GENERATIVE AI TRAINING IN HYDERABAD

What the challenge asks

You’re given cities on a straight line (their x-coordinates) and their populations.

Between every pair of cities 𝑖 , 𝑗 i,j, the “cable cost” is min ⁡ ( 𝑝 𝑜 𝑝 𝑖 , 𝑝 𝑜 𝑝 𝑗 ) × ∣ 𝑥 𝑖 − 𝑥 𝑗 ∣ min(pop i ​

,pop j ​

)×∣x i ​

−x j ​

∣.

Compute the sum over all pairs, typically mod 10 9 + 7 10 9 +7 (the problem page shows the setup and samples). HackerRank

What’s inside the forum

The Discussions page (currently showing ~62 discussions) includes a mix of hints and some spammy posts; you don’t need to read them to solve the problem, but a few comments outline efficient strategies. HackerRank

A helpful comment sketch suggests processing cities in decreasing population, and for each city, efficiently summing distances to cities already processed by splitting them into left and right of the current coordinate and using data structures to query counts and coordinate sums in 𝑂 ( log ⁡ 𝑛 ) O(logn). HackerRank +1

Standard solution approach (condensed)

Sort by population (desc). When visiting city 𝑖 i, all previously inserted cities have population ≥ ≥ current, so each pair contributes current population × × distance.

Coordinate-compress the x-positions.

Maintain two Fenwick trees / segment trees as you insert cities:

Count tree: number of cities at or before an index.

Sum tree: sum of x-coordinates at or before an index.

For a city at coordinate 𝑥 x:

Let 𝐿 L = count of cities to the left, 𝑆 𝐿 S L ​

= sum of their coordinates.

Left distance sum

𝐿 ⋅ 𝑥 − 𝑆 𝐿 =L⋅x−S L ​

.

Let 𝑅 R = count to the right, 𝑆 𝑅 S R ​

= sum of their coordinates.

Right distance sum

𝑆 𝑅 − 𝑅 ⋅ 𝑥 =S R ​

−R⋅x.

Contribution

𝑝 𝑜 𝑝 𝑖 × ( left dist + right dist ) =pop i ​

×(left dist+right dist).

Insert the current city into the trees and continue. Overall complexity: 𝑂 ( 𝑛 log ⁡ 𝑛 ) O(nlogn). HackerRank +1

Gotchas called out by solvers

Use 64-bit integers (or big ints) and take mod 10 9 + 7 10 9 +7 at safe points. HackerRank

Handle duplicate coordinates and duplicate populations carefully (coordinate compression avoids large indexes; stable or grouped processing handles equal populations).

The 'Direct Connections' challenge is an interesting problem with real-world applications in [data analytics](https://dataanalyticsmasters.in/ ). Optimizing network connections and analyzing data flow efficiently are crucial for handling large-scale datasets. Similar techniques are used in telecommunications and logistics to minimize costs and improve performance. Great problem for developing analytical thinking Data Analytics Course In Hyderabad

This blog is fantastic! Clear, informative, and filled with valuable insights. Your writing makes even complex topics easy to understand. Excited for the next post! https://dataanalyticsmasters.in/

Some thoughts on this problem:

• sorting the cities according to population, then traversing the cities in increasing order of population is useful. It guarantees that at city_i, all previous cities (cities from 0 to i-1) have less population and hence the cost should be increased by city_i's population * sum of distances from city_i to all previous cities

• We can do simple mathematics to get sum of distances from city_i to all previous cities efficiently. Assume that cities from 0 to i-1 are divided into to categories:

  1. left cities (cities to the left of city_i, i.e. having coordinate < city_i's coordinate), assume there're l left cities which are city_l1, city_l2, ... sum of distances from city_i to these cities would be (city_i's coordinate - city_l1's coordinate) + (city_i's coordinate - city_l2's coordinate) + ... As we see, city_i's coordinate term is added l times (l = left cities count) and all left cities coordinates are subtracted, So the expression can be simplified as: left cities distances sum = city_i's coordinate * left cities count - left cities coordinates sum
  2. right cities (cities to the right of city_i, i.e. having coordinate > city_i's coordinate), assume there're r right cities which are city_r1, city_r2, ... sum of distances from city_i to these cities would be (city_i's coordinate - city_r1's coordinate) + (city_i's coordinate - city_r2's coordinate) + ... As we see, city_i's coordinate term is subtracted r times (r = right cities count) and all right cities coordinates are added, So the expression can be simplified as: right cities distances sum = right cities coordinates sum - city_i's coordinate * right cities count

• To get right & left cities count & coordinate sum, we can use augmented BST to store visited cities coordinates, we can insert, get count of values > threshold, get sum of values > threshold, get count of values < threshold, get sum of values < threshold, all in O(log n) time complixity, hence the overall time complixity is O(n log n)

Latest information and resources are important factors to get success in any competitive exam. Keeping this in mind we have designed a dedicated Hindi blog as a solution for job seekers looking to participate in government job exams. https://www.kaiseonline.com