# Check whether for all pair (X, Y) of given Array floor of X/Y is also present

• Last Updated : 21 Apr, 2022

Given an array arr[] of size N and an integer K, denoting the maximum value an array element can have, the task is to check if for all possible pairs of elements (X, Y) where X≥Y, ⌊X/Y⌋ (X divided by Y with rounding down) is also present in this array.

Examples:

Input: N = 3, K = 5, arr[]={1, 2, 5}
Output: YES
Explanation: There are such pair of elements (checking all possible conditions)
⌊1/1⌋ = 1,  ⌊2/2⌋ = 1, ⌊5/5⌋ = 1
⌊2/1⌋ = 2, ⌊5/1⌋ = 5, ⌊5/2⌋ = 2
As all [X/Y] for any X and Y exists in the array. So print YES

Input: N = 4, K = 8, arr[] = {1, 3, 3, 7}
Output: NO
Explanation: As there is a pair (7, 3) where ⌊7/3⌋ = 2 is not present in the array.

Naive Approach: A naive approach for this problem is to use nested loops and take every value for X and Y and store in a resultant array. Then later on check every element of the resultant array is present or not in the original array.

Time Complexity: O(N*N)
Auxiliary Space: O(1)

Efficient Approach: This problem can be solved on the basis of the following idea:

For any number x all the values in range c*x + d [where c is a positive integer and d is in range [0, x) ] will have the same floor value c when divided by x

To implement the above observation, use a map to store the elements present in the array and a prefix array to store how many elements are present till x (x in the range [1, K]). For every such x check in its multiples if such a c (as mentioned in the observation ) exists in the array using the map and the prefix sum array.

Follow the below steps to solve this problem:

• Use a hash array (say b[]) to store the elements present in the array and another prefix sum array(say a[]) as mentioned above.
• Iterate in the array b[] for i > 0 to K:
• If i is present in array arr[]:
• Check for multiples of i (say x) if the floor value of x/i is also present in the array using b[](because of the reason mentioned in observation).
• If not present then check from the prefix sum array if any element was present in the range (x-i, x]
• If present then the condition for i and that element is not satisfied. So return false. Otherwise, continue iteration.
• Otherwise, continue for the next values of i
• If all possible values are present, return true.

Follow the illustration below for a better understanding

Illustration:

Consider arr[] = {1, 2, 5}, K = 5

b[] = {0, 1, 1, 0, 0, 1} [denotes 1, 2 and 5 are present]
a[] = {0, 1, 2, 2, 2, 5} [denotes how many element till i is present]

Traverse b[] from i = 1 to 5

For i = 1:
=> Multiple = 1. 1/1 = 1. b = 1, So this is present.
=> Multiple = 2. 2/1 = 2. b = 1, So this is present.
=> Multiple = 3. 3/1 = 3. b = 0, No element in range (2, 3].
=> Multiple = 4. 4/1 = 4. b = 0, No element in range (3, 4].
=> Multiple = 5. 5/1 = 5. b = 1, So this is present.

For i = 2:
=> Multiple = 1. 2/2 = 1. b = 1, So this is present.
=> Multiple = 4. 4/2 = 2. b = 1, So this is present.

For i = 3:
=> 3 is not present in the array. So continue iteration.

For i = 4:
=> 4 is not present in the array. So continue iteration.

For i = 5:
=> Multiple = 5. 5/5 = 5. b = 1, So this is present.

So all possible elements are present for all possible pairs.

