Check if an area P can be obtained from an area of N * M
Given integers N, M and P, the task is to check if is possible to find a rectangular area of P sq. units inside a rectangular area of N × M sq units.
Examples:
Input: N = 3, M = 3, P = 4
Output: YES
Explanation: Rectangle of 2 x 2 sq. unit areaInput: N = 4, M = 4, P = 7
Output: NO
Approach: Follow the steps below to solve the problem
- Find all the factors of p and store them in the vector, say factors.
- Maintain order N ≤ M.
- Traverse the vector factors.
- For each array element factors[i], check if factors[i] ≤ N and p / factors[i] ≤ M, where factors[i] and p / factors[i] represent dimensions of rectangular area.
- If found to be true, print YES and return.
- Otherwise, print NO
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if a rectangle// of p sq units can be formed from// an area of n * m sq unitsvoid splitArea(int n, int m, int p){ // Maintain order n <= m if (n > m) swap(n, m); vector<int> factors; // Iterate to find factors of p for (int i = 1; i * i <= p; i++) { // p is divisible by i if (p % i == 0) { factors.push_back(i); } } for (int i = 0; i < (int)factors.size(); i++) { // Check if dimensions // lie in given area if (factors[i] <= n && p / factors[i] <= m) { cout << "YES"; return; } } cout << "NO";}// Driver Codeint main(){ int n = 3, m = 3, p = 4; splitArea(n, m, p);} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG{ // Function to swap two numbers // Using temporary variable static void swap(int m, int n) { // Swapping the values int temp = m; m = n; n = temp; } // Function to check if a rectangle // of p sq units can be formed from // an area of n * m sq units static void splitArea(int n, int m, int p) { // Maintain order n <= m if (n > m) swap(n, m); ArrayList<Integer> factors = new ArrayList<Integer>(); // Iterate to find factors of p for (int i = 1; i * i <= p; i++) { // p is divisible by i if (p % i == 0) { factors.add(i); } } for (int i = 0; i < (int)factors.size(); i++) { // Check if dimensions // lie in given area if (factors.get(i) <= n && p / factors.get(i) <= m) { System.out.print("YES"); return; } } System.out.print("NO"); }// Driver codepublic static void main(String[] args){ int n = 3, m = 3, p = 4; splitArea(n, m, p);}}// This code is contributed by code_hunt. |
Python3
# Python3 program for the above approach# Function to check if a rectangle# of p sq units can be formed from# an area of n * m sq unitsdef splitArea(n, m, p): # Maintain order n <= m if (n > m): n, m = m, n factors = [] # Iterate to find factors of p for i in range(1, p + 1): # p is divisible by i if (p % i == 0): factors.append(i) for i in range(len(factors)): # Check if dimensions # lie in given area if (factors[i] <= n and p // factors[i] <= m): print("YES") return print("NO")# Driver Codeif __name__ == '__main__': n, m, p = 3, 3, 4 splitArea(n, m, p) # This code is contributed by mohit kumar 29. |
C#
// C# Program to implement// the above approachusing System;using System.Collections.Generic;class GFG{ // Function to swap two numbers // Using temporary variable static void swap(int m, int n) { // Swapping the values int temp = m; m = n; n = temp; } // Function to check if a rectangle // of p sq units can be formed from // an area of n * m sq units static void splitArea(int n, int m, int p) { // Maintain order n <= m if (n > m) swap(n, m); List<int> factors = new List<int>(); // Iterate to find factors of p for (int i = 1; i * i <= p; i++) { // p is divisible by i if (p % i == 0) { factors.Add(i); } } for (int i = 0; i < (int)factors.Count; i++) { // Check if dimensions // lie in given area if (factors[i] <= n && p / factors[i] <= m) { Console.Write("YES"); return; } } Console.Write("NO"); } // Driver Code public static void Main(String[] args) { int n = 3, m = 3, p = 4; splitArea(n, m, p); }}// This code is contributed by splevel62. |
Output:
YES
Time Complexity: O(√P)
Auxiliary Space: O(log(P))
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