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++ 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;
} |
cab
Time Complexity: O(N)
Auxiliary Space: O(N)
Please refer complete article on Modify a string by performing given shift operations for more details!