Sum of special triplets having elements from 3 arrays

Given three arrays A, B and C, the task is to find sum of values of all special triplets. A special triplet is defined as a triplet (X, Y, Z) where the condition :
X ≤ Y and Z ≤ Y always holds true. The value of each triplet (X, Y, Z) is given by:

f(X, Y, Z) = (X + Y) * (Y + Z)

Note: If a triplet is not ‘special’,
f(x, y, z) = 0 for that particular triplet.

Examples:

Input : A = {1, 4, 5}, B = {2, 3}, C = {2, 1, 3}
Output : 81
Explanation
The special triplets and their values are given below
Triplet      f(x, y, z) = (x + y) * (y + z)
(1, 2, 2)         (1 + 2) * (2 + 2)  = 12
(1, 2, 1)         (1 + 2) * (2 + 1)  =  9
(1, 3, 2)         (1 + 3) * (3 + 2)  = 20
(1, 3, 1)         (1 + 3) * (3 + 1)  = 16
(1, 3, 3)         (1 + 3) * (3 + 3)  = 24
-------------------------------------
                           Sum = 81

Method 1 (Brute Force) We generate all triplets and check if a triplet is a special triplet, we calculate the value of the triplet using f(x, y, z) where (x, y, z) is a special triplet, and add it to the final sum of all such special triplets

C++

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// C++ Program to find sum of values of 
// all special triplets
#include <bits/stdc++.h>
using namespace std;
  
/* Finding special triplets (x, y, z) where
   x belongs to A; y belongs to B and z 
   belongs to C; p, q and r are size of 
   A, B and C respectively */
int findSplTripletsSum(int a[], int b[], int c[], 
                             int p, int q, int r)
{
  
    int sum = 0;
    for (int i = 0; i < p; i++) {
        for (int j = 0; j < q; j++) {
            for (int k = 0; k < r; k++) {
  
                // (a[i], b[j], c[k]) is special if
                // a[i] <= b[j] and c[k] <= b[j];
                if (a[i] <= b[j] && c[k] <= b[j]) {
  
                    // calculate the value of this special
                    // triplet and add sum of all values 
                    // of such triplets
                    sum +=  (a[i] + b[j]) * (b[j] + c[k]);
                }
            }
        }
    }
    return sum;
}
  
// Driver Code
int main()
{
    int A[] = { 1, 4, 5 };
    int B[] = { 2, 3 };
    int C[] = { 2, 1, 3 };
  
    int p = sizeof(A) / sizeof(A[0]);
    int q = sizeof(B) / sizeof(B[0]);
    int r = sizeof(C) / sizeof(C[0]);
  
    cout << "Sum of values of all special triplets = "
         << findSplTripletsSum(A, B, C, p, q, r) << endl;
}

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Python3

# Python3 Program to find sum of values of
# all special triplets

# Finding special triplets (x, y, z) where
# x belongs to A y belongs to B and z
# belongs to C p, q and r are size of
# A, B and C respectively
def findSplTripletsSum(a, b, c, p, q, r):
summ = 0
for i in range(p):
for j in range(q):
for k in range(r):

# (a[i], b[j], c[k]) is special if
# a[i] <= b[j] and c[k] <= b[j] if (a[i] <= b[j] and c[k] <= b[j]): # calculate the value of this special # triplet and add sum of all values # of such triplets summ += (a[i] + b[j]) * (b[j] + c[k]) return summ # Driver Code A = [1, 4, 5 ] B = [2, 3 ] C = [2, 1, 3 ] p = len(A) q = len(B) r = len(C) print("Sum of values of all special triplets = ", findSplTripletsSum(A, B, C, p, q, r)) # This code is contributed by Mohit kumar 29 [tabbyending]

Output:

Sum of values of all special triplets = 81

Time Complexity of this approach is O(P * Q * R) where P, Q and R are the sizes of the three arrays A, B and C respectively.

Method 2 (Efficient)

Suppose,
Array A contains elements {a, b, c, d, e},
Array B contains elements {f, g, h, i} and
Array C contains elements {j, k, l, m}.

