Given an integer n, the task is to generate all the left shift numbers possible. A left shift number is a number that is generated when all the digits of the number are shifted one position to the left and the digit at the first position is shifted to the last.
Examples:
Input: n = 123
Output: 231 312
Input: n = 1445
Output: 4451 4514 5144
Approach:
- Assume n = 123.
- Multiply n with 10 i.e. n = n * 10 = 1230.
- Add the first digit to the resultant number i.e. 1230 + 1 = 1231.
- Subtract (first digit) * 10k from the resultant number where k is the number of digits in the original number (in this case, k = 3).
- 1231 – 1000 = 231 is the left shift number of the original number.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of digits of n int numberOfDigits(int n) { int cnt = 0; while (n > 0) { cnt++; n /= 10; } return cnt; } // Function to print the left shift numbers void cal(int num) { int digits = numberOfDigits(num); int powTen = pow(10, digits - 1); for (int i = 0; i < digits - 1; i++) { int firstDigit = num / powTen; // Formula to calculate left shift // from previous number int left = ((num * 10) + firstDigit) - (firstDigit * powTen * 10); cout << left << " "; // Update the original number num = left; } } // Driver Code int main() { int num = 1445; cal(num); return 0; } |
Output:
4451 4514 5144
Time Complexity: O(log10(num))
Auxiliary Space: O(1)
Please refer complete article on Generate all rotations of a number for more details!
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