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Javascript 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:

Javascript




<script>
// Javascript program to implement
// the above approach
  
    // Function to find the count of
    // digits in N
    function numberOfDigit(N)
    {
   
        // Stores count of
        // digits in N
        let digit = 0;
   
        // Calculate the count
        // of digits in N
        while (N > 0) {
   
            // Update digit
            digit++;
   
            // Update N
            N = Math.floor( N / 10);
        }
        return digit;
    }
   
    // Function to rotate the digits of N by K
    function rotateNumberByK(N, K)
    {
   
        // Stores count of digits in N
        let X = numberOfDigit(N);
   
        // Update K so that only need to
        // handle left rotation
        K = ((K % X) + X) % X;
   
        // Stores first K digits of N
        let left_no = Math.floor (N / Math.floor(Math.pow(10,
                                         X - K)));
   
        // Remove first K digits of N
        N = N % Math.floor(Math.pow(10, X - K));
   
        // Stores count of digits in left_no
        let left_digit = numberOfDigit(left_no);
   
        // Append left_no to the right of
        // digits of N
        N = (N * Math.floor(Math.pow(10, left_digit)))
            + left_no;
   
        document.write(N);
    }
  
// Driver Code
  
    let N = 12345, K = 7;
   
        // Function Call
        rotateNumberByK(N, K);
    
  // This code is contributed by souravghosh0416.
</script>


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!



Last Updated : 27 Jan, 2022
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