JLTi Code Jam

Solution – Scoring Weight Loss

29th Friday Fun Session – 4th Aug 2017

Given a sequence of weights (decimal numbers), we want to find the longest decreasing subsequence. And the length of that subsequence is what we are calling weight loss score. This is essentially the standard longest increasing subsequence (LIS) problem, just the other way.

This is the solution to JLTi Code Jam – Jul 2017 problem.

Let us walk through an example

Let us take the example as mentioned here: 95, 94, 97, 89, 99, 100, 101, 102, 103, 104, 105, 100, 95, 90. The subsequence can start at any value, and a value in a subsequence must be strictly lower than the previous value. Any value in the input can be skipped. The soul goal is to find the longest subsequence of decreasing values. Here one of the longest decreasing subsequences could be:  105, 100, 95, 90 and the length would be 4.

Even though, in our weight loss example, we have to find the length of longest decreasing subsequence, the standard problem is called longest increasing subsequence. Essentially the problems are the same. We can have a LIS solution and can pass it the negative of the input values. Alternatively, in the algorithm, we can alter the small to large, greater than to smaller than etc. We chose the former.

We will use two approaches to solve this problem: one is a dynamic programming based solution having O(n2) complexity, another is, let’s call it Skyline solution having O(n log n) complexity.

Dynamic Programming Solution

Let’s work with this example: 95, 96, 93, 101, 91, 90, 95, 100 – to see how LIS would work.

When the first value, 95 comes, we know it alone can make a subsequence of length 1. Well, each value can make a subsequence on its own of length 1.

When the second value 96 comes, we know it is greater than 95. Since 95 already made a subsequence of length 1, 96 can sit next to it and make a subsequence of length 2. And it would be longer than a subsequence of its own of length 1.

When the value 93 comes, it sees it cannot sit next to any value that appeared prior to it (95 and 96). Hence, it has to make a subsequence of its own.

When the value 101 comes, it knows that it can sit next to any prior values (95, 96 and 93). After all, it is bigger than each of them. It then computes the score it would make if it sits next to each of them, separately. The scores would be 2, 3, and 2, if it sits next to 95, 96 and 93 respectively. Of course, it would choose 96. The subsequence is 95, 96, 101 and the score is 3.

So we see, we can go from left to right of the input, and then for each of the previous values, it sees whether it can be placed after it. If yes, it computes the possible score. Finally, it chooses the one that gives it the highest score as its predecessor.

So we are using the solutions already found for existing sub-problems (the scores already computed for its preceding input values) and can easily compute its own best score from them. Hence, it is called a dynamic programming solution.

The following table summarizes it.

DP table.png

There are two longest subsequences each with length 3. For a certain value, if we need to know the preceding value, we can backtrace and find from which earlier value its score is computed. That way, we can complete the full subsequence ending with this value.

Since for each of the input values we are looping all the preceding values, the complexity is O(n2).

Skyline Solution

In this approach, we would retain all incompatible and hence promising subsequences since any of them could lead to the construction of one of the final longest subsequences. Only at the end of the input we would know which one is the longest. Since we are retaining all incompatible subsequences I am calling it Skyline, inspired by Skyline operator.

It is obvious but let me state here, all these solutions are standard, already found and used. However, Skyline is a name I am using as I find it an appropriate term to describe this method.

If there are two apples: one big and another small, and if you are asked to choose the better one, you would choose the big one. However, if you are given an apple and an orange, you cannot, as they are incomparable. Hence you need to retain both.

When a value comes it can be one of the below three types:

Smallest value (case 1)

  1. It won’t fit at the end of any existing subsequences. Because the value is smaller than all the end values for all existing subsequences.
  2. There is no other way but to create a new subsequence with this value.
  3. We can safely discard all single value subsequences existed so far. After all, the new subsequence with the smallest value can be compared with each of them and it is clearly superior to them (score for each such subsequence is 1 and the end (and only) value for the new one is the smallest – hence it can accept more future input values than the rests).
  4. In the list of subsequences we can retain the single value subsequence at first.

