Find 5 non-intersecting ranges with given constraints

Last Updated : 17 Feb, 2023

Given Five integers A, B, C, D and E. the task is to find five non-intersecting ranges [X1, Y1], [X2, Y2], [X3, Y3], [X4, Y4], [X5, Y5] such that, the following 5 conditions hold.

X1 + Y1 = min(A, B)
X2 * Y2 = max(A, B)
|X3 - Y3| = C
?X4 / Y4? = D
X5 % Y5 = E

Examples:

Input: A = 6, B = 36, C = 20, D = 12, E = 55
Output: YES
Explanation: Let's take, X1 = 3, Y1 = 3, X1+Y1 = 3+3 = 6.
And, X2 = 6, Y2 = 6, X2 *Y2 = 6*6 = 36.
And, X3 = 10, Y3 = 30, |X3 -Y3| = |10-30| = 20.
And, X4 = 480, Y4 = 40, ?X4 / Y4? = ?480/40? = 12.
And, X5 = 555, Y5 = 500, X5 % Y5 = 555 % 500 = 55.
So, all the 5 ranges [3, 3], [6, 6], [10, 30], [40, 480] and [500, 555] are non-intersecting.

Input: A = 6, B = 6, C = 6, D = 6, E = 6
Output: NO

Approach: The problem can be solved based on the following mathematical observation:

It is clear that 3rd, 4th and 5th condition always holds for any value of C, D and E as there are infinite non-intersecting ranges, so, we have to only check for sum and product.

The idea is that we have to reduce the size of both of the ranges as much as possible to minimize the chance of intersection. Because if the range with the minimum size is overlapping then it is sure that the other possible ranges should also overlap.

Follow the below steps to solve the problem:

Set, prod = max(X, Y) and Sum = min(X, Y)
If the Sum is even, then,
- Set L = sum/2 and R = sum/2
Otherwise set L = sum/2 and R = (sum/2 +1).
Now, run a loop from sqrt(prod) to 0:
- If, prod is divisible by i.
- Then, check if the range is overlapping then print No.
- Else print Yes.

Below is the implementation of the above approach:

C++

// C++ code for above approach  #include <bits/stdc++.h> using namespace std;  // Function to check whether there is any // non-intersecting ranges exist or not void solve(int X, int Y, int Z, int D, int E) {      // find Product     int prod = max(X, Y);      // find Sum     int sum = min(X, Y);     int L, R;      // Select smallest size range     if ((sum % 2) == 0) {         L = sum / 2;         R = sum / 2;     }     else {         L = sum / 2;         R = sum / 2 + 1;     }      for (int i = sqrt(prod); i >= 0; i--)         if (prod % i == 0) {             if (i > R || L > (prod / i)) {                 cout << "YES" << endl;             }             else {                 cout << "NO" << endl;             }             return;         } }  // Driver Code  int main() {      int A = 6;     int B = 36;     int C = 20;     int D = 12;     int E = 55;      solve(A, B, C, D, E);      return 0; }

Java

// Java code to implement the approach  import java.io.*; import java.util.*;  class GFG {  // Function to check whether there is any // non-intersecting ranges exist or not static void solve(int X, int Y, int Z, int D, int E) {      // find Product     int prod = Math.max(X, Y);      // find Sum     int sum = Math.min(X, Y);     int L, R;      // Select smallest size range     if ((sum % 2) == 0) {         L = sum / 2;         R = sum / 2;     }     else {         L = sum / 2;         R = sum / 2 + 1;     }      for (int i = (int)Math.sqrt(prod); i >= 0; i--)         if (prod % i == 0) {             if (i > R || L > (prod / i)) {                 System.out.println("YES");             }             else {                 System.out.println("NO");             }             return;         } }       // Driver code public static void main(String[] args) {     int A = 6;     int B = 36;     int C = 20;     int D = 12;     int E = 55;      solve(A, B, C, D, E); } }

Python3

# Python code for above approach  import math # Function to check whether there is any # non-intersecting ranges exist or not def solve(X, Y, Z, D, E):     # find Product     prod = max(X, Y);      # find Sum     sum = min(X, Y);     L=0;     R=0;      # Select smallest size range     if ((sum % 2) == 0) :         L = sum / 2;         R = sum / 2;      else :         L = sum / 2;         R = sum / 2 + 1;          j=(int)(math.sqrt(prod));     for i in range(j,0,-1):         if (prod % i == 0):             if (i > R or L > (prod / i)) :                 print("YES") ;                      else:                 print("NO") ;             return;              # Driver Code  A = 6; B = 36; C = 20; D = 12; E = 55;  solve(A, B, C, D, E);

// C# code to implement the approach using System; public class GFG {    // Function to check whether there is any   // non-intersecting ranges exist or not   static void solve(int X, int Y, int Z, int D, int E)   {      // find Product     int prod = Math.Max(X, Y);      // find Sum     int sum = Math.Min(X, Y);     int L, R;      // Select smallest size range     if ((sum % 2) == 0) {       L = sum / 2;       R = sum / 2;     }     else {       L = sum / 2;       R = sum / 2 + 1;     }      for (int i = (int)Math.Sqrt(prod); i >= 0; i--)       if (prod % i == 0) {         if (i > R || L > (prod / i)) {           Console.WriteLine("YES");         }         else {           Console.WriteLine("NO");         }         return;       }   }    static public void Main()   {      // Code     int A = 6;     int B = 36;     int C = 20;     int D = 12;     int E = 55;      solve(A, B, C, D, E);   } }  // This code is contributed by lokeshmvs21.

JavaScript

// JavaScript equivalent to the given C++ code  // Function to check whether there is any non-intersecting ranges exist or not function solve(X, Y, Z, D, E) {   // find Product   let prod = Math.max(X, Y);    // find Sum   let sum = Math.min(X, Y);   let L, R;    // Select smallest size range   if (sum % 2 === 0) {     L = sum / 2;     R = sum / 2;   } else {     L = Math.floor(sum / 2);     R = Math.floor(sum / 2) + 1;   }    for (let i = Math.floor(Math.sqrt(prod)); i >= 0; i--) {     if (prod % i === 0) {       if (i > R || L > prod / i) {         console.log("YES");       } else {         console.log("NO");       }       return;     }   } }  // Driver Code  let A = 6; let B = 36; let C = 20; let D = 12; let E = 55;  solve(A, B, C, D, E);  //code by ksam24000