Below is the implementation of the above approach:

## C++

 `// C++ code to implement the above approach`   `#include ` `using` `namespace` `std;`   `// Function to check whether X/Y is present or not` `int` `solve(``int` `n, ``int` `c, ``int` `arr[])` `{` `    ``// Creating hash array` `    ``// and prefix sum array` `    ``vector<``int``> a(c + 1, 0), b(c + 1, 0);` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``a[arr[i]]++;` `        ``b[arr[i]]++;` `    ``}`   `    ``// Performing prefix sum.` `    ``for` `(``int` `i = 1; i <= c; i++) {` `        ``a[i] += a[i - 1];` `    ``}` `    ``for` `(``int` `i = 1; i <= c; i++) {`   `        ``// Taking original array elements` `        ``if` `(b[i] > 0) {` `            ``for` `(``int` `j = i - 1; j <= c; j += i) {`   `                ``// If element already exist` `                ``// it will give 1 hence true case` `                ``// if doesnt exist` `                ``// we will move forward` `                ``if` `(b[(j + 1) / i] == 0) {`   `                    ``// we will take two indices` `                    ``// to check  whether item` `                    ``// is present.` `                    ``// If any element is present` `                    ``// between then a[id1]!=ad[id2]` `                    ``int` `id1 = j;` `                    ``int` `id2 = j + i;` `                    ``id2 = min(id2, c);` `                    ``if` `(a[id1] != a[id2]) {` `                        ``return` `false``;` `                    ``}` `                ``}` `            ``}` `        ``}` `    ``}`   `    ``// If all above cases turns true` `    ``return` `true``;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 3, K = 5;` `    ``int` `arr[] = { 1, 2, 5 };` `    ``bool` `flag = solve(N, K, arr);` `    ``if` `(flag)` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;` `    ``return` `0;` `}`

## Java

 `// Java code to implement the above approach`   `import` `java.io.*;`   `class` `GFG {`   `  ``// Function to check whether X/Y is present or not` `  ``static` `boolean` `solve(``int` `n, ``int` `c, ``int` `arr[])` `  ``{` `    ``// Creating hash array` `    ``// and prefix sum array` `    ``int``[] a = ``new` `int``;` `    ``int``[] b = ``new` `int``;` `    ``for` `(``int` `i = ``0``; i < n; i++) {` `      ``a[arr[i]]++;` `      ``b[arr[i]]++;` `    ``}`   `    ``// Performing prefix sum.` `    ``for` `(``int` `i = ``1``; i <= c; i++) {` `      ``a[i] += a[i - ``1``];` `    ``}` `    ``for` `(``int` `i = ``1``; i <= c; i++) {`   `      ``// Taking original array elements` `      ``if` `(b[i] > ``0``) {` `        ``for` `(``int` `j = i - ``1``; j <= c; j += i) {`   `          ``// If element already exist` `          ``// it will give 1 hence true case` `          ``// if doesnt exist` `          ``// we will move forward` `          ``if` `(b[(j + ``1``) / i] == ``0``) {`   `            ``// we will take two indices` `            ``// to check  whether item` `            ``// is present.` `            ``// If any element is present` `            ``// between then a[id1]!=ad[id2]` `            ``int` `id1 = j;` `            ``int` `id2 = j + i;` `            ``id2 = Math.min(id2, c);` `            ``if` `(a[id1] != a[id2]) {` `              ``return` `false``;` `            ``}` `          ``}` `        ``}` `      ``}` `    ``}`   `    ``// If all above cases turns true` `    ``return` `true``;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `main (String[] args) {` `    ``int` `N = ``3``, K = ``5``;` `    ``int` `arr[] = { ``1``, ``2``, ``5` `};` `    ``boolean` `flag = solve(N, K, arr);` `    ``if` `(flag)` `      ``System.out.println(``"Yes"``);` `    ``else` `      ``System.out.println(``"No"``);` `  ``}` `}`   `// This code is contributed by hrithikgarg03188.`

