Maximum number of parallelograms that can be made using the given length of line segments in C++

Given the task is to find the maximum number of parallelograms that can be made using given N number of line segments if each line segment can be used at most in one parallelogram.

Let’s now understand what we have to do using an example −

Input − Arr[] = {8, 3, 1, 3, 8, 7, 1, 3, 5, 3}

Output − 2

Explanation − With the above given line segments, the two parallelograms that can be formed are with sides 8, 1, 8, 1 and 3, 3, 3, 3 respectively.

Input − Arr[] = {7, 9, 9, 7}

Output − 1

Approach used in the below program as follows

• The maximum number of parallelograms that can be made will be = Parallelograms that can be made with 4 equal or similar sides plus parallelograms that can be made using 2 similar sides.

• In function MaxParr(), initialize a variable L = Arr[0] which will be used as the size of the array that will be used to store the frequencies of the line segments.

• Loop from i=1 till i<N and check if (Arr[i] > L), and inside the if statement put L=Arr[i]. Outside the loop increase the size of L by 1.

• Then initialize the frequency array int Freq[L] = {0}. Loop from i=0 till i<N and increase occurrence of each segment by 1.

• The initialize count = 0 of type int to store the final count of parallelograms.

• Loop from i=0 till i<L and check for parallelograms that can be made with 4 similar sides and if found then increase the value of count accordingly.

• Initialize left=0 of type int to store the number of parallelograms that can be formed using 2 similar sides.

• Finally loop from i=0 till i<L and check if(Freq[i] >= 2) and if so then add 1 to left.

• Put count+= left/2 and return count.

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; int MaxParr(int N, int Arr[]){    //Finding length of frequency array    int L = Arr[0];    for (int i = 1; i < N; i++){       if (Arr[i] > L)          L = Arr[i];    }    L = L + 1;    int Freq[L] = {0};    for (int i = 0; i < N; i++){       //Increasing occurrence of each line segment       Freq[Arr[i]] += 1;    }    // To store the number of parallelograms    int count = 0;    for (int i = 0; i < L; i++){       /*parallelograms that can be made using 4 same sides*/       count += int(Freq[i] / 4);       Freq[i] = Freq[i] % 4;    }    int left = 0;    for (int i = 0; i < L; i++){       //Counting segments with 2 or more occurrences left       if (Freq[i] >= 2)          left += 1;    }    /*Adding parallelograms that can be formed using using 2 similar sides into the final count*/    count += left / 2;    return count; } int main(){    int N = 10;    int Arr[] = { 8, 3, 1, 3, 8, 7, 1, 3, 5, 3};    cout<< MaxParr(N, Arr); }

Output

If we run the above code we will get the following output −

2