# Analysis of Algorithms | Set 5 (Practice Problems)

We have discussed Asymptotic Analysis, Worst, Average and Best Cases , Asymptotic Notations and Analysis of loops in previous posts.

In this post, practice problems on analysis of algorithms are discussed.

Problem 1: Find the complexity of 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 ]
T(n) = 1
Time Complexity is O(1). Note that while
the recurrence relation looks exponential
the solution to the recurrence relation
here gives a different result.
```

Problem 3: Find the complexity of the below program:

 `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.

 `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 of the above function O(n). Even though the inner loop is bounded by n, but due to break statement it is executing only once.

Problem 4: Find the complexity of the below program:

 `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++; ` `} `

Solution: Consider the comments in the following function.

 `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 of the above function O(n log2n).

Problem 5: Find the complexity of the below program:

 `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++; ` `} `

Solution: Consider the comments in the following function.

 `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 of the above function O(n2logn).

Problem 6: Find the complexity of the below program:

 `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 of the above function O(√n).

Problem 7: Find a tight upper bound on complexity of the below program:

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

Solution:Consider the comments in the following function.

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

Time Complexity of the above function O(n5).

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