Triple sum


Problem Statement :


Given 3 arrays a, b , c of different sizes, find the number of distinct triplets ( p , q , r )  where p is an element of a, written as , , and , satisfying the criteria: p <= q and q > = r .


Function Description

Complete the triplets function in the editor below. It must return the number of distinct triplets that can be formed from the given arrays.

triplets has the following parameter(s):

a, b, c: three arrays of integers .
Input Format

The first line contains 3 integers lena, lenb, lenc , the sizes of the three arrays.
The next 3 lines contain space-separated integers numbering lena, lenb, lenc  respectively.


constraints

1  <=   lena, lenb , lenc <= 10^5
1  <=  all elements in a, b , c  <= 10^8



Output Format

Print an integer representing the number of distinct triplets.


Sample Input 0

3 2 3
1 3 5
2 3
1 2 3
Sample Output 0

8


Solution :



title-img 




                        Solution in C++ :

In  C ++ :







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

int main()
{
    int p,q,r,i,x,y;
    cin>>p>>q>>r;

    int a[p],b[q],c[r];
    for(i=0;i<p;++i) {
        cin>>a[i];
    }
    for(i=0;i<q;++i) {
        cin>>b[i];
    }
    for(i=0;i<r;++i) {
        cin>>c[i];
    }
    sort(a,a+p);
    sort(b,b+q);
    sort(c,c+r);

    long distinct_a[p],distinct_b[q],distinct_c[r];
    set<int> s;
    for(i=0;i<p;++i) {
        s.insert(a[i]);
        distinct_a[i]=s.size();
    }
    s.clear();
    for(i=0;i<q;++i) {
        s.insert(b[i]);
        distinct_b[i]=s.size();
    }
    s.clear();
    for(i=0;i<r;++i) {
        s.insert(c[i]);
        distinct_c[i]=s.size();
    }

    long ans=0;

    x = upper_bound(a,a+p,b[0])-a;
    y = upper_bound(c,c+r,b[0])-c;

    x-=1,y-=1;
    if(x>=0 && y>=0) {
        ans += distinct_a[x]*distinct_c[y];
    }

    for(i=1;i<q;++i) {
        if(b[i]!=b[i-1]) {
            x = upper_bound(a,a+p,b[i])-a;
            y = upper_bound(c,c+r,b[i])-c;

            x-=1,y-=1;
            if(x>=0 && y>=0) {
                ans += distinct_a[x]*distinct_c[y];
            }
        }
    }
    cout<<ans<<endl;

    return 0;
}
                    


                        Solution in Java :

In    Java :






static long triplets(int[] a, int[] b, int[] c) {

        Arrays.sort(a);
        a=Arrays.stream(a).distinct().toArray();
        Arrays.sort(b);
        b=Arrays.stream(b).distinct().toArray();
        Arrays.sort(c);
        c=Arrays.stream(c).distinct().toArray();
        int aIndex = 0;
        int bIndex = 0;
        int cIndex = 0;
        long result = 0;

        while(bIndex<b.length){
            while(aIndex<a.length && a[aIndex]<=b[bIndex])
                aIndex++;

            while(cIndex<c.length && c[cIndex]<=b[bIndex])
                cIndex++;
						
            // you should convert int to long first, 
            // avoid int overflow and lead to fail the test case 4
            result += ((long)aIndex)*((long)cIndex);
            bIndex++;
        }
        return result;
    }
                    


                        Solution in Python : 
                            
In    Python3 :




    a = list(sorted(set(a)))
    b = list(sorted(set(b)))
    c = list(sorted(set(c)))
    
    ai = 0
    bi = 0
    ci = 0
    
    ans = 0
    
    while bi < len(b):
        while ai < len(a) and a[ai] <= b[bi]:
            ai += 1
        
        while ci < len(c) and c[ci] <= b[bi]:
            ci += 1
        
        ans += ai * ci
        bi += 1
    
    return ans


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