * n` => `O(n^2)`
$n^2$	$T(n) = \Theta(n^3)$	`n * n^2` => `O(n^3)`
$\log n$	$T(n) = \Theta(n \log n)$	`n * logn` => `O(nlogn)`

Case 3: $a > 1$ (Exponential Growth)

Behavior: Exponential term $a^{n/b}$ dominates
Recurrence becomes: $\boxed{T(n) = a^{n/b} \cdot T(0) + \sum_{i=0}^{n/b - 1} a^i \cdot f(n - i \cdot b)}$
If $f(n)$ is polynomial:

$T(n) = \Theta(a^{n/b})$

The exponential term always dominates polynomial $f(n)$ .

Note: Skipping $f(n)$ only work if $f(n)$ is polynomial (or smaller than exponential)

General Unrolled Formula for Decremental Recurrence

Full Formula $\boxed{T(n) = a^{n/b} \cdot T(0) + \sum_{i=0}^{n/b - 1} a^i \cdot f(n - i \cdot b)}$

Case 1: $a < 1$ (shrinking recursion)

Exponential term: $a^{n/b} \cdot T(0) \to 0$ (as $n \to \infty$ )
Skip $a^{n/b} T(0)$ Term
$T(n) \approx \sum\_{i=0}^{n/b - 1} a^i \cdot f(n - i \cdot b)$

Case 2: $a = 1$ (linear accumulation)

Exponential terms: $a^{n/b} = 1$ , $a^i = 1$ → no geometric growth
Skip $a^{n/b}.T(0)$ & $a^i$ Terms
$T(n) = \sum\_{i=0}^{n/b - 1} f(n - i \cdot b)$

Case 3: $a > 1$ (exponential growth)

Exponential term dominates → cannot skip anything
Use full formula
$T(n) = a^{n/b} \cdot T(0) + \sum\_{i=0}^{n/b - 1} a^i \cdot f(n - i \cdot b)$

Summary Table: $T(n) = a \cdot T(n-b) + f(n)$

Case	Condition	Dominant Term	Result
1	$a < 1$	Summation term	$O(f(n))$ if $f$ non-decreasing
2	$a = 1$	All terms equal weight	$\Theta\left(n \cdot f(n)\right)$ if $f$ constant $\Theta(n^{k+1})$ if $f(n) = n^k$
3	$a > 1$	Exponential term	$\Theta(a^{n/b})$

a < 1 → F(n) dominates
a = 1 → F(n) repeats
a > 1 → T(n) dominates

My Conclusion ⭐

Case 1: a < 1

Recursive term keeps shrinking
T(n-b) becomes very small
F(n) dominates

T(n) ≈ F(n)

Case 2: a = 1

Recursive term stays same size
F(n) added again and again
Both matter, but repetition dominates

T(n) ≈ n · F(n)

Case 3: a > 1 ⭐

Recursive term grows fast
T(n-b) dominates over F(n)
F(n) becomes negligible

T(n) ≈ a^(n/b)   exponential

Examples

Case 1

Example: $T(n) = 0.5 \cdot T(n-1) + n$ $\begin{align} a &= 0.5 < 1\ T(n) &= \sum\_{i=0}^{n-1} (0.5)^i \cdot (n-i)\ &\approx O(n) \quad \text{(geometric series with } a < 1\text{)} \end{align}$

Case 2

Example: $T(n) = T(n-1) + c$ $T(n) = c \cdot n = \Theta(n)$
Example: $T(n) = T(n-1) + n$ $T(n) = n + (n-1) + (n-2) + \cdots + 1 = \frac{n(n+1)}{2} = \Theta(n^2)$
Example 3: $T(n) = T(n-2) + n$ $T(n) = n + (n-2) + (n-4) + \cdots = \Theta(n^2)$