Biggest value (case 2)

  1. The opposite of the previous case is: the new value is bigger than the end values of each of the existing subsequences.
  2. So it can fit at the end of all existing subsequences. So which one to choose?
  3. Suppose, it is the end of the input. In that case, we would like it to go at the end of the longest subsequence found so far and make it longer by one more.
  4. However, if it is not the end of the input and suppose there are some future input values coming that are bigger than the end value of the present longest subsequence and smaller than the present input value. By placing the present input value at the end of the present longest subsequence we will jeopardize a more promising possibility in future.
  5. So we should rather copy the longest subsequence found so far and add this new value at the end of it, making it the new longest.
  6. At the same time, we retain the previous longest subsequence as it is, that by now is the second longest subsequence.
  7. We will add this new and longest subsequence at the end of the list.

Middle value (case 3)

  1. We have a third case where the input value can fit the end of some subsequences and cannot fit at the end of the rest subsequences.
  2. This is because this new value is bigger than the end values of some sun-sequences and smaller than the same for the rests.
  3. So which one to choose? Of course, we have to choose one where it can fit, meaning from those whose end values are smaller than the input value.
  4. And we would like to choose one with the largest end element (yet it is smaller than the input value).
  5. However, we cannot just over-write it for the same reason as stated earlier (case 2, promising reasoning). Rather we copy it, add the new value at the end of it and add it to the list.
  6. Where – at the end of the list?
  7. No, we would insert in next to the subsequence from which we copied and extended it.
  8. And we can safely discard all other subsequences with the same length as this newly created  subsequence. After all, the length is the same and it’s end element is smaller than the end elements of the rests having equal length of it.
  9. Shall we run a loop over the list to find those to be deleted? No, we just need to find the next subsequence and if its length is the same as the newly created subsequence we delete it. No more checking is required.
  10. Why so? Please read the second point as stated below.

So we have handled all possible input values. The list of subsequences that we have created would have some nice properties:

  1. As we go from the first subsequence to the last in the list of subsequences, the length will gradually increase.
  2. There would be a maximum of one subsequence with a certain length.
  3. To find whether the input value is a case 1 or case 2 or case 3 type, we can easily run a binary search with O(log n) complexity over the end elements of the subsequences in the list. Since we would like to do so for each of the n input values, the complexity of this approach would be O(n log n).
  4. For doing the above we can use the list, just that we need to look at the end elements. Then why are we retaining the complete list?
  5. The answer is: to output the longest subsequence as well.
  6. Could we do it without saving the complete subsequence?
  7. We leave it for another day.

Walking through an example

Let’s go through the same example as used earlier: 95, 96, 93, 101, 91, 90, 95, 100.

95 (case 1)

95

96 (case 2)

95

95, 96

93 (case 1)

93

95, 96

101 (case 2)

93

95, 96

95, 96, 101

91 (case 1)

91

95, 96

95, 96, 101

90 (case 1)

90

95, 96

95, 96, 101

95 (case 3)

90

90 95

95, 96 (deleted)

95, 96, 101

100 (case 3)

90

90 95

90 95 100

95, 96, 101 (deleted)

Once all the input values are treated, the last subsequence would be the longest one.

GitHub: Scoring Weight Loss

FaaS

6th JLTi Code Jam – Aug 2017

Threatened by the JLTi Weight Loss Competition where the participants are lining up in front of Salad shops, and the likes of me, who have entirely given up lunch (hopefully I can continue forever), food court shops who are selling oily, low-fibre and various other kinds of unhealthy food have come up with a novel idea.

Inspired from the software world, and more importantly, to attract the software people who sit in their chairs for long hours and are the primary victims of eating these junk, those food shops have chosen a name for this scheme – Food as a Service (FaaS), borrowed from the likes of SaaS, PaaS, IaaS – whatever that means, if that means anything at all.

Instead of paying on a daily basis, they are asking people to subscribe for food.

For example, without subscription, a set lunch would cost S$ 6, as usual, if you want to pay as you eat, just like as you are doing now. No strings attached.

