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

Input: a[] = { 2, 1, 5, 4, 3 }
Output: 1
Explanation:
The maximum possible size would be 1 – {1} and {2}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

• At 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 one 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:
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 dist – 1, so the other array can only contain dist – 1 number of elements. In this case the answer would be, ans1 = min(dist – 1, maxfrq)
2. If we take all the elements having frequency maxfrq except one element in one array(that contains same elements), then the total number of distinct elements left would be dist, so the other array can only contain dist number of elements. In this casethe answer would be, ans2 = min(dist, maxfrq – 1).
• Hence, the actual answer is

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

.

Below is the implementation of above approach:

 `// C++ implementation to Divide the array ` `// into two arrays having same size, ` `// one with distinct elements ` `// and other with same elements ` ` `  `#include ` `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 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 frq = ``new` `ArrayList(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 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 `

Output:
```3
```

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Improved By : coder001, divyamohan123

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