Java Program to Modify a string by performing given shift operations
Last Updated :
06 Jun, 2022
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:
Java
import java.io.*;
class GFG
{
static void stringShift(String s, int [][] shift)
{
int val = 0 ;
for ( int i = 0 ; i < shift.length; ++i)
if (shift[i][ 0 ] == 0 )
val -= shift[i][ 1 ];
else
val += shift[i][ 1 ];
int len = s.length();
val = val % len;
String result = "" ;
if (val > 0 )
result = s.substring(len - val, (len - val) + val)
+ s.substring( 0 , len - val);
else
result = s.substring(-val, len + val)
+ s.substring( 0 , -val);
System.out.println(result);
}
public static void main(String[] args)
{
String s = "abc" ;
int [][] shift
= new int [][] {{ 0 , 1 }, { 1 , 2 }};
stringShift(s, shift);
}
}
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Please refer complete article on Modify a string by performing given shift operations for more details!
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