However, if you subscribe for a week (5 meals, one meal one day, 5 consecutive days, not calendar week, can start at any day), instead of paying S$ 30, you can pay S$ 27.99 for five meals. Of course you have to eat from the same (chain of) shop.

And if you subscribe for a month (20 meals, one meal one day, 20 consecutive days, not calendar month, can start at any day) that they are vying for, you pay only S$ 99.99.

Input: 1, 2, 4, 5, 17, 18

Output: 36

Explanation: Input is a list of day numbers when you want to have a meal. The number can start at 1, and go up to any number.

A certain day number, say, 4, would not come more than once in the input, if it comes at all, assuming one can have only one lunch meal a day.

The above input says – you eat for 6 days. It makes no sense for you to go for a monthly subscription. Well, it also does not make sense to go for a weekly subscription. Paying daily basis for 6 days would be the best cost effective decision for you. You pay: S$ 36.

Input: 3, 4, 5, 6, 7, 17, 18

Output: 39.99

You subscribe for one week (first 5 days) and pay individually for the last 2 days. Your best decision cost you S$ 39.99.

Input: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 24

Output: 105.99

Here, a monthly subscription and S$ 6 for the last day would be the best deal for you.

Task: Given lunch calendar for some days (it can be 3 days, 10 days, 121 days or any number of days) as input, as explained above, I am planning to write a program that would output me the best price. Well, if I can find the best price, I also know what subscription plans etc. are. However, put that aside. Let’s find the best price, as shown and explained above.

Solution – Manipulating Money Exchange

25th Friday Fun Session – 7th Jul 2017

Given a set of currencies and some exchange rates among them, we want to find if there exists an arbitrage; meaning to exploit the discrepancies in the exchange rates and transform one unit of a certain currency to more than one unit of the same, thus making a profit.

This is the solution to JLTi Code Jam – Jun 2017 problem.

Let us walk through an example

Let us take the example as mentioned here. We can start with 1 USD, convert that to SGD (1.380 SGD), then convert that to MYR (1.380 * 3.080 MYR), then convert to INR (1.380 * 3.080 * 15.120 INR), then convert to GBP (1.380 * 3.080 * 15.120 * 0.012 GBP), then convert that back to USD (1.380 * 3.080 * 15.120 * 0.012 * 1.30 = 1.0025503488 USD).

We end up with more than 1 USD. That means, we have an arbitrage in this set of exchange rates. The profit making cycle here is: USD -> SGD -> MYR -> INR -> GBP. And since it is a cycle, we can start from any currency within it. For example, SGD -> MYR -> INR -> GBP -> USD also represents the same cycle.

The transformation

In general, if we have to make a profit, the respective rates in the cycle, when multiplied, should give more than 1, as we have seen in the above example.

Formula

Negative cycle in Bellman-Ford

After some simple transformation of the profit making condition, we see, if we take negative of log rate, and use that as the edge cost/distance, then finding profit making cycle is equivalent to finding negative cycle in the corresponding graph. And we can do so using Bellman-Ford algorithm.

To be precise, each of the currencies would be considered as a vertex. If there exists an exchange rate r between two currencies then there would be a directed edge between the corresponding vertices, and –log r would be the associated cost/distance of that edge.

Source of Bellman-Ford

The next question comes: using which vertex as source shall we run the Bellman-Ford? Let us see the below graph.

Souce.png

Suppose, we have a single profit making cycle here: GBP-> AUD -> CAD. In that case, if we start with USD as source vertex, we will never detect this cycle.

Add extra currency as source

To solve this problem, we need to add an extra currency, and then create edges from it to all the existing currencies with cost 0. Now using this extra vertex (EXT) as source we have to run Bellman-Ford and that would ensure that we can detect a cycle, if there exist one.

Extra Souce

GitHub: Manipulating Money Exchange

Scoring Weight Loss

5th JLTi Code Jam – Jul 2017

Now that weight loss competition is back, some people are more than excited about it. And why not? After all, only by running 10 km in the last 3 days, they can effortlessly shed 15 kg!

