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:
PHP
<?php // PHP implementation of the approach function CheckKCycles( $n , $s ) { $ff = true; $x = 0; for ( $i = 1; $i < $n ; $i ++) { // Splitting the number at index i // and adding to the front $x = strlen ( substr ( $s , $i ). substr ( $s , 0, $i )); // Checking if the value is greater than // or equal to the given value if ( $x >= 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 ?> |
Yes
Time Complexity: O(N2), where N is the length of the given string. First it runs a loop for traversing the string and inside that substring function is used. So, the code will run in O(N2) time.
Auxiliary Space: O(1), no extra space required, so it is a constant.
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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