# Length of array pair formed where one contains all distinct elements and other all same elements

Given an array **arr[]**, the task is to determine the maximum possible size of two arrays of the same size that can be created using the given array elements, such that in one array all the elements are **distinct** and in the other, all elements are the** same**. **Note:** It is not mandatory to use all the elements of the given array to build the two arrays.

**Examples:**

Input:a[] = {4, 2, 4, 1, 4, 3, 4}Output:3Explanation:

The maximum possible size would be 3 – {1 2 3} and {4 4 4}

Input:a[] = {2, 1, 5, 4, 3}Output:1Explanation:

The maximum possible size would be 1 – {1} and {2}

**Approach:** To solve the problem mentioned above, follow the steps given below:

- First, count the frequency of all the numbers in the given array and also count the number of distinct elements in that array. Then count the maximum frequency among all of them
*maxfrq* - It is clear, that the maximum possible size of an array(that contains the same elements) would be maxfrq because it is the highest possible value of frequency of a number. And, the maximum possible size of another array(that contains all distinct elements) would be dist, which is the total number of distinct elements. Now we have to consider 2 cases:
- If we take all the elements having frequency maxfrq in one array(that contains the same elements), then the total number of distinct elements left would be
*dist – 1*, so the other array can only contain a dist – 1 number of elements. In this case, the answer would be,**ans1 = min(dist – 1, maxfrq)** - If we take all the elements having frequency maxfrq except one element in one array(that contains the same elements), then the total number of distinct elements left would be dist, so the other array can only contain a dist number of elements. In this case, the answer would be,
**ans2 = min(dist, maxfrq – 1)**.

- If we take all the elements having frequency maxfrq in one array(that contains the same elements), then the total number of distinct elements left would be
- Hence, the actual answer is

max( min(dist – 1, maxfrq), min(dist, maxfrq – 1) )

- Below is the implementation of the above approach:

## C++

`// C++ implementation to Divide the array` `// into two arrays having same size,` `// one with distinct elements` `// and other with same elements` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the max size possible` `int` `findMaxSize(` `int` `a[], ` `int` `n)` `{` ` ` `vector<` `int` `> frq(n + 1);` ` ` `for` `(` `int` `i = 0; i < n; ++i)` ` ` `frq[a[i]]++;` ` ` `// Counting the maximum frequency` ` ` `int` `maxfrq` ` ` `= *max_element(` ` ` `frq.begin(), frq.end());` ` ` `// Counting total distinct elements` ` ` `int` `dist` ` ` `= n + 1 - count(` ` ` `frq.begin(),` ` ` `frq.end(), 0);` ` ` `int` `ans1 = min(maxfrq - 1, dist);` ` ` `int` `ans2 = min(maxfrq, dist - 1);` ` ` `// Find max of both the answer` ` ` `int` `ans = max(ans1, ans2);` ` ` `return` `ans;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 4, 2, 4, 1, 4, 3, 4 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << findMaxSize(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to Divide the array` `// into two arrays having same size,` `// one with distinct elements` `// and other with same elements` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG {` ` ` `// Function to find the max size possible` `static` `int` `findMaxSize(` `int` `a[], ` `int` `n)` `{` ` ` `ArrayList<Integer> frq = ` `new` `ArrayList<Integer>(n+` `1` `);` ` ` ` ` `for` `(` `int` `i = ` `0` `; i <= n; i++)` ` ` `frq.add(` `0` `);` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < n; ++i)` ` ` `frq.set(a[i], frq.get(a[i]) + ` `1` `);` ` ` ` ` `// Counting the maximum frequency` ` ` `int` `maxfrq = Collections.max(frq);` ` ` `// Counting total distinct elements` ` ` `int` `dist = n + ` `1` `- Collections.frequency(frq, ` `0` `);` ` ` ` ` `int` `ans1 = Math.min(maxfrq - ` `1` `, dist);` ` ` `int` `ans2 = Math.min(maxfrq, dist - ` `1` `);` ` ` `// Find max of both the answer` ` ` `int` `ans = Math.max(ans1, ans2);` ` ` `return` `ans;` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = { ` `4` `, ` `2` `, ` `4` `, ` `1` `, ` `4` `, ` `3` `, ` `4` `};` ` ` `int` `n = arr.length;` ` ` ` ` `System.out.println(findMaxSize(arr, n));` `}` `}` `// This code is contributed by coder001` |

