Check whether for all pair (X, Y) of given Array floor of X/Y is also present
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 YESInput: 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 i is present in array arr[]:
- 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> |
Yes
Time Complexity : O(K * LogK)
Auxiliary Space: O(K)
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