# Sum of division of the possible pairs for the given Array

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation to compute the ` `// sum of division of all the possible ` `// pairs for the given array ` ` `  `#include ` `#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 = { 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); ` ` `  `    ``cout << func(arr, n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// 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 `

## Python3

 `# Python3 program to compute the sum ` `# of divison 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 `

## C#

 `// 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``; ` `    ``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 `

Output:

```12
```

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

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Improved By : amit143katiyar, himanshu77