Find winner in the game of Binary Array

Find the transition point in a binary array

Last Updated : 15 Sep, 2023

Try it on GfG Practice

Given a sorted array containing only numbers 0 and 1, the task is to find the transition point efficiently. The transition point is the point where “0” ends and “1” begins.

Examples :

Input: 0 0 0 1 1
Output: 3
Explanation: Index of first 1 is 3

Input: 0 0 0 0 1 1 1 1
Output: 4
Explanation: Index of first 1 is 4

Naive Approach: Traverse the array and print the index of the first 1.

Traverse the array from the start to the end of the array.
If the current element is 1, print the index and terminate the program.

Below is the implementation of the above approach:

C++

// C++ implementation to find 
// the transition point 
#include<iostream> 
using namespace std; 
  
// Function to find the transition point 
int findTransitionPoint(int arr[], int n) 
{ 
    //perform a linear search and 
    // return the index of  
    //first 1 
    for(int i=0; i<n ;i++) 
      if(arr[i]==1) 
        return i; 
  
    //if no element is 1 
    return -1; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = {0, 0, 0, 0, 1, 1}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    int point = findTransitionPoint(arr, n); 
  
    point >= 0 ? cout << "Transition point is " 
                      << point 
        : cout<<"There is no transition point"; 
    return 0; 
}

Java

// Java implementation to find the transition point 
import java.util.*; 
  
class GFG 
{ 
  
// Function to find the transition point 
static int findTransitionPoint(int arr[], int n) 
{ 
    // perform a linear search and return the index of  
    // first 1 
    for(int i = 0; i < n ; i++) 
    if(arr[i] == 1) 
        return i; 
  
    // if no element is 1 
    return -1; 
} 
  
// Driver code 
public static void main (String[] args) 
{ 
    int arr[] = {0, 0, 0, 0, 1, 1}; 
    int n = arr.length; 
      
    int point = findTransitionPoint(arr, n); 
      
    if (point >= 0) 
        System.out.print("Transition point is " + point); 
    else
        System.out.print("There is no transition point"); 
} 
} 
  
// This code is contributed by shivanisinghss2110 

Python3

# Python3 implementation to find the transition point 
  
# Function to find the transition point 
def findTransitionPoint(arr, n): 
      
    # perform a linear search and return the index of  
    # first 1 
    for i in range(n): 
        if(arr[i] == 1): 
            return i 
      
    # if no element is 1 
    return -1
  
# Driver code 
arr = [0, 0, 0, 0, 1, 1] 
n = len(arr) 
  
point = findTransitionPoint(arr, n) 
  
if point >= 0: 
    print("Transition point is", point) 
else:  
    print("There is no transition point") 
      
# This code is contributed by shubhamsingh10 

C#

// C# implementation to find the transition point 
using System; 
  
class GFG  
{ 
  
// Function to find the transition point 
static int findTransitionPoint(int []arr ,int n) 
{ 
    // perform a linear search and return the index of  
    // first 1 
    for(int i = 0; i < n ; i++) 
    if(arr[i] == 1) 
        return i; 
  
    // if no element is 1 
    return -1; 
} 
  
 // Driver method 
    public static void Main()  
    { 
        int []arr = {0, 0, 0, 0, 1, 1}; 
        int point = findTransitionPoint(arr, arr.Length); 
       
        Console.Write(point >= 0 ? "Transition point is " + 
                   point : "There is no transition point"); 
    } 
} 
   
  
// This code is contributed by shivanisinghss2110 

Javascript

<script> 
  
    // Javascript implementation to  
    // find the transition point 
      
    // Function to find the transition point 
    function findTransitionPoint(arr, n) 
    { 
        // perform a linear search and  
        // return the index of 
        // first 1 
        for(let i = 0; i < n ; i++) 
            if(arr[i] == 1) 
                return i; 
  
        // if no element is 1 
        return -1; 
    } 
      
    let arr = [0, 0, 0, 0, 1, 1]; 
    let point = findTransitionPoint(arr, arr.length); 
  
    document.write(point >= 0 ? "Transition point is " + 
                  point : "There is no transition point"); 
      
</script>

Output

Transition point is 4

Complexity Analysis:

Time Complexity: O(n), Only one traversal is needed, so the time complexity is O(n)
Auxiliary Space: O(1), No extra space is required.

Efficient Approach: The idea is to use Binary Search, and find the smallest index of 1 in the array. As the array is sorted, binary search can be performed.

Create two variables, l and r, initialize l = 0 and r = n-1 and a variable ans = -1 to store the answer.
Iterate the steps below till l <= r, the lower-bound is less than the upper-bound.
Check if the element at middle index mid = (l+r)/2, is one or not.
If the element is one, then check for the least index of 1 element on the left side of the middle element, i.e. update r = mid – 1 and update ans = mid.
If the element is zero, then check for the least index of 1 element on the right side of the middle element, i.e. update l = mid + 1.