On the other hand, slim people, by any global standard, like me have to starve an entire month and still have to win this competition only in dream, in some rainy days.

Since the enthusiastic participants approached me to participate, I am thinking of a new scoring system that would remove this inherent bias in the existing scoring system – deducting final weight from first day weight.

So here I propose a new scoring system that would otherwise value the sustained effort and success of the participants, ignoring the total/final/absolute loss.

Input: 73, 72.9, 72.8, 72.8, 72.9, 72.7, 72.0, 71.6, 73, 72.5, 72.4, 71.3, 73.5, 74

Output: 7

Explanation: The above is an estimation of my performance, if the competition runs for 14 days. You can clearly see I start with 73 and end up with 74. In the old standard, I gain weight and penalized. In the new scoring system I score 7. How? Well, it computes how long I keep on decreasing weight, without seeing how much. In the above example, the longest stretch where I continue to lose weight (a value in a sequence must be smaller than its immediate predecessor) is shown below.

73, 72.9, 72.8, 72.7, 72.0, 71.6, 71.3

Let us also find the approximate score of the last time winner. A possible set of weights of him might look like the below:

Input: 95, 94, 97, 89, 99, 100, 101, 102, 103, 104, 105, 100, 95, 90

Output: 4

His success story lies in the last 3 days of blitzkrieg (the first weight in the input does not necessarily need to be the first value in the sequence): 105, 100, 95, 90

Let us also talk about a hypothetical participant who misunderstands this to be a weight gain competition and eats cheese all along.

Input: 53, 53.1, 53.2, 53.4, 53.5, 53.6, 53.9, 54, 54.1, 54.2, 54.2, 54.7, 55.8, 56

Output: 1

The scorer takes note of 53 and it never goes towards the right direction.

Task: A good scoring system indeed – nobody gets zero or below. And I am sure all of you would agree with this. Now let us quickly write a small program that takes an array of weights and computes the score.

Manipulating Money Exchange

4th JLTi Code Jam – Jun 2017

Input:

1 USD = 1.380 SGD

1 SGD = 3.080 MYR

1 MYR = 15.120 INR

1 INR = 0.012 GBP

1 GBP = 1.30 USD

I CAD = 0.57 GBP

Explanation: Now that we realize, we have to wait substantial amount of time to make any meaningful gain from stock market, we change our focus to money exchange. I am particularly very excited after collecting the above exchange rates using Google search today at 8th Jun 2017. If I start with 10, 000 USD, then convert them to MYR, then convert them to INR and then to GBP and then back to USD, I realize I will end up with 10, 025.50 USD, making USD 25.50 on the same spot within minutes.

Output: USD -> SGD -> MYR -> INR -> GBP -> USD

The next step that I am going to take is to get a list of all money exchanges available to me, somehow collect their exchange rates daily, and push it to a program that will tell me when it sees there is a chance to make some profit.

I might not always be lucky. To check whether my adrenaline rush is justified I looked at the same rates in an online money exchange. The rates from it look like below:

Input:

1 USD = 1.38295 SGD

1 SGD = 3.08614 MYR

1 MYR = 15.0996 INR

1 INR = 0.0119755 GBP

1 GBP = 1.295 USD

As you realize, we will end up losing money.

Output: No luck here

Task: I also realize, given a few hundred currencies and thousands of exchange rates among them, there is a possibility of having a number of ways we can make money. For example, given a set of rates (not as shown in the above example), we could make USD 10, starting with USD 10, 000, using this route: USD -> SGD -> MYR -> USD. Again, we could possibly make SGD 50, starting with SGD 10, 000, using another route: SGD -> MYR -> INR -> SGD. I am happy with just one such route, not necessarily the one providing the most profit.

Solution – Making Money at Stock Market

20th Friday Fun Session – 2nd Jun 2017

Given the stock prices for a number of days, in order, we have to buy one stock at one day and then sell it on a later day to maximize the profit.

This is the solution to JLTi Code Jam – May 2017  problem.

