# Minimum number of segments required such that each segment has distinct elements

Given an array of integers, the task is to find the minimum number of segments that the array elements can be divided into such that all the segments contain distinct elements.

Examples:

```Input: n = 6; Array: 1, 7, 4, 3, 3, 8
Output: 2
Explanation:
Optimal way to create segments here is {1, 7, 4, 3} {3, 8}
Clearly, the answer is the maximum frequency of any element within the array i.e. '2'.
as '3' is the element which appears the most in the array (twice).

Input : n = 6; Array: 2, 2, 3, 3, 3, 5
Output : 4```

Below is the implementation of the above approach:

• Count the number of segments required
• For counting the frequency
• Iterate over the array
• Increment the frequency
• Check if there is any duplicate in current segment.
• Increment the segment required by 1.
• Clear the map
• Increment the frequency of current element in map.
• Return the count.

## C++

 `#include ``using` `namespace` `std;` `// Function to count the minimum minimum number of segments``// that the array elements can be divided into such that all``// the segments contain distinct elements.``int` `countSegment(vector<``int``>& arr)``{``    ``int` `n = arr.size();` `    ``// count the number of segment required``    ``int` `count = 1;` `    ``// For counting the frequency``    ``unordered_map<``int``, ``int``> unmap;` `    ``// Iterate over the array``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Increment the frequency``        ``unmap[arr[i]]++;` `        ``// Check if there is any duplicate in current``        ``// segment.``        ``if` `(unmap[arr[i]] > 1) {` `            ``// Increment the segment required by 1.``            ``count++;` `            ``// Clear the map``            ``unmap.clear();` `            ``// Increment the frequency of current element in``            ``// map``            ``unmap[arr[i]]++;``        ``}``    ``}` `    ``// Return the count.``    ``return` `count;``}` `int` `main()``{` `    ``// Input array.``    ``vector<``int``> arr = { 2, 2, 3, 3, 3, 5 };` `    ``// Function call and store the result``    ``int` `result = countSegment(arr);` `    ``// Print the result``    ``cout << result << endl;` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {` `    ``// Function to count the minimum minimum number of``    ``// segments that the array elements can be divided into``    ``// such that all the segments contain distinct elements.``    ``public` `static` `int` `countSegment(ArrayList arr)``    ``{``        ``int` `n = arr.size();` `        ``// count the number of segments required``        ``int` `count = ``1``;` `        ``// For counting the frequency``        ``HashMap map``            ``= ``new` `HashMap();` `        ``// Iterate over the array``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// Increment the frequency``            ``map.put(arr.get(i),``                    ``map.getOrDefault(arr.get(i), ``0``) + ``1``);` `            ``// Check if there is any duplicate in current``            ``// segment.``            ``if` `(map.get(arr.get(i)) > ``1``) {` `                ``// Increment the segment required by 1.``                ``count++;` `                ``// Clear the map``                ``map.clear();` `                ``// Increment the frequency of current``                ``// element in map``                ``map.put(arr.get(i), ``1``);``            ``}``        ``}` `        ``// Return the count.``        ``return` `count;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{` `        ``// Input array.``        ``ArrayList arr = ``new` `ArrayList();``        ``arr.add(``2``);``        ``arr.add(``2``);``        ``arr.add(``3``);``        ``arr.add(``3``);``        ``arr.add(``3``);``        ``arr.add(``5``);` `        ``// Function call and store the result``        ``int` `result = countSegment(arr);` `        ``// Print the result``        ``System.out.println(result);``    ``}``}`

## Python3

 `# Function to count the minimum number of segments that the array elements``# can be divided into such that all the segments contain distinct elements.`  `def` `countSegment(arr):``    ``n ``=` `len``(arr)` `    ``# Count the number of segments required``    ``count ``=` `1` `    ``# For counting the frequency``    ``unmap ``=` `{}` `    ``# Iterate over the array``    ``for` `i ``in` `range``(n):` `        ``# Increment the frequency``        ``if` `arr[i] ``not` `in` `unmap:``            ``unmap[arr[i]] ``=` `1``        ``else``:``            ``unmap[arr[i]] ``+``=` `1` `        ``# Check if there is any duplicate in current segment.``        ``if` `unmap[arr[i]] > ``1``:` `            ``# Increment the segment required by 1.``            ``count ``+``=` `1` `            ``# Clear the map``            ``unmap ``=` `{}` `            ``# Increment the frequency of current element in map``            ``unmap[arr[i]] ``=` `1` `    ``# Return the count.``    ``return` `count`  `# Input array``arr ``=` `[``2``, ``2``, ``3``, ``3``, ``3``, ``5``]` `# Function call and store the result``result ``=` `countSegment(arr)` `# Print the result``print``(result)`

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `Program {``    ``// Function to count the minimum minimum number of``    ``// segments that the array elements can be divided into``    ``// such that all the segments contain distinct elements.``    ``static` `int` `CountSegment(List<``int``> arr)``    ``{``        ``int` `n = arr.Count;` `        ``// count the number of segment required``        ``int` `count = 1;` `        ``// For counting the frequency``        ``Dictionary<``int``, ``int``> dict``            ``= ``new` `Dictionary<``int``, ``int``>();` `        ``// Iterate over the array``        ``for` `(``int` `i = 0; i < n; i++) {``            ``// Increment the frequency``            ``if` `(dict.ContainsKey(arr[i]))``                ``dict[arr[i]]++;``            ``else``                ``dict.Add(arr[i], 1);` `            ``// Check if there is any duplicate in current``            ``// segment.``            ``if` `(dict[arr[i]] > 1) {``                ``// Increment the segment required by 1.``                ``count++;` `                ``// Clear the map``                ``dict.Clear();` `                ``// Increment the frequency of current``                ``// element in map``                ``dict[arr[i]] = 1;``            ``}``        ``}` `        ``// Return the count.``        ``return` `count;``    ``}` `    ``static` `void` `Main(``string``[] args)``    ``{``        ``// Input array.``        ``List<``int``> arr = ``new` `List<``int``>{ 2, 2, 3, 3, 3, 5 };` `        ``// Function call and store the result``        ``int` `result = CountSegment(arr);` `        ``// Print the result``        ``Console.WriteLine(result);``    ``}``}`

## Javascript

 `// Function to count the minimum number of segments that the array elements can be divided into such that all the segments contain distinct elements.``function` `countSegment(arr) {``    ``const n = arr.length;` `    ``// count the number of segment required``    ``let count = 1;` `    ``// For counting the frequency``    ``const unmap = ``new` `Map();` `    ``// Iterate over the array``    ``for` `(let i = 0; i < n; i++) {``        ``// Increment the frequency``        ``if` `(unmap.has(arr[i])) {``            ``unmap.set(arr[i], unmap.get(arr[i]) + 1);``        ``} ``else` `{``            ``unmap.set(arr[i], 1);``        ``}` `        ``// Check if there is any duplicate in current segment.``        ``if` `(unmap.get(arr[i]) > 1) {``            ``// Increment the segment required by 1.``            ``count++;` `            ``// Clear the map``            ``unmap.clear();` `            ``// Increment the frequency of current element in map``            ``unmap.set(arr[i], 1);``        ``}``    ``}` `    ``// Return the count.``    ``return` `count;``}` `// Input array.``const arr = [2, 2, 3, 3, 3, 5];` `// Function call and store the result``const result = countSegment(arr);` `// Print the result``console.log(result);`

Output
```4
```

Time Complexity: O(n)
Auxiliary Space: O(n)

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