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Modify a string by performing given shift operations
  • Difficulty Level : Basic
  • Last Updated : 08 Mar, 2021

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 ith 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++




// C++ implementataion
// 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 calcuation
    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




// Java implementataion
// 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 calcuation
    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#




// C# implementataion
// 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 calcuation
    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

 
 

Output: 
cab

 

 

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

 

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