Check If it is Possible to Convert Binary String into Unary String Last Updated : 11 Dec, 2023 Comments Improve Suggest changes Like Article Like Report Given a binary string S of length N. You can apply following operation on S any number of times. Choose two adjacent characters, such that both are 1's or 0's. Then invert them, Formally, if both are 0's then into 1's or vice-versa. Then your task is to output YES or NO, by checking that using given operations, Is it possible to convert S into either all 0's or 1's ? Examples: Input: N = 6, S = "110110"Output: YESExplanation: The operations are performed as: First operation: S1 = 1 and S2 = 1. Both are same, therefore invert them. Then updated S = 000110Second operation: S4 = 1 and S5 = 1. Both are same, therefore invert them. Then updated S = 000000 All the string is converted into 0s. Therefore, output is YES. Note that S also can be converted into all 1s by following the sequence: 110110 → 110000 → 110011 → 111111. Both conversions are valid. Input: N = 7, S = 0101010Output: NOExplanation: It can be verified that S can't be converted into all 1s or 0s by using give operation. Approach: To Solve this problem follow the below idea It's an observation-based problem. It must be observed that, All occurrence of either number (zero or one) must be existed in pair in order to change the entire S to 1 or 0. Let us take some examples: 001100 = Possible as both 0 and 1 are occurred in pair.00100 = Possible because 0 occurred in pair.00101 = Not possible as none of the element occurred in pair.1100111 = Possible as 0 occurred in pairNow, we can use Stack to solve this problem: First, we have to create a Stack.Iterate on String and follow below steps:If stack is empty or peek and current characters are different then put current character into Stack Else If current character and peek character of Stack is same, then pop out the peek element from stack.If number of elements in stack is either 0 or 1, Then output YES, else NO.Steps were taken to solve the problem: Create a Stack of Character data type let say Stk.Iterate over each character of S using loop and follow below mentioned steps under the scope of loop:If (Stk is not empty && Peek character == current character)Pop out the peek character from StkElseInsert character into StkIf (Size of Stk is less than or equal to 1)Output YESElseOutput NOBelow is the code to implement the approach: C++ // code added by flutterfly #include <iostream> #include <stack> using namespace std; // Function to check the possibility void Is_possible(long long N, string S) { // Initializing Stack stack<char> st; // Iterating over String S for (char x : S) { // If current and peek characters are same // Then removing that peek element from Stack if (!st.empty() && st.top() == x) { st.pop(); } // Else putting the current character into Stack else st.push(x); } // If there is at max one character in // Stack, Then output YES else NO if (st.size() <= 1) cout << "YES" << endl; else cout << "NO" << endl; } // Driver Function int main() { // Inputs long long N = 6; string S = "110110"; // Function call Is_possible(N, S); return 0; } Java // Java code to implement te approach import java.util.*; // Driver Class public class GFG { // Driver Function public static void main(String[] args) { // Inputs long N = 6; String S = "110110"; // Function_call Is_possible(N, S); } // Method to check the possibility public static void Is_possible(long N, String S) { // Initializing Stack Stack<Character> st = new Stack<>(); // Iterating over String S for (char x : S.toCharArray()) { // If current and peek characters are same // Then removing that peek element from // Stack if (!st.empty() && st.peek() == x) { st.pop(); } // Else putting current character // into Stack else st.push(x); } // If there is at max one character in // Stack, Then output YES else NO if (st.size() <= 1) System.out.println("YES"); else System.out.println("NO"); } } Python # code added by flutterfly # Python code to implement the approach # Method to check the possibility def is_possible(N, S): # Initializing Stack st = [] # Iterating over String S for x in S: # If current and peek characters are the same # Then removing that peek element from Stack if st and st[-1] == x: st.pop() # Else putting the current character into Stack else: st.append(x) # If there is at max one character in # Stack, Then output YES else NO if len(st) <= 1: print("YES") else: print("NO") # Inputs N = 6 S = "110110" # Function call is_possible(N, S) C# // code added by flutterfly using System; using System.Collections.Generic; public class GFG { // Method to check the possibility public static void Is_possible(long N, string S) { // Initializing Stack Stack<char> st = new Stack<char>(); // Iterating over String S foreach (char x in S) { // If current and peek characters are the same // Then removing that peek element from Stack if (st.Count > 0 && st.Peek() == x) { st.Pop(); } // Else putting the current character into Stack else { st.Push(x); } } // If there is at max one character in // Stack, Then output YES else NO if (st.Count <= 1) Console.WriteLine("YES"); else Console.WriteLine("NO"); } // Driver Function public static void Main() { // Inputs long N = 6; string S = "110110"; // Function call Is_possible(N, S); } } JavaScript // JavaScript code added by flutterfly // Method to check the possibility function isPossible(N, S) { // Initializing Stack let st = []; // Iterating over String S for (let x of S) { // If current and peek characters are the same // Then removing that peek element from Stack if (st.length > 0 && st[st.length - 1] === x) { st.pop(); } // Else putting the current character into Stack else { st.push(x); } } // If there is at max one character in // Stack, Then output YES else NO if (st.length <= 1) { console.log("YES"); } else { console.log("NO"); } } // Inputs let N = 6; let S = "110110"; // Function call isPossible(N, S); OutputYESTime Complexity: O(N).Auxiliary Space: O(N). Comment More infoAdvertise with us Next Article Analysis of Algorithms P pradeep6036ymca Follow Improve Article Tags : Strings Stack Geeks Premier League DSA Java-Stack Geeks Premier League 2023 +2 More Practice Tags : StackStrings Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. Itâs the heart of coding, enabling programmers to think, reason, and arrive at smart solutions just like we do.Here are some tips for improving your programming logic: Understand the pro 2 min read Analysis of AlgorithmsAnalysis of Algorithms is a fundamental aspect of computer science that involves evaluating performance of algorithms and programs. 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