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# C++ Program to 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 1: 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``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 */`

Output:

`Yes`

Time Complexity: O(N2), where N represents the length of the given string.

The time complexity of the program is O(N2) because first it runs a loop for traversing the string and inside that substring function is used.

Auxiliary Space: O(1), no extra space is required, so it is a constant.

Approach 2:

1. Define a function CheckKCycles that takes an integer n and a string s as input.
2. Initialize a boolean variable isKCycle as true.
3. Iterate over the string starting from index 1 up to n-1.
4. Compare characters s[i] and s[i % n].
5. If the characters are not equal, set isKCycle as false and break the loop.
6. After the loop, check the value of isKCycle.
7. If it is true, print Yes to indicate a K-Cycle.
8. If it is false, print No to indicate it is not a K-Cycle.

Below is the implementation of the above approach:

## C++

 `#include ``#include ` `void` `CheckKCycles(``int` `n, ``const` `std::string& s) {``    ``bool` `isKCycle = ``true``;``    ``for` `(``int` `i = 1; i < n; i++) {``        ``if` `(s[i] != s[i % n]) {``            ``isKCycle = ``false``;``            ``break``;``        ``}``    ``}``    ``if` `(isKCycle) {``        ``std::cout << ``"Yes"``;``    ``} ``else` `{``        ``std::cout << ``"No"``;``    ``}``}``// Nikunj Sonigara``int` `main() {``    ``int` `n = 3;``    ``std::string s = ``"123"``;``    ``CheckKCycles(n, s);``    ``return` `0;``}`

Output

`Yes`

Time Complexity: O(N)

Auxiliary Space: O(1)

Please refer complete article on Check whether all the rotations of a given number is greater than or equal to the given number or not for more details!

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