Let us walk through an example

Suppose, we have 8 days’ stock prices, starting at day 1, in order and they are 2, 10, 5, 12, 1, 3, 11, and 9 respectively. We can clearly see, if we buy a stock at day 1, at a price of 2 and then sell it on day 4, at a price of 12; we can make a profit of 10. We could make the same profit had we bought it on day 5, at a price of 1 and then sold it on day 7, at a price of 11. Since we want to make the most profit only once, we would choose say, the first one.

Accumulator pattern

We could simply run two loops, consider all possible buys and sells, calculate all possible profits (like buy on day 1 and sell it on day 2 with profit 8, buy on day 1 and sell it on day 3 with profit 3, and so on) and find the maximum profit. The below code using accumulator pattern  – traverse a sequence and accumulate a value (here it is maximum), as it goes, can be used to do so.

void StockN2(double price[], int n, double &maxProfit, int &buyDay, int &sellDay)
{
  for(int i=0; i<n; i++)
    for(int j=i+1; j<n-1; j++)     {       if(price[j] - price[i] > maxProfit)
      {
        maxProfit = price[j] - price[i];
        buyDay = i;
        sellDay = j;
      }
    }
}

For the outer loop that runs (n-1) times the inner loop runs (n-1) times resulting in O(n2), also known as quadratic complexity. Can we do better?

Divide and conquer solution

When we have an algorithm with O(n2) complexity, and we think about optimizing it, we ask ourselves – what could be better than this. And then O(n log n) comes to our mind. Even though there are many other complexities in between these two, usually O(n log n) is next best after O(n2). And when we see log n, divide and concur comes to our mind.

If we use divide and conquer, we have to divide the input into two sets, find some kind of solutions from both sides and then combine them. What decision can be found from two sides? We can get the maximum profit from each of the two sides. For example, suppose, we have only 4 days’ stock prices: 2, 10, 5 and 12. If we divide them into two, we would get left side: 2 and 10. Right side: 5 and 12. Left side profit would be 8 and the same for right side would be 7. The best between the two would be 8.

However, we can clearly see that the maximum profit is not 8, but 10 when the buy happens in left and sell happens in the right side. It is understandable that local solutions from left and right sides alone would not result in a global optimal solution. Somehow we have to compute a third profit combining the two sides. It is also obvious that buy happens in the left side and sell happens in the right side (after all, we cannot sell before we buy). So we see that the merge phase of the divide and conquer should consider the below 3 profits and find the best among them.

  • Maximum profit from left side
  • Maximum profit from right side
  • Profit by buying at the lowest from left and then selling at the highest in right

The below code is doing this. By the way, we are not tracking the buy day and sell day to keep the focus on the main points. Buy day and sell day can be easily accommodated.

void StockDivideAndConquerNlogN(double price[], int start, int end, double &maxProfit)
{
  if(start == end)
  {
    // just one value, return
    maxProfit = 0;
    return;
  }

  int mid = start + (end-start)/2;

  double leftMaxProfit;
  StockDivideAndConquerNlogN(price, start, mid, leftMaxProfit);

  double rightMaxProfit;
  StockDivideAndConquerNlogN(price, mid+1, end, rightMaxProfit);

  double minLeft = GetMin(price, start, end);
  double maxRight = GetMax(price, start, end);

  double minValue = GetMin(price, start, mid);
  double maxValue = GetMax(price, mid+1, end);
  maxProfit = maxOutOfThree(leftMaxProfit, rightMaxProfit, maxValue - minValue);
}

For our working example, with 8 days’ stock prices, it looks like below:

Divide and Conquer

The value inside the circle indicates the profit. It is coming from the best of the three as detailed earlier.

Computing complexity

How much is the cost? To compute it, we have to find two things – how many levels and how much work is done at each level.

How many levels? We have 8 items. At each level, it is halved. 8 -> 4 -> 2 -> 1. Suppose, in general, we are halving it k times. That means, n is divided by 2, k times. That means, n is divided by 2k and then it becomes 1. 1 = n/2k => n = 2k => log n = k log 2. Since, log 2 = 1, base being 2, k = log n. For n = 8, k = 3.