First of all we sort the arrays A and C so that we are able to find the number of elements in arrays A and C which are less than a particular Bi which can be done by applying binary search on each value of Bi.

Lets suppose that at a particular index i, the element of array B is Bi. Lets also suppose that after we are done sorting A and C, we have elements {a, b, c} belonging to array A which are less than or equal to Bi and elements {j, k} belonging to array C which are also less than Bi.

Lets take Bi = Y from here on.

Let, Total Sum of values of all special triplets = S 
We Know S = Σ f(x, y, z) for all possible (x, y, z)

Since elements {a, b, c} of Array A and elements {j, k} of array C are less than Y,
the Special Triplets formed consists of triplets formed only using these elements 
with Y always being the second element of every possible triplet

All the Special Triplets and their corresponding values are shown below:

Triplet   f(x, y, z) = (x + y) * (y + z)
(a, Y, j)         (a + Y)(Y + j)  
(a, Y, k)         (a + Y)(Y + k)  
(b, Y, j)         (b + Y)(Y + j)  
(b, Y, k)         (b + Y)(Y + k)  
(c, Y, j)         (c + Y)(Y + j)  
(c, Y, k)         (c + Y)(Y + k)

The sum of these triplets is
S = (a + Y)(Y + j) + (a + Y)(Y + k) + (b + Y)(Y + j) + (b + Y)(Y + k) 
    + (c + Y)(Y + j) + (c + Y)(Y + k)

Taking (a + X), (b + X) and (c + x) as common terms we have,

S = (a + Y)(Y + j + Y + k) + (b + Y)(Y + j + Y + k) + (c + Y)(Y + j + Y + k)
Taking (2Y + j + k) common from every term,
S = (a + Y + b + Y + c + Y)(2Y + j + k)  

∴ S = (3Y + a + b + c)(2Y + j + k)

Thus,
S = (N * Y + S1) * (M * Y + S2) 
where, 
N = Number of elements in A less than Y,
M = Number of elements in C less than Y,
S1 = Sum of elements in A less than Y and
S2 = Sum of elements in C less than Y

So for every element in B we can find the number of elements less than it in arrays A and C using Binary Search and the sum of these elements can be found using prefix sums

C++

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// C++ Program to find sum of values
// of all special triplets
#include <bits/stdc++.h>
using namespace std;
  
/* Utility function for findSplTripletsSum() 
finds total sum of values of all special 
triplets */
int findSplTripletsSumUtil(int A[], int B[], int C[],
                   int prefixSumA[], int prefixSumC[], 
                                 int p, int q, int r)
{
  
    int totalSum = 0;
  
    // Traverse through whole array B
    for (int i = 0; i < q; i++) {
  
        // store current element Bi
        int currentElement = B[i];
  
        // n = number of elements in A less than current
        // element
        int n = upper_bound(A, A + p, currentElement) - A;
  
        // m = number of elements in C less than current 
        // element
        int m = upper_bound(C, C + r, currentElement) - C;
  
        // if there are Elements neither in A nor C which 
        // are less than or equal to the current element
        if (n == 0 || m == 0)
            continue;
  
        /* total sum = (n * currentElement + sum of first
           n elements in A) + (m * currentElement + sum of 
           first m elements in C) */
        totalSum += 
            ((prefixSumA[n - 1] + (n * currentElement)) *
              (prefixSumC[m - 1] + (m * currentElement)));
    }
  
    return totalSum;
}
  
/* Builds prefix sum array for arr of size n
and returns a pointer to it */
int* buildPrefixSum(int* arr, int n)
{
  
    // Dynamically allocate memory tp Prefix Sum Array
    int* prefixSumArr = new int[n];   
  
    // building the prefix sum
    prefixSumArr[0] = arr[0];
    for (int i = 1; i < n; i++) 
        prefixSumArr[i] = prefixSumArr[i - 1] + arr[i];    
  
    return prefixSumArr;
}
  
/* Wrapper for Finding special triplets (x, y, z) 
   where x belongs to A; y belongs to B and z 
   belongs to C; p, q and r are size of 
   A, B and C respectively */
int findSplTripletsSum(int A[], int B[], int C[],
                             int p, int q, int r)
{
  
    int specialTripletSum = 0;
  