Case 3

Example: $T(n) = 2 \cdot T(n-1) + n$ $\begin{align} a = 2 > 1 \quad T(n) = 2^n \cdot T(0) + \sum\_{i=0}^{n-1} 2^i \cdot (n-i) = \Theta(2^n) \quad \text{(exponential dominates)} \end{align}$
Example: $T(n) = 3 \cdot T(n-2) + n^2$ $\begin{align} a &= 3 > 1 \quad T(n) = 3^{n/2} \cdot c + \text{lower order terms} = \Theta(3^{n/2}) = \Theta(\sqrt{3}^n) \end{align}$
Example: $2T(n-1) + n2^n$ $a=2 >1 \quad T(n) = 2^n + \sum\_{i=0}^{n-1} 2^i \cdot \[(n-i) \cdot 2^{n-i}] = 2^n \sum\_{i=0}^{n-1} (n-i) + 2^n$ $T(n) = 2^n \cdot \frac{n(n+1)}{2} + 2^n \sim \Theta(n^2 \cdot 2^n)$

Part B: Divide-and-Conquer Recurrences — Master Theorem

General Form

$\boxed{T(n) = a \cdot T(n/b) + f(n)}$

Constraints:

$a \geq 1$ (number of subproblems)
$b > 1$ (input size reduction factor)
$f(n)$ asymptotically positive

Critical Parameter ⭐

$\boxed {c = \log_b(a)}$

This represents the exponent of the recursive tree cost.
Interpretation: Compare $f(n)$ with $n^c$

Case 1: $f(n)$ is polynomially smaller than $n^c$

Condition: $f(n) = O(n^{c-\epsilon})$ for some $\epsilon > 0$
Result: $T(n) = \Theta(n^c)$
Intuition: Recursive calls dominate.

Case 2: $f(n)$ balances with $n^c$

Condition: $f(n) = \Theta(n^c \cdot \log^P n)$ for some $P$
Result:

Subcase	Condition	Solution
2a	$P > -1$	$T(n) = \Theta(n^c \cdot \log^{P+1} n)$
2b	$P = -1$	$T(n) = \Theta(n^c \cdot \log \log n)$
2c	$P < -1$	$T(n) = \Theta(n^c)$

Intuition: Work is evenly distributed across levels.

Case 3: $f(n)$ is polynomially larger than $n^c$

Condition: $f(n) = \Omega(n^{c+\epsilon})$ for some $\epsilon > 0$
AND regularity condition: $a \cdot f(n/b) \leq k \cdot f(n) \text{ for some } k < 1 \text{ and sufficiently large } n$
Result:

Subcase	Condition	Solution
3a	$P \geq 0$	$T(n) = \Theta(n^k \cdot \log^P n)$ where $f(n) = \Theta(n^k \cdot \log^P n)$
3b	$P < 0$	$T(n) = \Theta(n^k)$

Intuition: Root cost dominates.

Examples

Case 1

Example: $T(n) = 4T(n/2) + n$ $T (n) = 4 T (n /2) + n$ \begin{align} a = 4, \quad b = 2, \quad f(n) = n, \quad c = log\_2(4) = 2, \quad k = 1 \end{align}
- Compare: $c = 2 > k = 1$ → Case 1
- Solution: $T(n) = \Theta(n^2)$

Case 2a

Example: $T(n) = 2T(n/2) + n$ $T (n) = 2 T (n /2) + n$ \begin{align} a &= 2, \quad b = 2, \quad f(n) = n, \quad c = log\_2(2) = 1, \quad f(n) = n^1 \cdot \log^0 n \Rightarrow k = 1, P = 0 \end{align}
- Compare: $c = k = 1$ → Case 2a ( $P = 0 > -1$ )
- Solution: $T(n) = \Theta(n \log n)$

Case 2b

Example: $T(n) = 2T(n/2) + \frac{n}{\log n}$ $T (n) = 2 T (n /2) + \frac{n}{l o g n}$ \begin{align} a &= 2, \quad b = 2, \quad f(n) = n \cdot \log^{-1} n, \quad c = \log\_2(2) = 1, \quad k = 1, \quad P = -1 \end{align}
- Compare: $c = k = 1$ → Case 2b ( $P = -1$ )
- Solution: $T(n) = \Theta(n \log \log n)$

