Modify a string by performing given shift operations

Given a string S containing lowercase English alphabets, and a matrix shift[][] consisting of pairs of the form{direction, amount}, where the direction can be 0 (for left shift) or 1 (for right shift) and the amount is the number of indices by which the string S is required to be shifted. The task is to return the modified string that can be obtained after performing the given operations.

Note: A left shift by 1 refers to removing the first character of S and append it to the end. Similarly, a right shift by 1 refers to removing the last character of S and insert at the beginning.

Examples

Input: S = "abc", shift[][] = {{0, 1}, {1, 2}}
Output: cab
Explanation: 
[0, 1] refers to shifting  S[0] to the left by 1. Therefore, the string S modifies from "abc" to "bca".
[1, 2] refers to shifting  S[0] to the right by 1. Therefore, the string S modifies from "bca"to "cab".

Input: S = "abcdefg", shift[][] = { {1, 1}, {1, 1}, {0, 2}, {1, 3} }
Output: efgabcd
Explanation:  
[1, 1] refers to shifting S[0] to the right by 1. Therefore, the string S modifies from "abcdefg" to "gabcdef".
[1, 1] refers to shifting S[0] to the right by 1. Therefore, the string S modifies from "gabcdef" to "fgabcde".
[0, 2] refers to shifting S[0] to the left by 2. Therefore, the string S modifies from "fgabcde" to "abcdefg".
[1, 3] refers to shifting S[0] to the right by 3. Therefore, the string S modifies from "abcdefg" to "efgabcd".

Naive Approach: The simplest approach to solve the problem is to traverse the matrix shift[][] and shift S[0] by amount number of indices in the specified direction. After completing all shift operations, print the final string obtained.

Time Complexity: O(N²)
Auxiliary space: O(N)

Efficient Approach: To optimize the above approach, follow the steps below:

• Initialize a variable, say val, to store the effective shifts.
• Traverse the matrix shift[][] and perform the following operations on every i^th row:
• If shift[i][0] = 0 (left shift), then decrease val by -shift[i][1].
• Otherwise (left shift), increase val by shift[i][1].
• Update val =  val % len (for further optimizing the effective shifts).
• Initialize a string, result = "", to store the modified string.
• Now, check if val > 0. If found to be true, then perform the right rotation on the string by val.
• Otherwise, perform left rotation of the string by |val| amount.
• Print the result.

Below is the implementation of the above approach:

// C++ implementation
// of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the string obtained
// after performing given shift operations
void stringShift(string s,
                 vector<vector<int> >& shift)
{

    int val = 0;

    for (int i = 0; i < shift.size(); ++i)

        // If shift[i][0] = 0, then left shift
        // Otherwise, right shift
        val += shift[i][0] == 0
                   ? -shift[i][1]
                   : shift[i][1];

    // Stores length of the string
    int len = s.length();

    // Effective shift calculation
    val = val % len;

    // Stores modified string
    string result = "";

    // Right rotation
    if (val > 0)
        result = s.substr(len - val, val)
                 + s.substr(0, len - val);

    // Left rotation
    else
        result
            = s.substr(-val, len + val)
              + s.substr(0, -val);

    cout << result;
}

// Driver Code
int main()
{
    string s = "abc";
    vector<vector<int> > shift = {
        { 0, 1 },
        { 1, 2 }
    };

    stringShift(s, shift);

    return 0;
}

// Java implementation
// of above approach
import java.io.*;
class GFG 
{

  // Function to find the string obtained
  // after performing given shift operations
  static void stringShift(String s, int[][] shift)
  {
    int val = 0;
    for (int i = 0; i < shift.length; ++i)

      // If shift[i][0] = 0, then left shift
      // Otherwise, right shift
      if (shift[i][0] == 0)
        val -= shift[i][1];
    else
      val += shift[i][1];

    // Stores length of the string
    int len = s.length();

    // Effective shift calculation
    val = val % len;

    // Stores modified string
    String result = "";

    // Right rotation
    if (val > 0)
      result = s.substring(len - val, (len - val) + val)
      + s.substring(0, len - val);

    // Left rotation
    else
      result = s.substring(-val, len + val)
      + s.substring(0, -val);

    System.out.println(result);
  }

  // Driver Code
  public static void main(String[] args)
  {
    String s = "abc";
    int[][] shift
      = new int[][] {{ 0, 1 }, { 1, 2 }};

    stringShift(s, shift);
  }
}

// This code is contributed by Dharanendra L V

// C# implementation
// of above approach
using System;

public class GFG 
{

  // Function to find the string obtained
  // after performing given shift operations
  static void stringShift(String s, int[,] shift)
  {
    int val = 0;
    for (int i = 0; i < shift.GetLength(0); ++i)

      // If shift[i,0] = 0, then left shift
      // Otherwise, right shift
      if (shift[i,0] == 0)
        val -= shift[i, 1];
    else
      val += shift[i, 1];

    // Stores length of the string
    int len = s.Length;

    // Effective shift calculation
    val = val % len;

    // Stores modified string
    String result = "";

    // Right rotation
    if (val > 0)
      result = s.Substring(len - val, val)
      + s.Substring(0, len - val);

    // Left rotation
    else
      result = s.Substring(-val, len)
      + s.Substring(0, -val);

    Console.WriteLine(result);
  }

  // Driver Code
  public static void Main(String[] args)
  {
    String s = "abc";
    int[,] shift
      = new int[,] {{ 0, 1 }, { 1, 2 }};

    stringShift(s, shift);
  }
}

// This code contributed by shikhasingrajput 

# Python implementation
# of above approach

# Function to find the string obtained
# after performing given shift operations
def stringShift(s, shift):
    val = 0
    for i in range(len(shift)):

        # If shift[i][0] = 0, then left shift
        # Otherwise, right shift
        val += -shift[i][1] if shift[i][0] == 0 else shift[i][1]

    # Stores length of the string
    Len = len(s)

    # Effective shift calculation
    val = val % Len

    # Stores modified string
    result = ""

    # Right rotation
    if (val > 0):
        result = s[Len - val:Len] + s[0: Len - val]

    # Left rotation
    else:
        result = s[-val: Len] + s[0: -val]

    print(result)

# Driver Code
s = "abc"
shift = [
    [ 0, 1 ],
    [ 1, 2 ]
]
stringShift(s, shift)

# This code is contributed by shinjanpatra

<script>

// Javascript implementation
// of above approach

// Function to find the string obtained
// after performing given shift operations
function stringShift(s, shift)
{

    var val = 0;

    for (var i = 0; i < shift.length; ++i)

        // If shift[i][0] = 0, then left shift
        // Otherwise, right shift
        val += shift[i][0] == 0
                   ? -shift[i][1]
                   : shift[i][1];

    // Stores length of the string
    var len = s.length;

    // Effective shift calculation
    val = val % len;

    // Stores modified string
    var result = "";

    // Right rotation
    if (val > 0)
        result = s.substring(len - val,len)
                 + s.substring(0, len - val);

    // Left rotation
    else
        result
            = s.substring(-val, len )
              + s.substring(0, -val);

    document.write( result);
}

// Driver Code
var s = "abc";
var shift = [
    [ 0, 1 ],
    [ 1, 2 ]
];
stringShift(s, shift);

// This code is contributed by importantly.
</script> 

cab

Time Complexity: O(N)
Auxiliary Space: O(N)