# Sample Practice Problems on Complexity Analysis of Algorithms

• Difficulty Level : Medium
• Last Updated : 21 Sep, 2022

### Problem 1: Find the complexity of the below recurrence:

{ 3T(n-1), if  n>0,
T(n) =   { 1, otherwise

Solution:

Let us solve using substitution.

T(n) = 3T(n-1)
= 3(3T(n-2))
= 32T(n-2)
= 33T(n-3)
…
…
= 3nT(n-n)
= 3nT(0)
= 3n

This clearly shows that the complexity of this function is O(3n).

### Problem 2: Find the complexity of the recurrence:

{ 2T(n-1) – 1, if n>0,
T(n) =   { 1, otherwise

Solution:

Let us try solving this function with substitution.

T(n) = 2T(n-1) – 1
= 2(2T(n-2)-1)-1
= 22(T(n-2)) – 2 – 1
= 22(2T(n-3)-1) – 2 – 1
= 23T(n-3) – 22 – 21 – 20
…..
…..
= 2nT(n-n) – 2n-1 – 2n-2 – 2n-3
….. 22 – 21 – 20

= 2n – 2n-1 – 2n-2 – 2n-3
….. 22 – 21 – 20
= 2n – (2n-1)

[Note: 2n-1 + 2n-2 + …… +  20 = 2n – 1]

T(n) = 1

Time Complexity is O(1). Note that while the recurrence relation looks exponential
he solution to the recurrence relation here gives a different result.

## CPP

 `function(``int` `n)``{``    ``if` `(n==1)``       ``return``;``    ``for` `(``int` `i=1; i<=n; i++)``    ``{``        ``for` `(``int` `j=1; j<=n; j++)``        ``{``            ``printf``(``"*"``);``            ``break``;``        ``}``    ``}``}`

Solution: Consider the comments in the following function.

## CPP

 `function(``int` `n)``{``    ``if` `(n==1)``       ``return``;``    ``for` `(``int` `i=1; i<=n; i++)``    ``{``        ``// Inner loop executes only one``        ``// time due to break statement.``        ``for` `(``int` `j=1; j<=n; j++)``        ``{``            ``printf``(``"*"``);``            ``break``;``        ``}``    ``}``}`

Time Complexity: O(n), Even though the inner loop is bounded by n, but due to the break statement, it is executing only once.

## CPP

 `void` `function(``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i=n/2; i<=n; i++)``        ``for` `(``int` `j=1; j<=n; j = 2 * j)``            ``for` `(``int` `k=1; k<=n; k = k * 2)``                ``count++;``}`

## Java

 `static` `void` `function(``int` `n)``{``    ``int` `count = ``0``;``    ``for` `(``int` `i = n / ``2``; i <= n; i++)``        ``for` `(``int` `j = ``1``; j <= n; j = ``2` `* j)``            ``for` `(``int` `k = ``1``; k <= n; k = k * ``2``)``                ``count++;``}` `// This code is contributed by rutvik_56.`

## C#

 `static` `void` `function(``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i = n / 2; i <= n; i++)``        ``for` `(``int` `j = 1; j <= n; j = 2 * j)``            ``for` `(``int` `k = 1; k <= n; k = k * 2)``                ``count++;``}` `// This code is contributed by pratham76.`

## Javascript

 ``

Solution: Consider the comments in the following function.

## CPP

 `void` `function(``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i=n/2; i<=n; i++)` `        ``// Executes O(Log n) times``        ``for` `(``int` `j=1; j<=n; j = 2 * j)` `            ``// Executes O(Log n) times``            ``for` `(``int` `k=1; k<=n; k = k * 2)``                ``count++;``}`

Time Complexity: O(n log2n).

## CPP

 `void` `function(``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i=n/2; i<=n; i++)``        ``for` `(``int` `j=1; j+n/2<=n; j = j++)``            ``for` `(``int` `k=1; k<=n; k = k * 2)``                ``count++;``}`

## Java

 `static` `void` `function(``int` `n)``{``    ``int` `count = ``0``;``    ``for` `(``int` `i=n/``2``; i<=n; i++)``        ``for` `(``int` `j=``1``; j+n/``2``<=n; j = j++)``            ``for` `(``int` `k=``1``; k<=n; k = k * ``2``)``                ``count++;``}` `// This code is contributed by Pushpesh Raj.`

Solution: Consider the comments in the following function.

## CPP

 `void` `function(``int` `n)``{``    ``int` `count = 0;` `    ``// outer loop executes n/2 times``    ``for` `(``int` `i=n/2; i<=n; i++)` `        ``// middle loop executes  n/2 times``        ``for` `(``int` `j=1; j+n/2<=n; j = j++)` `            ``// inner loop executes logn times``            ``for` `(``int` `k=1; k<=n; k = k * 2)``                ``count++;``}`

Time Complexity: O(n2logn).

## CPP

 `void` `function(``int` `n)``{``    ``int` `i = 1, s =1;``    ``while` `(s <= n)``    ``{``        ``i++;``        ``s += i;``        ``printf``(``"*"``);``    ``}``}`

Solution: We can define the terms ‘s’ according to relation si = si-1 + i. The value of ‘i’ increases by one for each iteration. The value contained in ‘s’ at the ith iteration is the sum of the first ‘i’ positive integers. If k is total number of iterations taken by the program, then while loop terminates if: 1 + 2 + 3 ….+ k = [k(k+1)/2] > n So k = O(√n).
Time Complexity: O(√n).

## CPP

 `void` `function(``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i=0; i

Solution: Consider the comments in the following function.

## CPP

 `void` `function(``int` `n)``{``    ``int` `count = 0;` `    ``// executes n times``    ``for` `(``int` `i=0; i

Time Complexity: O(n5)

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