Cognizant Genc Next coding assessment(Help me in solving this question)

1

Picnic:
A school is organizing a picnic for all its students. There is a total of N students labeled from 1 to N in the school. Each student i has a compatibility factor of Xi

It is time for the picnic and all students to stand in a line. The student line is to be split into groups. A set of consecutive students standing in a line can form a group. For the picnic to be safe, each group must have at least 2 students with the same compatibility factor:

Finding the maximum number of groups that can be created.

Input Specification:

input1: N, denoting the number of students.

input2: An array of N elements where the ith element denotes the

compatibility factor of th student.

Output Specification:

Your function should return the maximum number of groups that can be

created.

Example 1:

Input 1: 2
Input 2:{1,1}
Output: 1

Explanation:

Only 1 group exists which consists of first 2 students.

Example 2:

input1: 8
input2: (1,2,1,1,1,1,1,1)

Output: 3

Explanation:

The following 3 groups can be formed

(1,2,1)
(1,1)
(1,1,1)

2

You can check the solution here

3

Extremely simple code, such a solution is very light, so I don’t think anyone will need it

4

Can I get the implementation in Java?

2 Likes
5

#include <bits/stdc++.h>
using namespace std;

int main()
{
long long n, val ,ans = 0;

cin >> n;
unordered_map<int, int> m;

for (int i = 0; i < n; i++){
    cin >> val;
    m[val]++;
    if( m[val] == 2 ) {
        ans++;
        m.clear();
    }
}
cout << ans << endl;

}

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int count=0;
HashSet set=new HashSet();
for (int x:
arr) {
if(set.contains(x)){
count++;
set.clear();
}else{
set.add(x);
}
}
System.out.println(count);