Check if an integer is rotation of another given integer
Given two integers A and B, the task is to check if the integer A is rotation of the digits of the integer B or not. If found to be true, then print “Yes”. Otherwise print “No”.
Examples:
Input: A= 976, B= 679
Output: YesInput: A= 974, B= 2345
Output: No
Approach: Follow the steps below to solve the problem:
- If A == B, then print “Yes”.
- Calculate the count of digits of integer present in the integer A and B in variables, say dig1 and dig2.
- If dig1 != dig2 is found ot be true, then print “No”.
- Initialize a variable, say temp. Assign temp = A.
- Now, iterate and perform the following operations:
- Print “Yes” if A is equal to B and break.
- Check if (A== temp) i.e A becomes equal to original integer temp. If found to be true, then print “No” and break.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to check if the integer// A is a rotation of the integer Bint check(int A, int B){ if (A == B) { return 1; } // Stores the count of digits in A int dig1 = floor(log10(A) + 1); // Stores the count of digits in B int dig2 = floor(log10(B) + 1); // If dig1 not equal to dig2 if (dig1 != dig2) { return 0; } int temp = A; while (1) { // Stores position of first digit int power = pow(10, dig1 - 1); // Stores the first digit int firstdigit = A / power; // Rotate the digits of the integer A = A - firstdigit * power; A = A * 10 + firstdigit; // If A is equal to B if (A == B) { return 1; } // If A is equal to the initial// value of integer A if (A == temp) { return 0; } }}// Driver Codeint main(){ int A = 967, B = 679; if (check(A, B)) cout << "Yes"; else cout << "No" << endl; return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG { // Function to check if the integer // A is a rotation of the integer B static int check(int A, int B) { if (A == B) { return 1; } // Stores the count of digits in A int dig1 = (int)Math.floor(Math.log10(A) + 1); // Stores the count of digits in B int dig2 = (int)Math.floor(Math.log10(B) + 1); // If dig1 not equal to dig2 if (dig1 != dig2) { return 0; } int temp = A; while (true) { // Stores position of first digit int power = (int)Math.pow(10, dig1 - 1); // Stores the first digit int firstdigit = A / power; // Rotate the digits of the integer A = A - firstdigit * power; A = A * 10 + firstdigit; // If A is equal to B if (A == B) { return 1; } // If A is equal to the initial // value of integer A if (A == temp) { return 0; } } } // Driver Code public static void main(String[] args) { int A = 967, B = 679; if (check(A, B) == 1) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Dharanendra L V. |
Python3
# Python3 implementation of the approachimport math# Function to check if the integer# A is a rotation of the integer Bdef check(A, B) : if (A == B) : return 1 # Stores the count of digits in A dig1 = math.floor(math.log10(A) + 1) # Stores the count of digits in B dig2 = math.floor(math.log10(B) + 1) # If dig1 not equal to dig2 if (dig1 != dig2) : return 0 temp = A while (True) : # Stores position of first digit power = pow(10, dig1 - 1) # Stores the first digit firstdigit = A // power # Rotate the digits of the integer A = A - firstdigit * power A = A * 10 + firstdigit # If A is equal to B if (A == B) : return 1 # If A is equal to the initial value of integer A if (A == temp) : return 0 # Driver codeA, B = 967, 679 if (check(A, B)) : print("Yes")else : print("No") # This code is contributed by divyesh072019. |
C#
// C# implementation of the approachusing System;public class GFG{ // Function to check if the integer // A is a rotation of the integer B static int check(int A, int B) { if (A == B) { return 1; } // Stores the count of digits in A int dig1 = (int)Math.Floor(Math.Log10(A) + 1); // Stores the count of digits in B int dig2 = (int)Math.Floor(Math.Log10(B) + 1); // If dig1 not equal to dig2 if (dig1 != dig2) { return 0; } int temp = A; while (true) { // Stores position of first digit int power = (int)Math.Pow(10, dig1 - 1); // Stores the first digit int firstdigit = A / power; // Rotate the digits of the integer A = A - firstdigit * power; A = A * 10 + firstdigit; // If A is equal to B if (A == B) { return 1; } // If A is equal to the initial // value of integer A if (A == temp) { return 0; } } } // Driver Code public static void Main(String[] args) { int A = 967, B = 679; if (check(A, B) == 1) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by 29AjayKumar |
Output:
Yes
Time Complexity: O(digit(N))
Auxiliary Space: O(1)
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