Kevin and the tower of coins

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Last Updated: 7 Mar, 2021

Kevin and the tower of coins

Moderate

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Problem statement

Kevin has ‘N’ coins. Each coin has a specific width and diameter. Kevin wants to build a tower using these coins such that each coin in the tower has strictly less width and diameter as compared to all coins placed below this coin.

You have to find the maximum height of the tower that Kevin can build by using these coins.

Note:

The height of the tower is calculated by adding the width of all the coins used in the formation of this tower.

Input Format:

The first line contains a single integer ‘T’ representing the number of test cases. 

The first line of each test case will contain a single integer ‘N’ which denotes the number of coins available.

The next ‘N’ lines contain the two space-separated integers “ARR[i][0]” and “ARR[i][1]”, where “ARR[i][0]” is the width of the ‘i-th’ coin and “ARR[i][1]” is the diameter of the ‘i-th’ coin.

Output Format:

For each test case, return the maximum possible height of the tower.

Output for every test case will be printed in a separate line.

Note:

You don’t need to print anything; It has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 50
1 <= N <= 10^4
1 <= ARR[i][0] and ARR[i][1] <= 10^5

Time limit: 1 sec

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Approaches

01Approach

The basic idea is to first sort the given array/list of integers in the increasing order of width and then just find the longest increasing subsequence of coins on the basis of their diameter.

We are here using a helper function that is recursive in nature and it is used to find the LIS (Longest Increasing Subsequence) of coins on the basis of diameter.

int helper(vector<vector<int>> &arr, int previous, int current)

Where ‘ARR’ is the vector/list of pairs of integers that contain all the given coins (width of the coins of first and diameter on second), ‘PREVIOUS’ is the maximum diameter of the previous coins, and ‘CURRENT’ is the index of the current coin.

There will be two types of recursive calls: first, in which we will consider picking the current coin and second, in which we drop the current coin.

The steps are as follows:

In the 'HELPER' function
- If ‘CURRENT’ becomes equal to the length of 'ARR' return 0;
- Create two variables ‘PICK’ and ‘DROP’ to store the result of two recursive calls. Initialise both of them by 0.
- If the current coin has a diameter greater than 'PREVIOUS' then store the result of HELPER(ARR, ARR[CURRENT][1], CURRENT + 1) in ‘PICK’ and increment it by 1.
- Store the result of HELPER(ARR, PREVIOUS, CURRENT + 1) in ‘DROP’.
- Return the maximum between ‘PICK’ and 'DROP'.
In the given function
- Sort the vector so that all the coins become arranged in the increasing order of their width.
- Return the result of HELPER(ARR, INT_MIN, 0).

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02Approach

The basic idea of this approach is to compute the LIS(Longest Increasing Subsequence) for the starting ‘i’ coins and by using this result compute LIS for the starting ‘i’ + 1 coins.

Reference: https://cpalgorithms.com/sequences/longest_increasing_subsequence.html

The steps are as follows:

Sort the vector so that all the coins become arranged in the increasing order of their width.
Create a vector ‘LIS’ to store LISs and for the first index set LIS 1 because the first coin in the ‘ARR’ will always have LIS as 1.
Iterate through the 'ARR' from 1 to ‘N’. (say, iterator = ‘i’)
- Make LIS[ i ] = 1
- Iterate again through ‘ARR’ from 0 to ‘i’. (say, iterator = ‘j’)
  - If found a coin with less diameter than the coin at ‘i’ and its LIS in ‘LIS’ + 1 is also greater than LIS[ i ] then set LIS[ i ] to LIS[ j ] + 1.
Return the maximum value stored in ‘LIS’.

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03Approach

The basic idea of this approach is to store the tail (or last and the least) element of the LIS of lengths 1, 2, 3, and so on. Whenever we found a new element that is greater than the last filled element in the ‘TAIL’ array then just append this to the array because it increases the previous longest length of LIS. Otherwise, replace this element with the element just greater than this in the already picked element ( in the ‘TAIL’ array).

The steps are as follows:

Sort the vector so that all the coins become arranged in the increasing order of their width.
Create a vector ‘TAIL’ to store the least possible element which makes the LIS of length ‘i’ + 1 where ‘i’ is the index in the ‘TAIL’ array.
Append the first element in the ‘TAIL’ as it is because it is the only element that just makes the length of the longest LIS equal to 1.
Iterate through the ‘ARR’ from 1 to ‘N’. (say, iterator = ‘i’)
- If the last added element in the tail is smaller than the element at ‘ARR[ i ][ 1 ]’ then just append it into the ‘TAIL’ array.
- Else,
- Found the lower bound of ARR[ i ][ 1 ] in the 'TAIL' and replace it with this coin.
Return size of 'TAIL'.

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