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

Last Updated : 20 May, 2021

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=5 ; Array: 2, 2, 3, 3, 3, 5
Output : 3

Approach:

The optimal approach is to put all the distinct elements in a single segment.
And then, put all other elements that have several occurrences one by one in new segments such that no segment contains repetitions of elements.
So, the answer is the maximum frequency of any element within the given array.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that counts the
// minimum segments required
void CountSegments(int N, int a[])
{
    // all values are '0' initially
    int frequency[10001] = { 0 };
 
    // count of segments
    int c = 0;
 
    // store frequency of every element
    for (int i = 0; i < N; i++) {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (int i = 0; i <= 10000; i++)
        c = max(c, frequency[i]);
 
    cout << c << "\n";
}
 
// Driver code
int main()
{
    int N = 6;
    int a[] = { 1, 3, 4, 3, 2, 3 };
 
    CountSegments(N, a);
 
    return 0;
}

Java

// Java implementation of the approach
 
import java.io.*;
 
class GFG {
 
// Function that counts the
// minimum segments required
static void CountSegments(int N, int a[])
{
    // all values are '0' initially
    int frequency[] =  new int[10001];
 
    // count of segments
    int c = 0;
 
    // store frequency of every element
    for (int i = 0; i < N; i++) {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (int i = 0; i <= 10000; i++)
        c = Math.max(c, frequency[i]);
 
    System.out.println( c);
}
 
       // Driver code
    public static void main (String[] args) {
        int N = 6;
    int []a = { 1, 3, 4, 3, 2, 3 };
 
    CountSegments(N, a);
    }
}
 
// This Code is contributed by inder_verma..

Python 3

# Python3 implementation of the approach
 
# Function that counts the
# minimum segments required
def CountSegments(N, a):
     
    # all values are '0' initially
    frequency = [0] * 10001
 
    # count of segments
    c = 0
 
    # store frequency of every element
    for i in range(N) :
        frequency[a[i]] += 1
 
    # find maximum frequency
    for i in range(10001):
        c = max(c, frequency[i])
 
    print(c)
 
# Driver code
if __name__ == "__main__":
    N = 6
    a = [ 1, 3, 4, 3, 2, 3 ]
    CountSegments(N, a)
 
# This code is contributed
# by ChitraNayal

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that counts the
// minimum segments required
static void CountSegments(int N, int []a)
{
    // all values are '0' initially
    int []frequency = new int[10001];
 
    // count of segments
    int c = 0;
 
    // store frequency of every element
    for (int i = 0; i < N; i++)
    {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (int i = 0; i <= 10000; i++)
        c = Math.Max(c, frequency[i]);
 
    Console.WriteLine( c);
}
 
// Driver code
public static void Main ()
{
    int N = 6;
    int []a = { 1, 3, 4, 3, 2, 3 };
 
    CountSegments(N, a);
}
}
 
// This code is contributed
// by inder_verma

Javascript

<script>
 
// js implementation of the approach
 
// Function that counts the
// minimum segments required
function CountSegments(N,a)
{
    // all values are '0' initially
    var frequency = new Array(10001).fill(0);
 
    // count of segments
    let c = 0;
 
    // store frequency of every element
    for (var i = 0; i < N; i++)
    {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (var i = 0; i <= 10000; i++)
        c = Math.max(c, frequency[i]);
 
    document.write(c);
}
 
// Driver code
 
var N = 6;
var a = [1, 3, 4, 3, 2, 3 ];
 
CountSegments(N, a);
 
 
</script>

Output:

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