Case 2c

Example: $T(n) = 2T(n/2) + \dfrac{n}{\log^2 n}$ $T (n) = 2 T (n /2) + \frac{n}{lo g ^{2} n}$ \begin{align} a &= 2,\quad b = 2,\quad f(n) = n\cdot \log^{-2} n, \quad c = \log\_2(2) = 1 \quad k = 1,\quad P = -2 \end{align}
- Compare: (c = k = 1) → Case 2c ((P < -1))
- Solution: $T(n) = \Theta(n)$

Case 3a

Example: $T(n) = 2T(n/2) + n^2$ $T (n) = 2 T (n /2) + n^{2}$ ) \begin{align} a &= 2,\quad b = 2,\quad f(n) = n^2 \quad c = \log\_2(2) = 1 \quad k = 2,\quad P = 0 \end{align}
- Compare: (c = 1 < k = 2) → Case 3a ((P \ge 0))
- Solution: $T(n) = \Theta(n^2)$

Case 3b

Example: $T(n) = 2T(n/2) + n^2 \log n$ $T (n) = 2 T (n /2) + n^{2} lo g n$ \begin{align} a &= 2,\quad b = 2,\quad f(n) = n^2 \log n \quad c = 1 \quad k = 2,\quad P = 1 \end{align}
- Compare: (c < k) → Case 3b
- Solution: $T(n) = \Theta(n^2 \log n)$

Case 3c

Example: $T(n) = 2T(n/2) + \dfrac{n^2}{\log n}$ $T (n) = 2 T (n /2) + \frac{n ^{2}}{lo g n}$ \begin{align} a &= 2,\quad b = 2,\quad f(n) = n^2 \log^{-1} n \quad c = 1 \quad k = 2,\quad P = -1 \end{align}
- Compare: (c < k) → Case 3c ((P < 0))
- Solution: $T(n) = \Theta(n^2)$

Key Differences

Feature	$T(n-b)$ (Decremental)	$T(n/b)$ (Divide-and-Conquer)
Growth type	Arithmetic reduction	Geometric reduction
Number of levels	$\Theta(n)$	$\Theta(\log n)$
When $a=1$	$\Theta(n \cdot f(n))$	Use Master Theorem Case 2
When $a>1$	$\Theta(a^n)$ exponential	Depends on $f(n)$ vs $n^{\log\_b a}$
Master Theorem	Does NOT apply	Applies

Time Complexity in Competitive Programming

Summary ⭐

Most competitive programming platforms allow ~10^8 operations per second.
Generally allowed time limit: 1–2 seconds.
So, our algorithm should have ≤ 10^8 steps/operations.

(Assume 1 input = 1 operation)

$O(1) ≤ 10^8$ -----→ N can be very large (practically infinite)
$O(log N) ≤ 10^8$ -----→ $N ≤ 2^{10^{8}}$ (extremely large, practically infinite)
$O(N) ≤ 10^8$ -----→ $N ≤ 10^8$
$O(N log N) ≤ 10^8$ -----→ $N ≤ 10^8 / log2(10^8) ≈ 3.7 × 10^6$
$O(N^2) ≤ 10^8$ -----→ $N ≤ √(10^8) ≈ 10^4$
$O(N^3) ≤ 10^8$ -----→ $N ≤ ∛(10^8) ≈ 464$
$O(2^N) ≤ 10^8$ -----→ $N ≤ log2(10^8) ≈ 27$
$O(N!) ≤ 10^8$ -----→ $N ≤ 12$

How to Analyze Required Time Complexity for a Problem

You need to estimate what time complexity is acceptable based on the input size n. Here’s a practical guide:

Note: 1 In second ~ 10^7 to 10^8 operations in C++ (for competitive programming).