Below is the implementation of the above approach:

C++

// C++ implementation to find the transition point 
#include<iostream> 
using namespace std; 
  
// Function to find the transition point 
int findTransitionPoint(int arr[], int n) 
{ 
    // Initialise lower and upper bounds 
    int lb = 0, ub = n-1; 
  
    // Perform Binary search 
    while (lb <= ub) 
    { 
        // Find mid 
        int mid = (lb+ub)/2; 
  
        // update lower_bound if mid contains 0 
        if (arr[mid] == 0) 
            lb = mid+1; 
  
        // If mid contains 1 
        else if (arr[mid] == 1) 
        { 
            // Check if it is the left most 1 
            // Return mid, if yes 
            if (mid == 0 
                    || (mid > 0 &&  
                       arr[mid - 1] == 0)) 
                return mid; 
  
            // Else update upper_bound 
            ub = mid-1; 
        } 
    } 
    return -1; 
} 
  
// Driver Code 
int main() 
{ 
    int arr[] = {0, 0, 0, 0, 1, 1}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    int point = findTransitionPoint(arr, n); 
  
    point >= 0 ? cout<<"Transition point is " << point 
               : cout<<"There is no transition point"; 
    return 0; 
} 

Java

// Java implementation to find the transition point 
  
class Test { 
    // Method to find the transition point 
    static int findTransitionPoint(int arr[], int n) 
    { 
        // Initialise lower and upper bounds 
        int lb = 0, ub = n - 1; 
  
        // Perform Binary search 
        while (lb <= ub) { 
            // Find mid 
            int mid = (lb + ub) / 2; 
  
            // update lower_bound if mid contains 0 
            if (arr[mid] == 0) 
                lb = mid + 1; 
            // If mid contains 1 
            else if (arr[mid] == 1) { 
                // Check if it is the left most 1 
                // Return mid, if yes 
                if (mid == 0
                    || (mid > 0 &&  
                       arr[mid - 1] == 0)) 
                    return mid; 
                // Else update upper_bound 
                ub = mid - 1; 
            } 
        } 
        return -1; 
    } 
  
    // Driver method 
    public static void main(String args[]) 
    { 
        int arr[] = { 0, 0, 0, 0, 1, 1 }; 
  
        int point = findTransitionPoint(arr, arr.length); 
  
        System.out.println( 
            point >= 0 ? "Transition point is " + point 
                       : "There is no transition point"); 
    } 
}

Python3

# python implementation to find the 
# transition point 
  
# Function to find the transition 
# point 
def findTransitionPoint(arr, n): 
    # Initialise lower and upper 
    # bounds 
    lb = 0
    ub = n - 1
  
    # Perform Binary search 
    while (lb <= ub): 
        # Find mid 
        mid = (int)((lb + ub) / 2) 
  
        # update lower_bound if 
        # mid contains 0 
        if (arr[mid] == 0): 
            lb = mid + 1
  
        # If mid contains 1 
        else if (arr[mid] == 1): 
              
            # Check if it is the  
            # left most 1 Return 
            # mid, if yes 
            if (mid == 0 \ 
                or (mid > 0 and\ 
                arr[mid - 1] == 0)): 
                return mid 
  
            # Else update  
            # upper_bound 
            ub = mid-1
      
    return -1
  
# Driver code 
arr = [0, 0, 0, 0, 1, 1] 
n = len(arr) 
point = findTransitionPoint(arr, n); 
if(point >= 0): 
    print("Transition point is ", point) 
else: 
    print("There is no transition point") 
  
# This code is contributed by Sam007 

C#

// C# implementation to find the transition point 
using System; 
          
class GFG  
{ 
    // Method to find the transition point 
    static int findTransitionPoint(int []arr, int n) 
    { 
        // Initialise lower and upper bounds 
        int lb = 0, ub = n-1; 
      
        // Perform Binary search 
        while (lb <= ub) 
        { 
            // Find mid 
            int mid = (lb+ub)/2; 
      
            // update lower_bound if mid contains 0 
            if (arr[mid] == 0) 
                lb = mid+1; 
      
            // If mid contains 1 
            else if (arr[mid] == 1) 
            { 
                // Check if it is the left most 1 
                // Return mid, if yes 
                if (mid == 0 
                    || (mid > 0 &&  
                       arr[mid - 1] == 0)) 
                    return mid; 
      
                // Else update upper_bound 
                ub = mid-1; 
            } 
        } 
        return -1; 
    } 
      
      
    // Driver method 
    public static void Main()  
    { 
        int []arr = {0, 0, 0, 0, 1, 1}; 
        int point = findTransitionPoint(arr, arr.Length); 
      
