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# TCS Codevita | Philaland Coin

In this blog, we will discuss a problem asked in TCS Codevita 2019.

Before running into the solution please try it by yourself first.

### Problem Description:

The problem solvers have found a new Island for coding and named it as Philaland.

These smart people were given a task to make a purchase of items at the Island easier by distributing various coins with different values.

Manish has come up with a solution that if we make coins category starting from \$1 till the maximum price of the item present on Island, then we can purchase any item easily. He added the following example to prove his point.

Let's suppose the maximum price of an item is 5\$ then we can make coins of {\$1, \$2, \$3, \$4, \$5} to purchase any item ranging from \$1 till \$5.

Now Manisha, being a keen observer suggested that we could actually minimize the number of coins required and gave following distribution {\$1, \$2, \$3}. According to him, any item can be purchased one time ranging from \$1 to \$5. Everyone was impressed with both of them.

Your task is to help Manisha come up with a minimum number of denominations for any arbitrary max price in Philaland.

1<=T<=100

1<=N<=5000

### Input Format:

First-line contains an integer T denoting the number of test cases.

Next T lines contain an integer N denoting the maximum price of the item present on Philaland.

### Output:

For each test case print a single line denoting the minimum number of denominations of coins required.

1

### Example 1:

Input:

```2
10
5```

Output:

```4
3```

Explanation:

For test case 1, N=10.

According to Manish {\$1, \$2, \$3,... \$10} must be distributed.

But as per Manisha only {\$1, \$2, \$3, \$4} coins are enough to purchase any item ranging from \$1 to \$10. Hence the minimum is 4. Likewise denominations could also be {\$1, \$2, \$3, \$5}. Hence answer is still 4.

For test case 2, N=5.

According to Manish {\$1, \$2, \$3, \$4, \$5} must be distributed.

But as per Manisha only {\$1, \$2, \$3} coins are enough to purchase any item ranging from \$1 to \$5. Hence the minimum is 3. Likewise denominations could also be {\$1, \$2, \$4}. Hence answer is still 3.

### Implementation using Python:

```#get input of number of cases
cases=int(input())

#for each case, find the coins required
for i in range(1,cases+1):
#get input of value
value=int(input())

#set coin counter at 0
coincount = 0

#count the number of coins
#logic number of coins required will be one more than the earlier rupee value that was a power of 2
while value>=1:
value=value//2
coincount=coincount+1

print (coincount)```

Happy Coding!

Please write comments if you find any bug in the above code/algorithm, or find other ways to solve the same problem.

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