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

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  • Last Updated : 21 Apr, 2022
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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] = 1, So this is present.
        => Multiple = 2. 2/1 = 2. b[2] = 1, So this is present.
        => Multiple = 3. 3/1 = 3. b[3] = 0, No element in range (2, 3].
        => Multiple = 4. 4/1 = 4. b[4] = 0, No element in range (3, 4].
        => Multiple = 5. 5/1 = 5. b[5] = 1, So this is present.

For i = 2:
        => Multiple = 1. 2/2 = 1. b[1] = 1, So this is present.
        => Multiple = 4. 4/2 = 2. b[2] = 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] = 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 <bits/stdc++.h>
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




<script>
 
// JavaScript code to implement the above approach
 
 
// Function to check whether X/Y is present or not
function solve(n, c, arr)
{
    // Creating hash array
    // and prefix sum array
    let a = new Array(c + 1).fill(0), b = new Array(c + 1).fill(0);
    for (let i = 0; i < n; i++) {
        a[arr[i]]++;
        b[arr[i]]++;
    }
 
    // Performing prefix sum.
    for (let i = 1; i <= c; i++) {
        a[i] += a[i - 1];
    }
    for (let i = 1; i <= c; i++) {
 
        // Taking original array elements
        if (b[i] > 0) {
            for (let 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[Math.floor((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]
                    let id1 = j;
                    let 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
 
let N = 3, K = 5;
let arr = [ 1, 2, 5 ];
let flag = solve(N, K, arr);
if (flag)
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by Shinjanpatra
 
</script>


Output

Yes

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


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