Find the last player to be able to flip a character in a Binary String Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a binary string S of length N, the task is to find the winner of the game if two players A and B plays optimally as per the following rules: Player A always starts the game.In a player's first turn, he can move to any index (1-based indexing) consisting of '0' and make it '1'.For the subsequent turns, if any player is at index i, then he can move to one of it's adjacent indice, if it contains 0, and convert it to '1' after moving.If any player is unable to move to any position during his turn, then the player loses the game. The task is to find the winner of the game. Examples: Input: S = "1100011"Output: Player AExplanation:The indices 3, 4 and 5 consists of 0s and indices 1, 2, 6 and 7 consists of 1s. A starts by flipping the character at index 4..B flips either the index 3 or 5.A is now left with only one index adjacent to 4, which B did not pick. After A flips the character at that index, B does not have any character to flip. Since B has no moves, A wins.Hence, print "Player A". Input: S = "11111"Output: Player B Approach: The idea is to store the length of all the substrings consisting only of 0s from the given array arr[] in another array, say V[]. Now, the following cases arise: If the size of V is 0: In this case, the array does not contain any 0s. Therefore, Player A can't make any move and loses the game. Hence, print Player B.If the size of V is 1: In this case, there is 1 substring consisting only of 0s, say of length L. If the value of L is odd, then Player A wins the game. Otherwise, Player B wins the game.In all other cases: Store the length of the largest and the second-largest consecutive segment of 0s in first and second respectively. Player A can win the game if and only if the value of first is odd and (first + 1)/2 > second. Otherwise, Player B wins the game. Below is the implementation of the above approach: C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if player A wins // the game or not void findWinner(string a, int n) { // Stores size of the groups of 0s vector<int> v; // Stores size of the group of 0s int c = 0; // Traverse the array for (int i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a[i] == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.push_back(c); c = 0; } } if (c != 0) v.push_back(c); // If there is no substring of // odd length consisting only of 0s if (v.size() == 0) { cout << "Player B"; return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.size() == 1) { if (v[0] & 1) cout << "Player A"; // Otherwise else cout << "Player B"; return; } // Stores the size of the largest // and second largest substrings of 0s int first = INT_MIN; int second = INT_MIN; // Traverse the array v[] for (int i = 0; i < v.size(); i++) { // If current element is greater // than first, then update both // first and second if (a[i] > first) { second = first; first = a[i]; } // If arr[i] is in between // first and second, then // update second else if (a[i] > second && a[i] != first) second = a[i]; } // If the condition is satisfied if ((first & 1) && (first + 1) / 2 > second) cout << "Player A"; else cout << "Player B"; } // Driver Code int main() { string S = "1100011"; int N = S.length(); findWinner(S, N); return 0; } Java // Java program for the above approach import java.util.*; public class GFG { // Function to check if player A wins // the game or not static void findWinner(String a, int n) { // Stores size of the groups of 0s Vector<Integer> v = new Vector<Integer>(); // Stores size of the group of 0s int c = 0; // Traverse the array for (int i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a.charAt(i) == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.add(c); c = 0; } } if (c != 0) v.add(c); // If there is no substring of // odd length consisting only of 0s if (v.size() == 0) { System.out.print("Player B"); return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.size() == 1) { if ((v.get(0) & 1) != 0) System.out.print("Player A"); // Otherwise else System.out.print("Player B"); return; } // Stores the size of the largest // and second largest substrings of 0s int first = Integer.MIN_VALUE; int second = Integer.MIN_VALUE; // Traverse the array v[] for (int i = 0; i < v.size(); i++) { // If current element is greater // than first, then update both // first and second if (a.charAt(i) > first) { second = first; first = a.charAt(i); } // If arr[i] is in between // first and second, then // update second else if (a.charAt(i) > second && a.charAt(i) != first) second = a.charAt(i); } // If the condition is satisfied if ((first & 1) != 0 && (first + 1) / 2 > second) System.out.print("Player A"); else System.out.print("Player B"); } // Driver code public static void main(String[] args) { String S = "1100011"; int N = S.length(); findWinner(S, N); } } // This code is contributed by divyeshrabadiya07. Python3 # Python3 program for the above approach import sys # Function to check if player A wins # the game or not def findWinner(a, n) : # Stores size of the groups of 0s v = [] # Stores size of the group of 0s c = 0 # Traverse the array for i in range(0, n) : # Increment c by 1 if a[i] is 0 if (a[i] == '0') : c += 1 # Otherwise, push the size # in array and reset c to 0 else : if (c != 0) : v.append(c) c = 0 if (c != 0) : v.append(c) # If there is no substring of # odd length consisting only of 0s if (len(v) == 0) : print("Player B", end = "") return # If there is only 1 substring of # odd length consisting only of 0s if (len(v) == 1) : if ((v[0] & 1) != 0) : print("Player A", end = "") # Otherwise else : print("Player B", end = "") return # Stores the size of the largest # and second largest substrings of 0s first = sys.minsize second = sys.minsize # Traverse the array v[] for i in range(len(v)) : # If current element is greater # than first, then update both # first and second if (a[i] > first) : second = first first = a[i] # If arr[i] is in between # first and second, then # update second elif (a[i] > second and a[i] != first) : second = a[i] # If the condition is satisfied if (((first & 1) != 0) and (first + 1) // 2 > second) : print("Player A", end = "") else : print("Player B", end = "") S = "1100011" N = len(S) findWinner(S, N) # This code is contributed by divyesh072019. C# // C# program to implement // the