Given a string str, and two integers L and R, the task is to reverse the string in the range [L, R] i.e. str[L…R].
Examples:
Input: str = “neveropen”, L = 5, R = 7
Output: neveropenrofneveropen
Reverse the characters in the range str[5…7] = “neveropenforneveropen”
and the new string will be “neveropenrofneveropen”Input: str = “ijklmn”, L = 1, R = 2
Output: ikjlmn
Approach:
- If the range is invalid i.e. either L < 0 or R ? len or L > R then print the original string.
- If the range is valid then keep swapping the characters str[L] and str[R] while L < R and update L = L + 1 and R = R – 1 after every swap operation. Print the updated string in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the string after// reversing characters in the range [L, R]string reverse(string str, int len, int l, int r){ // Invalid range if (l < 0 || r >= len || l > r) return str; // While there are characters to swap while (l < r) { // Swap(str[l], str[r]) char c = str[l]; str[l] = str[r]; str[r] = c; l++; r--; } return str;}// Driver codeint main(){ string str = "neveropen"; int len = str.length(); int l = 5, r = 7; cout << reverse(str, len, l, r); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{ // Function to return the string after // reversing characters in the range [L, R] static String reverse(char[] str, int len, int l, int r) { // Invalid range if (l < 0 || r >= len || l > r) return "Invalid range!"; // While there are characters to swap while (l < r) { // Swap(str[l], str[r]) char c = str[l]; str[l] = str[r]; str[r] = c; l++; r--; } String string = new String(str); return string; } // Driver code public static void main (String[] args) { String str = "neveropen"; int len = str.length(); int l = 5, r = 7; System.out.println(reverse(str.toCharArray(), len, l, r)); }}// This code is contributed by Ashutosh450 |
Python3
# Python3 implementation of the approach # Function to return the string after # reversing characters in the range [L, R] def reverse(string, length, l, r) : # Invalid range if (l < 0 or r >= length or l > r) : return string; string = list(string) # While there are characters to swap while (l < r) : # Swap(str[l], str[r]) c = string[l]; string[l] = string[r]; string[r] = c; l += 1; r -= 1; return "".join(string); # Driver code if __name__ == "__main__" : string = "neveropen"; length = len(string); l = 5; r = 7; print(reverse(string, length, l, r)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG{ // Function to return the string after // reversing characters in the range [L, R] static String reverse(char[] str, int len, int l, int r) { // Invalid range if (l < 0 || r >= len || l > r) return "Invalid range!"; // While there are characters to swap while (l < r) { // Swap(str[l], str[r]) char c = str[l]; str[l] = str[r]; str[r] = c; l++; r--; } return String.Join("",str); } // Driver code public static void Main (String[] args) { String str = "neveropen"; int len = str.Length; int l = 5, r = 7; Console.WriteLine(reverse(str.ToCharArray(), len, l, r)); }}// This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach // Function to return the string after // reversing characters in the range [L, R] function reverse(str, len, l, r) { // Invalid range if (l < 0 || r >= len || l > r) return "Invalid range!"; // While there are characters to swap while (l < r) { // Swap(str[l], str[r]) let c = str[l]; str[l] = str[r]; str[r] = c; l++; r--; } return str.join(""); } let str = "neveropen"; let len = str.length; let l = 5, r = 7; document.write(reverse(str.split(''), len, l, r));// This code is contributed by divyeshrabadiya07.</script> |
Output:
neveropenrofneveropen
Time complexity: O(N), where N = (r – l)/2
Auxiliary space: O(1)
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