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<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 */ |
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:
- Define a function CheckKCycles that takes an integer n and a string s as input.
- Initialize a boolean variable isKCycle as true.
- Iterate over the string starting from index 1 up to n-1.
- Compare characters s[i] and s[i % n].
- If the characters are not equal, set isKCycle as false and break the loop.
- After the loop, check the value of isKCycle.
- If it is true, print Yes to indicate a K-Cycle.
- If it is false, print No to indicate it is not a K-Cycle.
Below is the implementation of the above approach:
C++
#include <iostream> #include <string> 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; } |
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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