Program to find most visited sector in a circular track using Python

Suppose we have a number n and an array called rounds. We have a circular track which consists of n different sectors labeled from 1 to n. Now consider a race will be held on this track, the race consists of m different rounds. The ith round starts at sector rounds[i - 1] and ends at sector rounds[i]. For example, if the round 1 starts at sector rounds[0] and ends at sector rounds[1]. So we have to find the most visited sectors sorted in ascending order. (The track numbers are in ascending order of sector numbers in the counter-clockwise direction)

So, if the input is like n = 4 rounds = [1,3,1,2], then the output will be [1, 2]

because the race starts from sector 1. The order of the visited sectors is as follows: [1,2,3 (end of first round),4,1 (end of second round), 2 (end of 3rd round)]. Here both of the sectors 1 and 2 are visited twice and they are the most visited sectors. And sectors 3 and 4 are visited only once.

To solve this, we will follow these steps −

• d := a new map

• for j in range 1 to n, do

  • d[j] := 0

• d[rounds[0]] := 1

• for i in range 1 to size of rounds - 1, do

  • if rounds[i] > rounds[i-1], then

    • for j in range rounds[i-1]+1 to rounds[i]+1, do

      • d[j] := d[j] + 1

  • otherwise,

    • for j in range rounds[i-1]+1 to n, do

      • d[j] := d[j] + 1

    • for j in range 1 to rounds[i], do

      • d[j] := d[j] + 1

• curr := d[rounds[0]]

• out := [rounds[0]]

• for i in range 1 to n, do

  • if i is not same as rounds[0], then

    • if d[i] > curr, then

      • curr := d[i]

      • out := [i]

    • otherwise when d[i] is same as curr, then

      • append i with out

• return out after sorting

Example (Python)

Let us see the following implementation to get better understanding −

 Live Demo

def solve(n, rounds):
   d = {}
   for j in range(1,n+1):
      d[j] = 0
   d[rounds[0]] = 1
   for i in range(1, len(rounds)):
      if rounds[i] > rounds[i-1]:
         for j in range(rounds[i-1]+1, rounds[i]+1):
            d[j] += 1
      else:
         for j in range(rounds[i-1]+1,n+1):
            d[j] += 1
         for j in range(1,rounds[i]+1):
            d[j] += 1

   curr = d[rounds[0]]
   out = [rounds[0]]
   for i in range(1,n+1):
      if i != rounds[0]:
         if d[i] > curr:
            curr = d[i]
            out = [i]
         elif d[i] == curr:
            out = out + [i]
   return(sorted(out))

n = 4
rounds = [1,3,1,2]
print(solve(n, rounds))

Input

4, [1,3,1,2]

Output

[1, 2]