    // sort arrays A and C
    sort(A, A + p);
    sort(C, C + r);
  
    // build prefix arrays for A and C
    int* prefixSumA = buildPrefixSum(A, p);
    int* prefixSumC = buildPrefixSum(C, r);
  
    return findSplTripletsSumUtil(A, B, C, 
           prefixSumA, prefixSumC, p, q, r);
}
  
// Driver Code
int main()
{
    int A[] = { 1, 4, 5 };
    int B[] = { 2, 3 };
    int C[] = { 2, 1, 3 };
    int p = sizeof(A) / sizeof(A[0]);
    int q = sizeof(B) / sizeof(B[0]);
    int r = sizeof(C) / sizeof(C[0]);
  
    cout << "Sum of values of all special triplets = "
         << findSplTripletsSum(A, B, C, p, q, r);
}

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Java

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// Java Program to find sum of values of 
// all special triplets
import java.io.*;
import java.util.*;
  
public class GFG {
      
    /* Finding special triplets (x, y, z)
    where x belongs to A; y belongs to B
    and z belongs to C; p, q and r are 
    size of A, B and C respectively */
    static int findSplTripletsSum(int []a, 
                  int []b, int []c, int p,
                             int q, int r)
    {
        int sum = 0;
        for (int i = 0; i < p; i++) {
            for (int j = 0; j < q; j++) {
                for (int k = 0; k < r; k++)
                {
      
                    // (a[i], b[j], c[k]) is
                    // special if a[i] <= b[j]
                    // and c[k] <= b[j];
                    if (a[i] <= b[j] && 
                                 c[k] <= b[j])
                    {
      
                        // calculate the value
                        // of this special
                        // triplet and add sum
                        // of all values 
                        // of such triplets
                        sum += (a[i] + b[j]) 
                               * (b[j] + c[k]);
                    }
                }
            }
        }
        return sum;
    }
      
    // Driver Code
    public static void main(String args[])
    {
        int []A = { 1, 4, 5 };
        int []B = { 2, 3 };
        int []C = { 2, 1, 3 };
      
        int p = A.length;
        int q = B.length;
        int r = C.length;
      
        System.out.print("Sum of values of all"
                       + " special triplets = "
        + findSplTripletsSum(A, B, C, p, q, r));
    }
}
  
// This code is contributed by Manish Shaw
// (manishshaw1)

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C#

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// C# Program to find sum of values of 
// all special triplets
using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
      
    /* Finding special triplets (x, y, z) where
       x belongs to A; y belongs to B and z 
       belongs to C; p, q and r are size of 
       A, B and C respectively */
    static int findSplTripletsSum(int []a, int []b, int []c, 
                                       int p, int q, int r)
    {
        int sum = 0;
        for (int i = 0; i < p; i++) {
            for (int j = 0; j < q; j++) {
                for (int k = 0; k < r; k++) {
      
                    // (a[i], b[j], c[k]) is special if
                    // a[i] <= b[j] and c[k] <= b[j];
                    if (a[i] <= b[j] && c[k] <= b[j]) {
      
                        // calculate the value of this special
                        // triplet and add sum of all values 
                        // of such triplets
                        sum += (a[i] + b[j]) * (b[j] + c[k]);
                    }
                }
            }
        }
        return sum;
    }
      
    // Driver Code
    public static void Main()
    {
        int []A = { 1, 4, 5 };
        int []B = { 2, 3 };
        int []C = { 2, 1, 3 };
      
        int p = A.Length;
        int q = B.Length;
        int r = C.Length;
      
        Console.WriteLine("Sum of values of all special triplets = "
                            + findSplTripletsSum(A, B, C, p, q, r));
    }
}
  
// This code is contributed by 
// Manish Shaw (manishshaw1)

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Output:

Sum of values of all special triplets = 81

Since we need to iterate through the entire array B and for every element apply binary searches in array A and C, the time Complexity of this approach is O(Q * (logP + logR)) where P, Q and R are the sizes of the three arrays A, B and C respectively.



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