Given two integers N and K, the task is to rotate the digits of N by K. If K is a positive integer, left rotate its digits. Otherwise, right rotate its digits.
Examples:
Input: N = 12345, K = 2
Output: 34512
Explanation:
Left rotating N(= 12345) by K(= 2) modifies N to 34512.
Therefore, the required output is 34512Input: N = 12345, K = -3
Output: 34512
Explanation:
Right rotating N(= 12345) by K( = -3) modifies N to 34512.
Therefore, the required output is 34512
Approach: Follow the steps below to solve the problem:
- Initialize a variable, say X, to store the count of digits in N.
- Update K = (K + X) % X to reduce it to a case of left rotation.
- Remove the first K digits of N and append all the removed digits to the right of the digits of N.
- Finally, print the value of N.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to find the count of // digits in N int numberOfDigit(int N) { // Stores count of // digits in N int digit = 0; // Calculate the count // of digits in N while (N > 0) { // Update digit digit++; // Update N N /= 10; } return digit; } // Function to rotate the digits of N by K void rotateNumberByK(int N, int K) { // Stores count of digits in N int X = numberOfDigit(N); // Update K so that only need to // handle left rotation K = ((K % X) + X) % X; // Stores first K digits of N int left_no = N / (int)(pow(10, X - K)); // Remove first K digits of N N = N % (int)(pow(10, X - K)); // Stores count of digits in left_no int left_digit = numberOfDigit(left_no); // Append left_no to the right of // digits of N N = (N * (int)(pow(10, left_digit))) + left_no; cout << N; } // Driver code int main() { int N = 12345, K = 7; // Function Call rotateNumberByK(N, K); return 0; } // The code is contributed by Dharanendra L V |
Output:
34512
Time Complexity: O(log10N)
Auxiliary Space: O(1)
Please refer complete article on Rotate digits of a given number by K for more details!
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