Given a string str, the task is to perform the following type of queries on the given string:
- (1, K): Left rotate the string by K characters.
- (2, K): Print the Kth character of the string.
Examples:
Input: str = “abcdefgh”, q[][] = {{1, 2}, {2, 2}, {1, 4}, {2, 7}}
Output:
d
e
Query 1: str = “cdefghab”
Query 2: 2nd character is d
Query 3: str = “ghabcdef”
Query 4: 7th character is e
Input: str = “abc”, q[][] = {{1, 2}, {2, 2}}
Output:
a
Approach: The main observation here is that the string doesn’t need to be rotated in every query instead we can create a pointer ptr pointing to the first character of the string and which can be updated for every rotation as ptr = (ptr + K) % N where K the integer by which the string needs to be rotated and N is the length of the string. Now for every query of the second type, the Kth character can be found by str[(ptr + K – 1) % N].
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define size 2// Function to perform the required// queries on the given stringvoid performQueries(string str, int n, int queries[][size], int q){ // Pointer pointing to the current starting // character of the string int ptr = 0; // For every query for (int i = 0; i < q; i++) { // If the query is to rotate the string if (queries[i][0] == 1) { // Update the pointer pointing to the // starting character of the string ptr = (ptr + queries[i][1]) % n; } else { int k = queries[i][1]; // Index of the kth character in the // current rotation of the string int index = (ptr + k - 1) % n; // Print the kth character cout << str[index] << "\n"; } }}// Driver codeint main(){ string str = "abcdefgh"; int n = str.length(); int queries[][size] = { { 1, 2 }, { 2, 2 }, { 1, 4 }, { 2, 7 } }; int q = sizeof(queries) / sizeof(queries[0]); performQueries(str, n, queries, q); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG {static int size = 2;// Function to perform the required// queries on the given stringstatic void performQueries(String str, int n, int queries[][], int q){ // Pointer pointing to the current // starting character of the string int ptr = 0; // For every query for (int i = 0; i < q; i++) { // If the query is to rotate the string if (queries[i][0] == 1) { // Update the pointer pointing to the // starting character of the string ptr = (ptr + queries[i][1]) % n; } else { int k = queries[i][1]; // Index of the kth character in the // current rotation of the string int index = (ptr + k - 1) % n; // Print the kth character System.out.println(str.charAt(index)); } }}// Driver codepublic static void main(String[] args){ String str = "abcdefgh"; int n = str.length(); int queries[][] = { { 1, 2 }, { 2, 2 }, { 1, 4 }, { 2, 7 } }; int q = queries.length; performQueries(str, n, queries, q);}} // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach size = 2# Function to perform the required # queries on the given string def performQueries(string, n, queries, q) : # Pointer pointing to the current starting # character of the string ptr = 0; # For every query for i in range(q) : # If the query is to rotate the string if (queries[i][0] == 1) : # Update the pointer pointing to the # starting character of the string ptr = (ptr + queries[i][1]) % n; else : k = queries[i][1]; # Index of the kth character in the # current rotation of the string index = (ptr + k - 1) % n; # Print the kth character print(string[index]); # Driver code if __name__ == "__main__" : string = "abcdefgh"; n = len(string); queries = [[ 1, 2 ], [ 2, 2 ], [ 1, 4 ], [ 2, 7 ]]; q = len(queries); performQueries(string, n, queries, q); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG {static int size = 2;// Function to perform the required// queries on the given stringstatic void performQueries(String str, int n, int [,]queries, int q){ // Pointer pointing to the current // starting character of the string int ptr = 0; // For every query for (int i = 0; i < q; i++) { // If the query is to rotate the string if (queries[i, 0] == 1) { // Update the pointer pointing to the // starting character of the string ptr = (ptr + queries[i, 1]) % n; } else { int k = queries[i, 1]; // Index of the kth character in the // current rotation of the string int index = (ptr + k - 1) % n; // Print the kth character Console.WriteLine(str[index]); } }}// Driver codepublic static void Main(String[] args){ String str = "abcdefgh"; int n = str.Length; int [,]queries = { { 1, 2 }, { 2, 2 }, { 1, 4 }, { 2, 7 } }; int q = queries.GetLength(0); performQueries(str, n, queries, q);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approachvar size = 2;// Function to perform the required// queries on the given stringfunction performQueries(str, n, queries, q){ // Pointer pointing to the current starting // character of the string var ptr = 0; // For every query for (var i = 0; i < q; i++) { // If the query is to rotate the string if (queries[i][0] == 1) { // Update the pointer pointing to the // starting character of the string ptr = (ptr + queries[i][1]) % n; } else { var k = queries[i][1]; // Index of the kth character in the // current rotation of the string var index = (ptr + k - 1) % n; // Print the kth character document.write( str[index] + "<br>"); } }}// Driver codevar str = "abcdefgh";var n = str.length;var queries = [ [ 1, 2 ], [ 2, 2 ], [ 1, 4 ], [ 2, 7 ] ];var q = queries.length;performQueries(str, n, queries, q);</script> |
Output:
d e
Time Complexity: O(Q), Where Q is the number of queries
Auxiliary Space: O(1)
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