above approach using System; using System.Collections.Generic; using System.Linq; class GFG{ // Function to check if player A wins // the game or not static void findWinner(string a, int n) { // Stores size of the groups of 0s List<int> v = new List<int>(); // Stores size of the group of 0s int c = 0; // Traverse the array for (int i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a[i] == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.Add(c); c = 0; } } if (c != 0) v.Add(c); // If there is no substring of // odd length consisting only of 0s if (v.Count == 0) { Console.Write("Player B"); return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.Count == 1) { if ((v[0] & 1) != 0) Console.Write("Player A"); // Otherwise else Console.Write("Player B"); return; } // Stores the size of the largest // and second largest substrings of 0s int first = Int32.MinValue; int second = Int32.MinValue; // Traverse the array v[] for (int i = 0; i < v.Count; i++) { // If current element is greater // than first, then update both // first and second if (a[i] > first) { second = first; first = a[i]; } // If arr[i] is in between // first and second, then // update second else if (a[i] > second && a[i] != first) second = a[i]; } // If the condition is satisfied if ((first & 1) != 0 && (first + 1) / 2 > second) Console.Write("Player A"); else Console.Write("Player B"); } // Driver Code public static void Main(String[] args) { string S = "1100011"; int N = S.Length; findWinner(S, N); } } // This code is contributed by splevel62. JavaScript <script> // Javascript program to implement the above approach // Function to check if player A wins // the game or not function findWinner(a, n) { // Stores size of the groups of 0s let v = []; // Stores size of the group of 0s let c = 0; // Traverse the array for (let i = 0; i < n; i++) { // Increment c by 1 if a[i] is 0 if (a[i] == '0') { c++; } // Otherwise, push the size // in array and reset c to 0 else { if (c != 0) v.push(c); c = 0; } } if (c != 0) v.push(c); // If there is no substring of // odd length consisting only of 0s if (v.length == 0) { document.write("Player B"); return; } // If there is only 1 substring of // odd length consisting only of 0s if (v.length == 1) { if ((v[0] & 1) != 0) document.write("Player A"); // Otherwise else document.write("Player B"); return; } // Stores the size of the largest // and second largest substrings of 0s let first = Number.MIN_VALUE; let second = Number.MIN_VALUE; // Traverse the array v[] for (let i = 0; i < v.length; i++) { // If current element is greater // than first, then update both // first and second if (a[i] > first) { second = first; first = a[i]; } // If arr[i] is in between // first and second, then // update second else if (a[i] > second && a[i] != first) second = a[i]; } // If the condition is satisfied if ((first & 1) != 0 && parseInt((first + 1) / 2, 10) > second) document.write("Player A"); else document.write("Player B"); } let S = "1100011"; let N = S.length; findWinner(S, N); // This code is contributed by suresh07. </script> Output: Player A Time Complexity: O(N) Auxiliary Space: O(N) Comment More infoAdvertise with us Next Article Analysis of Algorithms V vermashivani543 Follow Improve Article Tags : DSA binary-string substring Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. It’s 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. Efficiency is measured in terms of time and space.BasicsWhy is Analysis Important?Order of GrowthAsymptotic Analysis Worst, Average and Best Cases Asymptotic NotationsB 1 min read Data StructuresArray Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous 3 min read String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut 2 min read Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The 2 min read Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List: 2 min read Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first 2 min read Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems 2 min read Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most 4 min read Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of 3 min read Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this 15+ min read AlgorithmsSearching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input 2 min read Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ 3 min read Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution 14 min read Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get 3 min read Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net 3 min read Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of 3 min read Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit 4 min read AdvancedSegment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree 3 min read Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i 2 min read GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br 2 min read Interview PreparationInterview Corner: All Resources To Crack Any Tech InterviewThis article serves as your one-stop guide to interview preparation, designed to help you succeed across different experience levels and company expectations. Here is what you should expect in a Tech Interview, please remember the following points:Tech Interview Preparation does not have any fixed s 3 min read GfG160 - 160 Days of Problem SolvingAre you preparing for technical interviews and would like to be well-structured to improve your problem-solving skills? Well, we have good news for you! GeeksforGeeks proudly presents GfG160, a 160-day coding challenge starting on 15th November 2024. In this event, we will provide daily coding probl 3 min read Practice ProblemGeeksforGeeks Practice - Leading Online Coding PlatformGeeksforGeeks Practice is an online coding platform designed to help developers and students practice coding online and sharpen their programming skills with the following features. GfG 160: This consists of most popular interview problems organized topic wise and difficulty with with well written e 6 min read Problem of The Day - Develop the Habit of CodingDo you find it difficult to develop a habit of Coding? If yes, then we have a most effective solution for you - all you geeks need to do is solve one programming problem each day without any break, and BOOM, the results will surprise you! Let us tell you how:Suppose you commit to improve yourself an 5 min read Like