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++ implementation// of above approach#include <bits/stdc++.h>using namespace std;// Function to find the string obtained// after performing given shift operationsvoid 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 Codeint 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!
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