# Reverse given Range of String for M queries

In this problem, we will perform M reverse queries on the given string according to the array values.

The naïve approach to solving the problem is to reverse each string segment according to the given array value.

The optimized approach uses the logic that when we reverse the same substring two times, we get the original string.

Problem statement − We have given an alpha string containing the alphabetical characters. Also, we have given an arr[] array of size M containing the positive integers. We need to perform the M operations on the given string and return the final string.

In each operation, we need to take the arr[i] and reveres the substring arr[i] to N − arr[i] + 1.

Sample examples

Input

alpha = "pqrst"; arr = {2, 1};


Output

tqrsp


Explanation

• After performing the first query, the string becomes ‘psrqt’.

• After performing the second query, we get the ‘tqrsp’.

Input

−  alpha = "pqrst"; arr = {1, 1};


Output

 − ‘pqrst’


Explanation − If we perform the same query for even number of times, we get the same string.

Input

 −  alpha = "pqrst"; arr = {1, 1, 1};


Output

 − ‘tsrqp’


Explanation − If we perform the same query for an odd number of times, we get the reverse of the string.

## Approach 1

In this approach, we will use the reverse() method to reverse the substring. We will take the starting and ending pointers using the given query and reverse the substring of the given string.

### Algorithm

Step 1 − Start traversing the array of queries.

Step 2 − Initialize the ‘left’ variable with arr[p] − 1.

Step 3 − Initialize the ‘right’ variable with str_len − arr[p] + 1.

Step 4 − Use the reverse() method to reverse the substring from left pointer to right pointer.

### Example

#include <bits/stdc++.h>
using namespace std;

void reverseStrings(string &alpha, int str_len, vector<int> &arr, int arr_len){
// Traverse all queries
for (int p = 0; p < arr_len; p++){
// Get starting pointer
int left = arr[p] - 1;
// Ending pointer
int right = str_len - arr[p] + 1;
// Reverse the string
reverse(alpha.begin() + left, alpha.begin() + right);
}
}
int main(){
int str_len = 5;
string alpha = "pqrst";
int arr_len = 2;
vector<int> arr = {2, 1};
reverseStrings(alpha, str_len, arr, arr_len);
cout << "The string after performing queries is " << alpha << endl;
return 0;
}


### Output

The string after performing queries is tqrsp


Time complexity − O(N*M) for reversing substring M times.

Space complexity − O(1) as we don’t use any dynamic space.

## Approach 2

In this approach, we will calculate that particular index and how many times included in the reversal using given queries. If any index is included for an even number of times, we don’t need to reverse it. If any index is included for an odd number of times in all given queries, we need to reverse the character at particular indexes.

### Algorithm

Step 1 − Initialize the ‘cnt’ list of length equal to the string’s length with 0 to store the number of times a particular index is included in the reversal.

Step 2 − Traverse the array of given queries, and take a left and right pointer of the string according to the current query.

Step 3 − Also execute the changeRange() function to update the ‘cnt’ list according to the current query's left and right pointers.

Step 3.1 − In the changeRange() function, increment the value at the ‘left’ index in the ‘cnt’ list.

Step 3.2 − Decrement all values in the ‘cnt’ list, which are at the right of the ‘right + 1’ pointer.

Here, we needed to increment all the values of the ‘cnt’ list by 1 in the range [left, right]. So, we incremented only cnt[left] by 1 because taking the prefix sum will increment all values by 1, which is at right to the ‘left’ index. Also, we don’t want to increment the cnt values between [right, str_len] indexes, so we have decremented it by 1 already as the prefix sum will increase it by 1.

Step 4 − Next, execute the getPrefixSum() function to calculate the prefix sum of the ‘cnt’ list.

Step 4.1 − In the getPrefixSum() function, traverse the string and add the previous element’s value to the current element.

Step 5 − Next, traverse the ‘cnt’ list in reverse order. If the current element is odd, append to the ‘tmp’ string.

Step 6 − Initialize the ‘p’ and ‘q’ with 0, traversing the ‘cnt’ list in the original order.

Step 7 − If the current element is odd in the ‘cnt’ list, update the alpha[p] with the tmp[q].

Step 8 − At the end, return the alpha string.

### Example

#include <iostream>
#include <vector>
using namespace std;

void changeRange(vector<int>& cnt, int left, int right) {
// Increase the value of the left index
cnt[left]++;
// Decrement value for all indexes after the right index
if (right + 1 < cnt.size())
cnt[right + 1]--;
}
void getPrefixSum(vector<int>& cnt) {
// Calculate prefix sum
for (int p = 1; p < cnt.size(); p++) {
cnt[p] += cnt[p - 1];
}
}
string reverseStrings(string alpha, int str_len, vector<int>& arr, int arr_len) {
vector<int> cnt(str_len, 0);
// Traverse the array
for (int p = 0; p < arr_len; p++) {
int left = arr[p] <= (str_len + 1) / 2 ? arr[p] - 1 : str_len - arr[p];
int right = arr[p] <= (str_len + 1) / 2 ? str_len - arr[p] : arr[p] - 1;
// Changes index ranges between left and right
changeRange(cnt, left, right);
}
getPrefixSum(cnt);
string tmp;
// Store characters with the odd reversal in the reverse order in the tmp string
for (int p = cnt.size() - 1; p >= 0; p--) {
if (cnt[p] % 2 != 0)
tmp.push_back(alpha[p]);
}
int p = 0, q = 0;
// For even reversal, pick the character from the original string.
// For odd reversal, pick the character from the temp string.
for (p = 0; p < cnt.size(); p++) {
if (cnt[p] % 2 != 0)
alpha[p] = tmp[q++];
}
return alpha;
}
int main() {
int str_len = 5;
string alpha = "pqrst";
int arr_len = 2;
vector<int> arr = { 2, 1 };
alpha = reverseStrings(alpha, str_len, arr, arr_len);
cout << "The string after performing queries is: " <<alpha << endl;
return 0;
}


### Output

The string after performing queries is: tqrsp


Time complexity − O(M*N + N), where O(M*N) is to update the ‘cnt’ list according to the query, and O(N) is to update the given string.

Space complexity − O(N) to use the ‘cnt’ list.

In the first approach, we used the reveres() method to perform all queries on the given string. In the second approach, we used the prefix sum technique to count how many times a particular index is included in the reversal.

Updated on: 17-Jul-2023

80 Views 