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Sum of division of the possible pairs for the given Array
  • Last Updated : 26 Aug, 2020

Given an array arr[] of N positive integers. For all the possible pairs (x, y) the task is to find the summation of x/y.

Note: If decimal part of (x/y) is &ge 0.5 then add ceil of (x/y), else add floor of (x/y).

Examples:

Input: arr[] = {1, 2, 3}
Output: 12
Explanation:
All possible pairs with division are:
(1/1) = 1, (1/2) = 1, (1/3) = 0
(2/1) = 2, (2/2) = 1, (2/3) = 1
(3/1) = 3, (3/2) = 2, (3/3) = 1
Sum = 1 + 1 + 0 + 2 + 1 + 1 + 3 + 2 + 1 = 12.

Input: arr[] = {1, 2, 3, 4}
Output: 22
Explanation:
All possible pairs with division are:
(1/1) = 1, (1/2) = 1, (1/3) = 0, (1/4) = 0
(2/1) = 2, (2/2) = 1, (2/3) = 1, (2/4) = 1
(3/1) = 3, (3/2) = 2, (3/3) = 1, (3/4) = 1
(4/1) = 4, (4/2) = 2, (4/3) = 1, (4/4) = 1
Sum = 1 + 1 + 0 + 0 + 2 + 1 + 1 + 1 + 3 + 2 + 1 + 1 + 4 + 2 + 1 + 1 = 22.



Naive Approach: The idea is to generate all the possible pairs in the given array and find the summation of (x/y) for each pair (x, y).

Time Complexity: O(N2)

Efficient Approach:

To optimize the above method we have to compute the frequency array where freq[i] denotes the number of occurrences of number i.

  • For any given number X, all the numbers ranging from [0.5X, 1.5X] would result in contributing 1 to the answer when divided by X. Similarly all the numbers ranging from [1.5X, 2.5X] would result in contributing 2 to the answer when divided by X.
  • Generalizing this fact all the numbers ranging from [(n-0.5)X, (n+0.5)X] would result in contributing n to the answer when divided by X.
  • Thus for every number P in the range 1 to N we can get the count of the numbers which lie in the given range [L, R] by just computing a prefix sum of frequency array.
  • For a number P we need to query on the ranges at most N/P times.

Below is the implementation of the above approach:

C++

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// C++ implementation to compute the
// sum of division of all the possible
// pairs for the given array
  
#include <bits/stdc++.h>
#define ll long long
using namespace std;
  
// Function to compute the sum
int func(int arr[], int n)
{
  
    double ans = 0;
    int maxx = 0;
    double freq[100005] = { 0 };
    int temp;
  
    // counting frequency
    // of each term
    // and finding maximum
    // among it
    for (int i = 0; i < n; i++) {
        temp = arr[i];
        freq[temp]++;
        maxx = max(maxx, temp);
    }
  
    // Making cumulative frequency
    for (int i = 1; i <= maxx; i++) {
        freq[i] += freq[i - 1];
    }
  
    for (int i = 1; i <= maxx; i++) {
        if (freq[i]) {
            i = (double)i;
            double j;
            ll value = 0;
  
            // Taking the ceil value
            double cur = ceil(0.5 * i) - 1.0;
  
            for (j = 1.5;; j++) {
                int val = min(maxx, (int)(ceil(i * j) - 1.0));
                int times = (freq[i] - freq[i - 1]), con = j - 0.5;
  
                // nos. in [(n-0.5)X, (n+0.5)X)
                // range will add n to the ans
  
                ans += times * con * (freq[(int)val] - freq[(int)cur]);
                cur = val;
  
                if (val == maxx)
                    break;
            }
        }
    }
  
    // Return the final result
    return (ll)ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << func(arr, n) << endl;
  
    return 0;
}

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Java

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// Java implementation to compute the
// sum of division of all the possible
// pairs for the given array
class GFG{
  
// Function to compute the sum
static long func(int arr[], int n)
{
    double ans = 0;
    int maxx = 0;
    double freq[] = new double[100005];
    int temp;
  