Level starts at 0 (the root) and ends at k. Hence, the actual number of levels is k+1. For simplicity, the rest of the post would consider the total number of levels to be log n, ignoring the small constant issue of 1.

Next thing to find: how much work is done at each level. We see, at each level, minimum is found from left, costing n/2 and maximum is found from right, also costing n/2. At each level, merging them (computing the third profit and then finding the best among the three) costs iterating n items and a few constant comparisons. Note that, at each level, all n items are present. It is the number of total processed items from all the sub-problems present in that level, not necessarily just from two sub-problems. So for log n levels the total cost is (n * log n) = n log n.

This calculation is explained/performed using master theorem.

Optimized divide and conquer solution

At each level, we are running loops and iterating n items to get the minimum from left and maximum from right. This loop is not necessary. The minimum and maximum can be computed bottom up doing a constant number of comparisons, at each level.

The below code shows this optimized version:

void StockDivideAndConquerOptimizedN(double price[], int start, int end, double &maxProfit, double &minValue, double &maxValue)
{
  if(start == end)
  {
    // just one value, return
    maxProfit = 0;
    minValue = maxValue = price[end];
    return;
  }

  int mid = start + (end-start)/2;

  double leftMaxProfit, leftMinValue, leftMaxValue;
  StockDivideAndConquerOptimizedN(price, start, mid, leftMaxProfit, leftMinValue, leftMaxValue);

  double rightMaxProfit, rightMinValue, rightMaxValue;
  StockDivideAndConquerOptimizedN(price, mid + 1, end, rightMaxProfit, rightMinValue, rightMaxValue);

  maxProfit = maxOutOfThree(leftMaxProfit, rightMaxProfit, rightMaxValue - leftMinValue);

  minValue = leftMinValue > rightMinValue ? rightMinValue : leftMinValue;
  maxValue = leftMaxValue > rightMaxValue ? leftMaxValue : rightMaxValue;
}

Computing complexity for this optimized version

In this optimized version, for each of the log n levels, we are still doing some processing. It is no longer n items. Rather, the number of items is decreasing by half at each level, upwards. At the bottom-most level there are n items. One level up, it is reduced by half (due to the merging), shrinking to n/2 items. As it goes up, it gets reduced and at the topmost level it becomes only one item. Let us add the items from each level that we we processing.

n + n/2 + n/22 + n/2+ n/24 + .  .  .  . + n/2k)

=> n (1 + 1/2 + 1/4 + 1/8 + . . .)

=> n * 2 (the convergent series gives 2)

=> 2 n

By discarding the constant terms, we get a complexity of O(n), meaning we can get the maximum profit in linear time.

GitHub: Stock Profit

JLTi Code Jam

At JLTi, I manage a monthly programming exercise. On the first week of every month, I set a programming problem and release it for all to solve by the end of the same month. We call it JLTi Code Jam, inspired by Google Code Jam.

We started it from Mar 2017 and so far we made it every month.

The programming problem is set in a way so that it can be solved using the data structures/algorithms discussed in the already conducted Friday Fun Sessions. The focus is on correctness, execution efficiency (time/space) and code quality.

Every JLTi Code Jam problem is published in this blog. The solution of a certain month’s JLTi Code Jam problem is discussed on the first Friday Fun Session on the following month.

JLTi Code Jam along with Friday Fun session is one of many endeavours as to how, we, mostly the engineers at JLTi, continuously learn, re-skill ourselves and sharpen our technical, programming and problem solving skills.

Finally, thank you all so much who participate in the JLTi Code Jam exercise, and encourage me to continue it. It is only you who made it a success so far.

