Cutting Rectangles | Mathematical | JAVA | C++ | GFG POTD 03 Mar 2023

Given a rectangle of dimensions L x B find the minimum number (N) of identical squares of maximum side that can be cut out from that rectangle so that no residue remains in the rectangle. Also find the dimension K of that square. Example 1: Input: L = 2, B = 4 Output: N = 2, K = 2 Explaination: 2 squares of 2x2 dimension. Example 2: Input: L = 6, B = 3 Output: N = 2, K = 3 Explaintion: 2 squares of 3x3 dimension. Your Task: You do not need to read input or print anything. Your task is to complete the function minimumSquares() which takes L and B as input parameters and returns a list of 2 integers containing the values of N and K respectively. Expected Time Complexity: O(log min(L, B)) Expected Auxiliary Space: O(1) Constraints: 1 ≤ L, B ≤ 109 ---------------------------------- SOLUTION OF THE QUESTION ---------------------------------- #ad Learn DSA In most affordable price Use CODE : ISRP14182585 / CLick Me to Avail MAX Discount See DSA Course : https://trainings.internshala.com/data-structures-algorithms-course/?tracking_source=ISRP14182585 SOLUTION in CPP/JAVA #CPP:- class Solution{

public:

vector<long long int> minimumSquares(long long int L, long long int B)

{

// code here

long long int x=__gcd(L,B);

vector<long long int>ans;

ans.push_back((L*B)/(x*x));

ans.push_back(x);

return ans;

}

};

JAVA :-

//{ Driver Code Starts

//Initial Template for Java

import java.io.*;

import java.util.*;

import java.lang.*;

class GFG{

public static void main(String args[])throws IOException

{

BufferedReader read = new BufferedReader(new InputStreamReader(System.in));

int t = Integer.parseInt(read.readLine());

while(t-- > 0){

String input_line[] = read.readLine().trim().split("\\s+");

long L = Long.parseLong(input_line[0]);

long B = Long.parseLong(input_line[1]);

Solution ob = new Solution();

List<Long> ans = new ArrayList<Long>();

ans = ob.minimumSquares(L, B);

System.out.print(ans.get(0)+" ");

System.out.println(ans.get(1));

}

}

}

// } Driver Code Ends

//User function Template for Java

class Solution{

static List<Long> minimumSquares(long L, long B)

{

// code here

List<Long> al = new ArrayList<>();

long len = gcd(L,B);

long no_of_sq = (L*B)/(len*len);

al.add(no_of_sq);

al.add(len);

return al;

}

static long gcd (long a, long b)

{

if(b==0) return a;

return gcd(b, a%b);

}

}

#tags : POTD , Problem of the day, code , c++ ,potd solution, 3rd march 2023,Java

Learn DSA In most affordable price
Use CODE : ISRP14182585 / CLick Me to Avail MAX Discount

See DSA Course : https://trainings.internshala.com/data-structures-algorithms-course/?tracking_source=ISRP14182585