Check whether all the rotations of a given number is greater than or equal to the given number or not
Given an integer x, the task is to find if every k-cycle shift on the element produces a number greater than or equal to the same element.
A k-cyclic shift of an integer x is a function that removes the last k digits of x and inserts them in its beginning.
For example, the k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2. Print Yes if the given condition is satisfied else print No.
Examples:
Input: x = 123
Output : Yes
The k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2.
Both 312 and 231 are greater than 123.Input: 2214
Output: No
The k-cyclic shift of 2214 when k=2 is 1422 which is smaller than 2214
Approach: Simply find all the possible k cyclic shifts of the number and check if all are greater than the given number or not.
Below is the implementation of the above approach:
C++
// CPP implementation of the approach #include<bits/stdc++.h> using namespace std; void CheckKCycles(int n, string s) { bool ff = true; int x = 0; for (int i = 1; i < n; i++) { // Splitting the number at index i // and adding to the front x = (s.substr(i) + s.substr(0, i)).length(); // Checking if the value is greater than // or equal to the given value if (x >= s.length()) { continue; } ff = false; break; } if (ff) { cout << ("Yes"); } else { cout << ("No"); } } // Driver code int main() { int n = 3; string s = "123"; CheckKCycles(n, s); return 0; } /* This code contributed by Rajput-Ji */ |
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Java
// Java implementation of the approach class GFG { static void CheckKCycles(int n, String s) { boolean ff = true; int x = 0; for (int i = 1; i < n; i++) { // Splitting the number at index i // and adding to the front x = (s.substring(i) + s.substring(0, i)).length(); // Checking if the value is greater than // or equal to the given value if (x >= s.length()) { continue; } ff = false; break; } if (ff) { System.out.println("Yes"); } else { System.out.println("No"); } } // Driver code public static void main(String[] args) { int n = 3; String s = "123"; CheckKCycles(n, s); } } /* This code contributed by PrinciRaj1992 */ |
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Python
# Python3 implementation of the approach def CheckKCycles(n, s): ff = True for i in range(1, n): # Splitting the number at index i # and adding to the front x = int(s[i:] + s[0:i]) # Checking if the value is greater than # or equal to the given value if (x >= int(s)): continue ff = False break if (ff): print("Yes") else: print("No") n = 3s = "123"CheckKCycles(n, s) |
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C#
// C# implementation of the approach using System; class GFG { static void CheckKCycles(int n, String s) { bool ff = true; int x = 0; for (int i = 1; i < n; i++) { // Splitting the number at index i // and adding to the front x = (s.Substring(i) + s.Substring(0, i)).Length; // Checking if the value is greater than // or equal to the given value if (x >= s.Length) { continue; } ff = false; break; } if (ff) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } // Driver code public static void Main(String[] args) { int n = 3; String s = "123"; CheckKCycles(n, s); } } // This code has been contributed by 29AjayKumar |
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PHP
= strlen($s))
{
continue;
}
$ff = false;
break;
}
if ($ff)
{
print(“Yes”);
}
else
{
print(“No”);
}
}
// Driver code
$n = 3;
$s = “123”;
CheckKCycles($n, $s);
// This code contributed by mits
?>
Output:
Yes
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