Making Money at Stock Market

3rd JLTi Code Jam – May 2017

Input: price = [961, 984.5, 965, 988.5, 956.5]

Explanation: Now that your bank accounts are flooding with April bonus money and you are contemplating to invest in stock market why not join me in doing some analysis first? After all, we are engineers flooded with data (quite a lot of them are free) and the capability to analyse them. I opened JLT’s historical stock price from yahoo finance. It’s amazing! As part of the analysis, you know, we have to do a tremendous amount of work. As a start, I wanted to focus on when to buy and when to sell a stock so that I can make the most profit. For example, I took the data from 19th Dec 2016 to 23rd Dec 2016 as specified above. I did some manual calculation and found that had I bought on 19th Dec 2016 at £961 and sold on 22nd Dec 2016 £988.5, I could make the most profit, £27.5.

I also checked if I had to make the most profit by buying and selling a single JLT share once within 2016 , I could do so by buying on 9th Feb 2016 at £776.5 and selling on 11th Apr 2016 at £1070, making a whopping £293.5 profit! I have not put the data here as one year’s data is too huge to fit here. You can collect and verify it by downloading it as an excel file from the above yahoo link.

Output: Buy on day 1 at £961 and sell on day 4 at £988.5, making £27.5

I also checked the values from 15th Jan 2016 to 21st Jan 2016 (excluding 16th Jan 2016 and 17th Jan 2016 when stock market was closed) and the price looked like the below.

Input: price = [890, 890, 853.5, 828.5, 809]

You can see it kept on dropping and there was no way to make money in this period.

Output: Don’t buy stock

Task: As you realize, as part of the bigger data analysis work that we need to do, it is a small part. We got so much data for hundreds of companies. Hence, it is essential that we do it efficiently. To be precise, I am looking for a solution more time efficient than O(n2).

No Two Team Members Next to Each Other

1st JLTi Code Jam – Mar 2017

Input: 1, 1, 2, 2, 2, 567, 567, 10000076, 4, 2, 3, 3

Explanation: There are 12 people listed above. They belong to 6 teams (Team 1, Team 2, Team 3, Team 567, Team 10000076, and Team 4). As you can see people are identified in the list by the team number.

Output: 1, 2, 1, 2, 4, 2, 567, 3, 567, 100076, 2, 3

As you can see, the output has rearranged the team members in way that no two members from the same team standing next to each other.

Input: 1, 1, 1, 1, 2

Output: It is not possible to rearrange them.

Task: You have to write a program that can rearrange even billions of such team members belonging to millions of teams very fast. If the input is such that it is not possible to rearrange then the output should be: It is not possible to rearrange them. A correct solution is not sufficient. The algorithm has to be efficient, otherwise the output for big data 🙂 will not come.

GitHub: https://github.com/gopalcdas/NoTeamMemberNextToEachOther

Company Tour 2017 to Noland

2nd JLTi Code Jam – Apr 2017

Input: Capacity = 125, w = [45, 25, 80, 100, 125]

Explanation: This year, RC has taken all JLT Asia employees to Noland for the company trip. As the name implies there is not much land in Noland, it is river everywhere. When we have to cross such a river having only one boat with a certain capacity (in the above example 125 Kg), Warren Downey, our Deputy CEO approaches RC and asks us to quickly divide the people so that each trip of the boat carries people exactly to its maximum capacity, 125 Kg in this example. He shows the example above and works out the below output that he desires.

Output: {45, 80}, {25, 100}, {120}

When RC team pointed out what would happen for a scenario when we have a case like Capacity = 120, w = [40, 20, 80, 100, 120, 70]. Warren informs us we always utilize our resources to its maximum capacity. No compromise. We will not cross the river and will change the tour itinerary.

Output: No crossing, change itinerary.

Task: When I woke up an hour ago from my afternoon nap with a lot of stress, I realized that the tour was just a bad dream. I started feeling relaxed. But the problem got into my head and now it is itching everywhere inside it. In this situation, I realize, I can spread the itching to my JLTi friends in Singapore and Mumbai as well.

You can imagine there is a boatman and his weight is out of consideration. The input capacity is only applicable for the passengers. The input w array is holding only the passenger weights. In short, you can ignore the boatman.

GitHub: https://github.com/gopalcdas/TourToNoLand