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Problem Brief

Get Mean Rank Count 🍓

INTERNOA
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Amazon Academy recently organized a scholaship test on its platform.

There are nstudents with roll numbers 1, 2, ..., n who appeared for the test, where the rank secured by the ithstudent is denoted by rank[i]. Thus, the array rank is a permutation of length n. Groups can only be formed with students having consecutive roll numbers, in other words, a subarray of the original array. For each value x (1 <= x <= n), find the number of groups that can be formed such that they have a mean rank equal to x.

More formally, given a permutaion of length n, find the number of subarrays of the given array having a mean value equal to x, for each xin the range [1, n].

Notes

  1. 1. The mean value of an array of kelements is defined as the sum of elements divided by k.
  2. 2. A permutation of leangth nis a sequence where each number from qtonappears exactly once.
  3. 3. A subarray of an array is a contiguous section of the array.

Function Description

Complete the function getMeanRankCount in the editor. getMeanRankCount has the following parameter: int rank[n]: the ranks of the students.

Returns

int[n]: the ith integer (where 1 <= i <= n) denotes the number of groups with a mean rank of i.

*** Credit to precurewalker 👑 ***

1Example 1

Input
rank = [1, 2, 3, 4, 5]
Output
[1, 2, 3, 2, 1]
Explanation
Example 1 illustration
Read the above as 'For the mean x = 1, the group [1] has mean value 1. There is 1 group'. and so on. The full answer is [1, 2, 3, 2, 1].

2Example 2

Input
rank = [4, 3, 2, 1]
Output
[1, 2, 2, 1]
Explanation
  • x = 1 -> [1]
  • x = 2 -> [3, 2, 1], [2]
  • x = 3 -> [3], [4, 3, 2]
  • x = 4 -> [4]

    • 3Example 3

    Input
    rank = [4, 7, 3, 6, 5, 2, 1]
    Output
    [1, 1, 1, 4, 4, 1, 1]
    Explanation
  • x = 1 -> [1]
  • x = 2 -> [2]
  • x = 3 -> [3]
  • x = 4 -> [4], [3, 6, 5, 2], [7, 3, 6, 5, 2, 1], [4, 7, 3, 6, 5, 2 ,1]
  • x = 5 -> [5], [7, 3], [4, 7, 3, 6] and [4, 7, 3, 6, 5]
  • x = 6 -> [6]
  • x = 7 -> [7]

    • Constraints

    Limits and guarantees your solution can rely on.

  • 1 <= n <= 103
  • 1 <= rank[i] <= n
  • The array rank contains all distinct elemens, and thus, is a permutation of {1..n}.
    • public int[] getMeanRankCount(int[] rank) {
  // write your code here
}
    Input

    rank

    [1, 2, 3, 4, 5]

    Output

    [1, 2, 3, 2, 1]

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