## Python3

 `# Python code to implement the above approach`   `# Function to check whether X/Y is present or not` `def` `solve(n, c, arr):` `  `  `    ``# Creating hash array` `    ``# and prefix sum array` `    ``a ``=` `[]` `    ``b ``=` `[]` `    ``a ``=` `[``0` `for` `i ``in` `range``(c ``+` `1``)]` `    ``b ``=` `[``0` `for` `i ``in` `range``(c ``+` `1``)]` `    ``for` `i ``in` `range``(``0``, n):` `        ``a[arr[i]] ``+``=` `1` `        ``b[arr[i]] ``+``=` `1`   `    ``# Performing prefix sum.` `    ``for` `i ``in` `range``(``0``, c ``+` `1``):` `        ``a[i] ``+``=` `a[i ``-` `1``]`   `    ``for` `i ``in` `range``(``0``, c ``+` `1``):` `      `  `        ``# Taking original array elements` `        ``if` `b[i] > ``0``:` `            ``for` `j ``in` `range``(i ``-` `1``, c ``+` `1``, i):` `              `  `                ``# If element already exist` `                ``# it will give 1 hence true case` `                ``# if doesnt exist` `                ``# we will move forward` `                ``if` `b[(j ``+` `1``) ``/` `i] ``=``=` `0``:` `                  `  `                    ``# we will take two indices` `                    ``# to check  whether item` `                    ``# is present.` `                    ``# If any element is present` `                    ``# between then a[id1]!=ad[id2]` `                    ``id1 ``=` `j` `                    ``id2 ``=` `j ``+` `i` `                    ``id2 ``=` `min``(id2, c)` `                    ``if` `a[id1] !``=` `a[id2]:` `                        ``return` `False` `        ``# If all above cases turns true` `        ``return` `True`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``N ``=` `3` `    ``K ``=` `5` `    ``arr ``=` `[``1``, ``2``, ``5``]` `    ``flag ``=` `solve(N, K, arr)` `    ``if` `(flag ``=``=` `True``):` `        ``print``(``"Yes"``)` `    ``else``:` `        ``print``(``"No"``)`   `# This code is contributed by Rohit Pradhan`

## C#

 `// C# code to implement the above approach` `using` `System;`   `class` `GFG ` `{` `  `  `  ``// Function to check whether X/Y is present or not` `  ``static` `bool` `solve(``int` `n, ``int` `c, ``int``[] arr)` `  ``{` `    ``// Creating hash array` `    ``// and prefix sum array` `    ``int``[] a = ``new` `int``;` `    ``int``[] b = ``new` `int``;` `    ``for` `(``int` `i = 0; i < n; i++) {` `      ``a[arr[i]]++;` `      ``b[arr[i]]++;` `    ``}`   `    ``// Performing prefix sum.` `    ``for` `(``int` `i = 1; i <= c; i++) {` `      ``a[i] += a[i - 1];` `    ``}` `    ``for` `(``int` `i = 1; i <= c; i++) {`   `      ``// Taking original array elements` `      ``if` `(b[i] > 0) {` `        ``for` `(``int` `j = i - 1; j <= c; j += i) {`   `          ``// If element already exist` `          ``// it will give 1 hence true case` `          ``// if doesnt exist` `          ``// we will move forward` `          ``if` `(b[(j + 1) / i] == 0) {`   `            ``// we will take two indices` `            ``// to check  whether item` `            ``// is present.` `            ``// If any element is present` `            ``// between then a[id1]!=ad[id2]` `            ``int` `id1 = j;` `            ``int` `id2 = j + i;` `            ``id2 = Math.Min(id2, c);` `            ``if` `(a[id1] != a[id2]) {` `              ``return` `false``;` `            ``}` `          ``}` `        ``}` `      ``}` `    ``}`   `    ``// If all above cases turns true` `    ``return` `true``;` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `Main()` `  ``{` `    ``int` `N = 3, K = 5;` `    ``int``[] arr = { 1, 2, 5 };` `    ``bool` `flag = solve(N, K, arr);` `    ``if` `(flag)` `      ``Console.Write(``"Yes"``);` `    ``else` `      ``Console.Write(``"No"``);` `  ``}` `}`   `// This code is contributed by sanjoy_62.`

## Javascript

 ``

Output

`Yes`

Time Complexity : O(K * LogK)
Auxiliary Space: O(K)

My Personal Notes arrow_drop_up
Recommended Articles
Page :