1 sec = ~10^8 operations
For a time complexity function
Code within this limit is likely to run in time

Input Size (`n`)	Acceptable Time Complexity
≤ 10	`O(n!)`, `O(2^n)`
≤ 15–20	`O(2^n)` or `O(n * 2^n)`
≤ 100	`O(n^4)`
≤ 1,000	`O(n^3)`
≤ 10,000	`O(n^2)`
≤ 1e5 (10^5)	`O(n log n)`
≤ 1e6 (10^6)	`O(n)`
~1e9 (10^9)	`O(log n)` or `O(1)`

1. Given Input Size → Find Suitable Time Complexity

If input size is x
Then choose a time complexity f(n) such that:
&#xNAN;f(x) < 10^8 (operations in 1 second limit)

2. Given Time Complexity → Find Maximum Input Size

If time complexity is f(n)
Then choose the maximum input size x such that:
&#xNAN;f(x) < 10^8

Note: f(x) represents the number of operations your program performs for input size x, and it should satisfy , f(x) < 10^8 to run within 1 second (standard time limit in competitive programming).

Why 10^8 Operations Matter in Competitive Programming ?

When you write code for competitive programming, the judge gives you 1 second to run your solution for each test case. Now, a modern computer can perform around 10⁷ to 10⁸ basic C++ operations in 1 second. So you should always make sure that:

For a given input size, your code should not do more than 10^8 operations.
If it does, the program will likely TLE (Time Limit Exceeded).
So you would need to optimise

This is why we estimate time complexity f(n) and ensure:

f(n) < 10^8
So that your code completes within the time limit.

Example:
If your algorithm is O(n^2), then n must be ≤ 10^3, because:
→ 10^3 * 10^3 = 10^6 operations (safe)
But if n = 10^5, then → 10^10 operations (TLE)

Examples of Time Complexities and Max Input Sizes

Brute-force Search (2 nested loops)
- Time Complexity: O(n^2)
- Max n: ~10^3
- Example: Check all pairs (i, j) in an array
Triple Nested Loop
- Time Complexity: O(n^3)
- Max n: ~100
- Example: Matrix multiplication (naive)
Backtracking / Subset Generation
- Time Complexity: O(2^n)
- Max n: ~20
- Example: Generating all subsets, permutations
Factorial Time
- Time Complexity: O(n!)
- Max n: ~10
- Example: Traveling Salesman Problem (TSP)
Sorting Algorithms (Merge Sort, Quick Sort)
- Time Complexity: O(n log n)
- Max n: ~10^5 to 10^6
- Example: std::sort() in C++
Binary Search
- Time Complexity: O(log n)
- Max n: up to 10^9
- Example: Find element in sorted array
Linear Search / Prefix Sum / One-pass Scan
- Time Complexity: O(n)
- Max n: ~10^7
- Example: Finding max in array
Map / Set Insert-Search (Unordered)
- Time Complexity: O(1) average
- Max n: ~10^6
- Example: unordered_map in C++
Heap / Priority Queue Operations
- Time Complexity: O(log n)
- Max n: ~10^6
- Example: Dijkstra’s algorithm
Dijkstra / BFS / DFS
- Time Complexity: O(V + E)
- Max V or E: ~10^5
- Example: Graph traversal
Segment Tree / BIT (Fenwick Tree)
- Time Complexity: O(log n) per operation
- Max n: ~10^6
- Example: Range queries and updates
Matrix Exponentiation / Fast Power
- Time Complexity: O(log n)
- Max n: ~10^18
- Example: Compute a^b mod m
Sieve of Eratosthenes
- Time Complexity: O(n log log n)
- Max n: ~10^7
- Example: Prime numbers up to n
DP with 1D array
- Time Complexity: O(n)
- Max n: ~10^6
- Example: Fibonacci with memoization
DP with 2D array
- Time Complexity: O(n^2)
- Max n: ~1000
- Example: Longest Common Subsequence (LCS)
Knapsack (DP O(n*W))
- Time Complexity: O(n * W)
- Max n*W: ~10^7
- Example: 0/1 Knapsack

Map Time Complexity

Map :

Implemented by self-balancing bst (like a Red-Black Tree)
Ordered Map : The time complexity for insertion, deletion, and access is O(log n) Unordered Map :
uses hash tables for faster lookups.
The average time complexity for insertion, deletion, and access is O(1), but in the worst case, it can degrade to O(n) if there are many hash collisions.