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++
#include <bits/stdc++.h>
using namespace std;
void stringShift(string s,
vector<vector<int> >& shift)
{
int val = 0;
for (int i = 0; i < shift.size(); ++i)
val += shift[i][0] == 0
? -shift[i][1]
: shift[i][1];
int len = s.length();
val = val % len;
string result = "";
if (val > 0)
result = s.substr(len - val, val)
+ s.substr(0, len - val);
else
result
= s.substr(-val, len + val)
+ s.substr(0, -val);
cout << result;
}
int main()
{
string s = "abc";
vector<vector<int> > shift = {
{ 0, 1 },
{ 1, 2 }
};
stringShift(s, shift);
return 0;
}
|
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);
}
}
|
C#
using System;
public class GFG
{
static void stringShift(String s, int[,] shift)
{
int val = 0;
for (int i = 0; i < shift.GetLength(0); ++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, val)
+ s.Substring(0, len - val);
else
result = s.Substring(-val, len)
+ s.Substring(0, -val);
Console.WriteLine(result);
}
public static void Main(String[] args)
{
String s = "abc";
int[,] shift
= new int[,] {{ 0, 1 }, { 1, 2 }};
stringShift(s, shift);
}
}
|
Python3
def stringShift(s, shift):
val = 0
for i in range(len(shift)):
val += -shift[i][1] if shift[i][0] == 0 else shift[i][1]
Len = len(s)
val = val % Len
result = ""
if (val > 0):
result = s[Len - val:Len] + s[0: Len - val]
else:
result = s[-val: Len] + s[0: -val]
print(result)
s = "abc"
shift = [
[ 0, 1 ],
[ 1, 2 ]
]
stringShift(s, shift)
|
Javascript
<script>
function stringShift(s, shift)
{
var val = 0;
for (var i = 0; i < shift.length; ++i)
val += shift[i][0] == 0
? -shift[i][1]
: shift[i][1];
var len = s.length;
val = val % len;
var result = "";
if (val > 0)
result = s.substring(len - val,len)
+ s.substring(0, len - val);
else
result
= s.substring(-val, len )
+ s.substring(0, -val);
document.write( result);
}
var s = "abc";
var shift = [
[ 0, 1 ],
[ 1, 2 ]
];
stringShift(s, shift);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)