## Python3

`# Python3 implementation to divide the ` `# array into two arrays having same ` `# size, one with distinct elements` `# and other with same elements` `# Function to find the max size possible` `def` `findMaxSize(a, n):` ` ` `frq ` `=` `[` `0` `] ` `*` `(n ` `+` `1` `)` ` ` `for` `i ` `in` `range` `(n):` ` ` `frq[a[i]] ` `+` `=` `1` ` ` `# Counting the maximum frequency` ` ` `maxfrq ` `=` `max` `(frq)` ` ` ` ` `# Counting total distinct elements` ` ` `dist ` `=` `n ` `+` `1` `-` `frq.count(` `0` `)` ` ` `ans1 ` `=` `min` `(maxfrq ` `-` `1` `, dist)` ` ` `ans2 ` `=` `min` `(maxfrq, dist ` `-` `1` `)` ` ` `# Find max of both the answer` ` ` `ans ` `=` `max` `(ans1, ans2)` ` ` `return` `ans` `# Driver code` `arr ` `=` `[ ` `4` `, ` `2` `, ` `4` `, ` `1` `, ` `4` `, ` `3` `, ` `4` `]` `n ` `=` `len` `(arr)` `print` `(findMaxSize(arr, n))` `# This code is contributed by divyamohan123` |

## C#

`// C# implementation to Divide the array` `// into two arrays having same size,` `// one with distinct elements` `// and other with same elements` `using` `System;` `using` `System.Collections;` `class` `GFG{` ` ` `// Function to find the max size possible` `static` `int` `findMaxSize(` `int` `[]a, ` `int` `n)` `{` ` ` `ArrayList frq = ` `new` `ArrayList(n + 1);` ` ` `for` `(` `int` `i = 0; i <= n; i++)` ` ` `frq.Add(0);` ` ` `for` `(` `int` `i = 0; i < n; ++i)` ` ` `frq[a[i]] = (` `int` `)frq[a[i]] + 1;` ` ` `// Counting the maximum frequency` ` ` `int` `maxfrq = ` `int` `.MinValue;` ` ` `for` `(` `int` `i = 0; i < frq.Count; i++)` ` ` `{` ` ` `if` `(maxfrq < (` `int` `)frq[i])` ` ` `{` ` ` `maxfrq = (` `int` `)frq[i];` ` ` `}` ` ` `}` ` ` `// Counting total distinct elements` ` ` `int` `dist = n + 1;` ` ` `for` `(` `int` `i = 0; i < frq.Count; i++)` ` ` `{` ` ` `if` `((` `int` `)frq[i] == 0)` ` ` `{` ` ` `dist--;` ` ` `}` ` ` `}` ` ` `int` `ans1 = Math.Min(maxfrq - 1, dist);` ` ` `int` `ans2 = Math.Min(maxfrq, dist - 1);` ` ` `// Find max of both the answer` ` ` `int` `ans = Math.Max(ans1, ans2);` ` ` `return` `ans;` `}` ` ` `// Driver code` `public` `static` `void` `Main(` `string` `[] args)` `{` ` ` `int` `[]arr = {4, 2, 4, 1, 4, 3, 4};` ` ` `int` `n = arr.Length;` ` ` `Console.Write(findMaxSize(arr, n));` `}` `}` `// This code is contributed by Rutvik_56` |

## Javascript

`<script>` `// Javascript implementation to Divide the array` `// into two arrays having same size,` `// one with distinct elements` `// and other with same elements` `// Function to find the max size possible` `function` `findMaxSize(a, n)` `{` ` ` `var` `frq = Array(n + 1).fill(0);` ` ` `for` `(` `var` `i = 0; i < n; ++i)` ` ` `frq[a[i]]++;` ` ` `// Counting the maximum frequency` ` ` `var` `maxfrq = frq.reduce((a,b)=>Math.max(a,b))` ` ` `// Counting total distinct elements` ` ` `var` `dist` ` ` `= n + 1 - frq.filter(x=> x ==0).length;` ` ` `var` `ans1 = Math.min(maxfrq - 1, dist);` ` ` `var` `ans2 = Math.min(maxfrq, dist - 1);` ` ` `// Find max of both the answer` ` ` `var` `ans = Math.max(ans1, ans2);` ` ` `return` `ans;` `}` `// Driver code` `var` `arr = [4, 2, 4, 1, 4, 3, 4];` `var` `n = arr.length;` `document.write( findMaxSize(arr, n));` `</script>` |

**Output:**

3

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.