    // Counting frequency of each term
    // and finding maximum among it
    for(int i = 0; i < n; i++)
    {
       temp = arr[i];
       freq[temp]++;
       maxx = Math.max(maxx, temp);
    }
  
    // Making cumulative frequency
    for(int i = 1; i <= maxx; i++)
    {
       freq[i] += freq[i - 1];
    }
  
    for(int i = 1; i <= maxx; i++)
    {
       if (freq[i] != 0)
       {
           double j;
             
           // Taking the ceil value
           double cur = Math.ceil(0.5 * i) - 1.0;
             
           for(j = 1.5;; j++) 
           {
              int val = Math.min(maxx, 
                  (int)(Math.ceil(i * j) - 1.0));
              int times = (int)(freq[i] - 
                                freq[i - 1]), 
                    con = (int)(j - 0.5);
                
              // nos. in [(n-0.5)X, (n+0.5)X)
              // range will add n to the ans
              ans += times * con * (freq[(int)val] - 
                                    freq[(int)cur]);
              cur = val;
               
              if (val == maxx)
                  break;
           }
       }
    }
      
    // Return the final result
    return (long)ans;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
  
    System.out.print(func(arr, n) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

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Python3

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# Python3 program to compute the sum
# of division of all the possible
# pairs for the given array
from math import *
  
# Function to compute the sum
def func (arr, n):
  
    ans = 0
    maxx = 0
    freq = [0] * 100005
    temp = 0
  
    # Counting frequency of each term
    # and finding maximum among it
    for i in range(n):
        temp = arr[i]
        freq[temp] += 1
        maxx = max(maxx, temp)
  
    # Making cumulative frequency
    for i in range(1, maxx + 1):
        freq[i] += freq[i - 1]
  
    for i in range(1, maxx + 1):
        if (freq[i]):
            value = 0
  
            # Taking the ceil value
            cur = ceil(0.5 * i) - 1.0
  
            j = 1.5
            while (1):
                val = min(maxx, (ceil(i * j) - 1.0))
                times = (freq[i] - freq[i - 1])
                con = j - 0.5
  
                # nos. in [(n-0.5)X , (n+0.5)X)
                # range will add n to the ans
                ans += times * con * (freq[int(val)] - 
                                      freq[int(cur)])
                cur = val
  
                if (val == maxx):
                    break
                j += 1
  
    return int(ans)
  
# Driver code
if __name__ == '__main__':
  
    arr = [ 1, 2, 3 ]
    n = len(arr)
  
    print(func(arr, n))
  
# This code is contributed by himanshu77

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C#

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// C# implementation to compute the
// sum of division of all the possible
// pairs for the given array
using System;
  
class GFG{
  
// Function to compute the sum
static long func(int []arr, int n)
{
    double ans = 0;
    int maxx = 0;
    double []freq = new double[100005];
    int temp;
  
    // Counting frequency of each term
    // and finding maximum among it
    for(int i = 0; i < n; i++)
    {
       temp = arr[i];
       freq[temp]++;
       maxx = Math.Max(maxx, temp);
    }
  
    // Making cumulative frequency
    for(int i = 1; i <= maxx; i++)
    {
       freq[i] += freq[i - 1];
    }
  
    for(int i = 1; i <= maxx; i++)
    {
       if (freq[i] != 0)
       {
           double j;
             
           // Taking the ceil value
           double cur = Math.Ceiling(0.5 * i) - 1.0;
             
           for(j = 1.5;; j++) 
           {
              int val = Math.Min(maxx, 
                  (int)(Math.Ceiling(i * j) - 1.0));
              int times = (int)(freq[i] - 
                                freq[i - 1]), 
                    con = (int)(j - 0.5);
                      
              // nos. in [(n-0.5)X, (n+0.5)X)
              // range will add n to the ans
              ans += times * con * (freq[(int)val] - 
                                    freq[(int)cur]);
              cur = val;
                
              if (val == maxx)
                  break;
           }
       }
    }
      
    // Return the readonly result
    return (long)ans;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
  
    Console.Write(func(arr, n) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

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Output:

12

Time Complexity: O(N * log (N) )

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