# 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<n; i++)
for (int j=i; j< i*i; j++)
if (j%i == 0)
{
for (int k=0; k<j; k++)
printf("*");
}
}
```

Solution:Consider the comments in the following function.

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

// executes n times
for (int i=0; i<n; i++)

// executes O(n*n) times.
for (int j=i; j< i*i; j++)
if (j%i == 0)
{
// executes j times = O(n*n) times
for (int k=0; k<j; k++)
printf("*");
}
}
```

Time Complexity of the above function O(n5).

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