Common_elements.cpp

/*
Problem Statement:
------------------
Given three arrays sorted in increasing order. Find the elements that are common in all three arrays.
Note: can you take care of the duplicates without using any additional Data Structure?

Example 1:
---------
Input:
n1 = 6; A = {1, 5, 10, 20, 40, 80}
n2 = 5; B = {6, 7, 20, 80, 100}
n3 = 8; C = {3, 4, 15, 20, 30, 70, 80, 120}
Output: 20 80

Explanation: 20 and 80 are the only common elements in A, B and C.

Your Task: You don't need to read input or print anything. Your task is to complete the function commonElements() which take the 3 arrays A[], B[], C[] 
and their respective sizes n1, n2 and n3 as inputs and returns an array containing the common element present in all the 3 arrays in sorted order. 
If there are no such elements return an empty array. In this case the output will be printed as -1.

Expected Time Complexity: O(n1 + n2 + n3)
Expected Auxiliary Space: O(n1 + n2 + n3)
*/

// Link --> https://practice.geeksforgeeks.org/problems/common-elements1132/1

// Code:
class Solution
{
public: 
    vector <int> result;
    vector <int> commonElements(int A[] , int B[] , int C[] , int n1 , int n2 , int n3)
    {
        result.clear();
        int i=0 , j=0 , k=0;
        int last = INT_MIN;
        while(i < n1 and j < n2 and k < n3)
        {
            if(A[i] == B[j] && A[i] == C[k] && A[i] != last) 
            {
                result.push_back(A[i]);
                last = A[i];
                i++;
                j++;
                k++;
            }
            
            else if(min ({A[i], B[j], C[k]}) == A[i]) 
                i++;
                
            else if(min ({A[i], B[j], C[k]}) == B[j]) 
                j++;
                
            else 
                k++;
        }
        return result;
    }
};