C++ Program to Rotate digits of a given number by K

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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:

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:

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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