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Rotate digits of a given number by K

 4 years ago
source link: https://www.geeksforgeeks.org/rotate-digits-of-a-given-number-by-k/
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Rotate digits of a given number by K

  • Difficulty Level : Easy
  • Last Updated : 29 Apr, 2021

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 34512

Input: 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:

  • Python3
  • Javascript
// 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)

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