String character swap problem Last Updated : 15 Sep, 2023 Comments Improve Suggest changes Like Article Like Report Given a string S of length N consisting of characters 'L' (sees every character lying to its left) or 'R' (sees every character lying to its right). In one operation, any two consecutive characters can be swapped. The operation can be performed at most once. Determine whether or not each character is being seen by at least one character. Print -1 if it is impossible, 0 if no swap is needed, or the index (1 - indexed) of the first friend if a swap is needed. Note: A character cannot see itself. Examples: Input: N = 2, S = "LL"Output: -1Explanation: It is impossible even after swapping the two. Input: N = 2, S = "LR"Output: 1Explanation: It is necessary to swap friends 1 and 2. The string becomes "RL". Friend 1 is looking at friend 2 and friend 2 is looking at friend 1. Input: N = 2, S = "RL"Output: 0Explanation: Every friend is already being seen. No swap is needed. Approach: This can be solved with the following idea: If there is an index i such that si = R and si+1 = L, then friend i will see friends i+1, i+2, …, n, and friend i+1 will see 1, 2, …, i. In this case, if L appears right after some R, then all friends are already seen. Therefore, strings like LLRRLL, LRLRLR, RRRLLL, etc. do not require any further operations.However, if there is no such index i, we need to perform an operation to see all the friends. We can do this by transforming LR to RL if it appears in the string since we have already concluded that all friends are seen in that case.An edge case arises when L and R are never adjacent (neither LR nor RL appears). In this case, si = si+1 must hold for i=1, 2, …, n−1, which means that LL…L and RR…R are the only impossible strings for which the answer is -1.Steps involved in the implementation of code: Initialize Boolean variables 'left', 'right', and 'check' to false.Loop through the characters of the string 's' from 'left' to 'right'. For each character, if it is 'L', set 'left' to true, set 'ind' to the index of the character plus one, and if 'right' is also true, set 'check' to true. If the character is 'R', set right to true.After looping through the string 's', check the value of 'check'. If it is true, output 0 and exit the function.Otherwise output '-1' and exit the function.Below is the implementation of the code: C++ // C++ code for the above approach: #include <bits/stdc++.h> using namespace std; // Function to check whether swap is // needed or not void left_right(string s, int n) { // Initialize boolean variables bool left = 0, right = 0, check = 0; // Initialize index to -1 int ind = -1; // Loop through string s for (int i = 0; i < n; i++) { // If character is L if (s[i] == 'L') { // Set left to true left = 1; // Set index to current // index + 1 ind = i + 1; // If right is also true, // set check to true if (right) check = 1; } // If character is R else right = 1; } // If there are consecutive L and // R characters, output 0 if (check) cout << 0 << endl; else cout << ind << endl; } // Driver Code int main() { // Sample inputs int n = 5; string s = "RRLRL"; // Function call left_right(s, n); return 0; } Java // Nikunj Sonigara public class GFG { // Function to check whether swap is needed or not static void left_right(String s, int n) { // Initialize boolean variables boolean left = false, right = false, check = false; // Initialize index to -1 int ind = -1; // Loop through string s for (int i = 0; i < n; i++) { // If character is L if (s.charAt(i) == 'L') { // Set left to true left = true; // Set index to current index + 1 ind = i + 1; // If right is also true, set check to true if (right) check = true; } // If character is R else right = true; } // If there are consecutive L and R characters, output 0 if (check) System.out.println(0); else System.out.println(ind); } // Driver Code public static void main(String[] args) { // Sample inputs int n = 5; String s = "RRLRL"; // Function call left_right(s, n); } } Python3 def left_right(s, n): left = False right = False check = False ind = -1 for i in range(n): if s[i] == 'L': left = True ind = i + 1 if right: check = True else: right = True if check: print(0) else: print(ind) n = 5 s = "RRLRL" left_right(s, n) C# // C# Implementation using System; public class GFG { // Function to check whether swap is needed or not static void left_right(string s, int n) { // Initialize boolean variables bool right = false, check = false; // Initialize index to -1 int ind = -1; // Loop through string s for (int i = 0; i < n; i++) { // If character is L if (s[i] == 'L') { // Set left to true // left = true; // Set index to current index + 1 ind = i + 1; // If right is also true, set check to true if (right) check = true; } // If character is R else right = true; } // If there are consecutive L and R characters, output 0 if (check) Console.WriteLine(0); else Console.WriteLine(ind); } // Driver Code public static void Main(string[] args) { // Sample inputs int n = 5; string s = "RRLRL"; // Function call left_right(s, n); } } // This code is contributed by Sakshi JavaScript function left_right(s, n) { let left = false; let right = false; let check = false; let ind = -1; for (let i = 0; i < n; i++) { if (s[i] === 'L') { left = true; ind = i + 1; if (right) { check = true; } } else { right = true; } } if (check) { console.log(0); } else { console.log(ind); } } const n = 5; const s = "RRLRL"; left_right(s, n); Output0Time Complexity: O(N)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms R rohityadavcdece19 Follow Improve Article Tags : Strings DSA Data Structures Practice Tags : Data StructuresStrings Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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