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 <bits/stdc++.h> 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<Integer> arr)
{
int n = arr.size();
// count the number of segments required
int count = 1 ;
// For counting the frequency
HashMap<Integer, Integer> map
= new HashMap<Integer, Integer>();
// 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<Integer> arr = new ArrayList<Integer>();
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)