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Wednesday, 27 July 2022

[Solution] High Frequency CodeChef Solution


Problem

Chef has an array A of length N.

Let F(A) denote the maximum frequency of any element present in the array.

For example:

  • If A = [1, 2, 3, 2, 2, 1], then F(A) = 3 since element 2 has the highest frequency = 3.
  • If A = [1, 2, 3, 3, 2, 1], then F(A) = 2 since highest frequency of any element is 2.

Chef can perform the following operation at most once:

  • Choose any subsequence S of the array such that every element of S is the same, say x. Then, choose an integer y and replace every element in this subsequence with y.

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For example, let A = [1, 2, 2, 1, 2, 2]. A few examples of the operation that Chef can perform are:

  • [1, \textcolor{red}{2, 2}, 1, 2, 2] \to [1, \textcolor{blue}{5, 5}, 1, 2, 2]
  • [1, \textcolor{red}{2}, 2, 1, \textcolor{red}{2, 2}] \to [1, \textcolor{blue}{19}, 2, 1, \textcolor{blue}{19, 19}]
  • [\textcolor{red}{1}, 2, 2, 1, 2, 2] \to [\textcolor{blue}{2}, 2, 2, 1, 2, 2]

Determine the minimum possible value of F(A) Chef can get by performing the given operation at most once.

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Input Format

  • The first line of input will contain a single integer T, denoting the number of test cases.
  • Each test case consists of two lines of input.
    • The first line of each test case contains a single integer N denoting the length of array A.
    • The second line contains N space-separated integers denoting the array A.

Output Format

For each test case, output the minimum value of F(A) Chef can get if he performs the operation optimally.


Explanation:

Test case 1: Chef cannot reduce the value of F(A) by performing any operation.

Test case 2: Chef can perform the operation [\textcolor{red}{1}, 1, \textcolor{red}{1}, 1, \textcolor{red}{1}] \to [\textcolor{blue}{2}, 1, \textcolor{blue}{2}, 1, \textcolor{blue}{2}]. The value of F(A) in this case is 3, which is the minimum possible.

Test case 3: Chef can perform the operation [1, \textcolor{red}{2, 2}, 1, 2, 2] \to [1, \textcolor{blue}{5, 5}, 1, 2, 2]. The value of F(A) in this case is 2, which is the minimum possible.

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