        Console.Write(point >= 0 ? "Transition point is " + 
                   point : "There is no transition point"); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP implementation to find 
// the transition point 
  
// Function to find the  
// transition point 
function findTransitionPoint($arr, $n) 
{ 
      
    // Initialise lower and 
    // upper bounds 
    $lb = 0; $ub = $n-1; 
  
    // Perform Binary search 
    while ($lb <= $ub) 
    { 
          
        // Find mid 
        $mid = floor(($lb + $ub) / 2); 
  
        // update lower_bound 
        // if mid contains 0 
        if ($arr[$mid] == 0) 
            $lb = $mid + 1; 
  
        // If mid contains 1 
        else if ($arr[$mid] == 1) 
        { 
              
            // Check if it is the  
            // left most 1 
            // Return mid, if yes 
            if ($mid == 0 or 
                ($mid > 0 and 
                 $arr[$mid - 1] == 0)) 
                return $mid; 
  
            // Else update upper_bound 
            $ub = $mid - 1; 
        } 
    } 
    return -1; 
} 
  
    // Driver Code 
    $arr = array(0, 0, 0, 0, 1, 1); 
    $n = sizeof($arr); 
    $point = findTransitionPoint($arr, $n); 
  
    if($point >= 0)  
        echo "Transition point is " , $point; 
    else
        echo"There is no transition point"; 
    return 0; 
  
// This code is contributed by nitin mittal. 
?> 

Javascript

<script> 
    // Javascript implementation to find the transition point 
      
    // Method to find the transition point 
    function findTransitionPoint(arr, n) 
    { 
        // Initialise lower and upper bounds 
        let lb = 0, ub = n-1; 
       
        // Perform Binary search 
        while (lb <= ub) 
        { 
            // Find mid 
            let mid = parseInt((lb+ub)/2, 10); 
       
            // update lower_bound if mid contains 0 
            if (arr[mid] == 0) 
                lb = mid+1; 
       
            // If mid contains 1 
            else if (arr[mid] == 1) 
            { 
                // Check if it is the left most 1 
                // Return mid, if yes 
                if (mid == 0 
                    || (mid > 0 && 
                       arr[mid - 1] == 0)) 
                    return mid; 
       
                // Else update upper_bound 
                ub = mid-1; 
            } 
        } 
        return -1; 
    } 
      
    let arr = [0, 0, 0, 0, 1, 1]; 
    let point = findTransitionPoint(arr, arr.length); 
  
    document.write(point >= 0 ? "Transition point is " + 
                  point : "There is no transition point"); 
        
</script>

Output

Transition point is 4

Complexity Analysis:

Time Complexity: O(log n). The time complexity for binary search is O(log n).
Auxiliary Space: O(1). No extra space is required.

Find winner in the game of Binary Array

S

Sahil Chhabra

News

Improve

Article Tags :

Practice Tags :

Similar Reads

Find the transition point in a binary array

Given a sorted array containing only numbers 0 and 1, the task is to find the transition point efficiently. The transition point is the point where "0" ends and "1" begins. Examples : Input: 0 0 0 1 1 Output: 3 Explanation: Index of first 1 is 3 Input: 0 0 0 0 1 1 1 1 Output: 4 Explanation: Index of

Find winner in the game of Binary Array

Given a binary array X[] of size N. Two players A and B are playing games having the following rules, then the task is to determine the winner if B starts first and both players play optimally. The rules are: In the first turn, a player selects any index i (1 â‰¤ i â‰¤ N) such that A[i] = 0.After their

Find a String in given Array of Strings using Binary Search

Given a sorted array of Strings arr and a String x, find an index of x if it is present in the array, using Binary Search Examples: Input: arr[] = {"contribute", "geeks", "ide", "practice"}, x = "ide"Output: 2Explanation: The String x is present at index 2. Input : arr[] = {"contribute", "geeks", "i

Find the unvisited positions in Array traversal

Given an array A[] of N positive integers where A[i] represent the units each i can traverse in one step. You can start from position 0 and need to reach destination d. The goal is to find the number of positions that are not visited when all of them have reached position d. Examples: Input: N = 3,

Find a partition point in array

Given an unsorted array of integers. Find an element such that all the elements to its left are smaller and to its right are greater. Print -1 if no such element exists. Note that there can be more than one such elements. For example an array which is sorted in increasing order all elements follow t

Find element with the maximum set bits in an array

Given an array arr[]. The task is to find an element from arr[] which has the maximum count of set bits.Examples: Input: arr[] = {10, 100, 1000, 10000} Output: 1000 Binary(10) = 1010 (2 set bits) Binary(100) = 1100100 (3 set bits) Binary(1000) = 1111101000 (6 set bits) Binary(10000) = 10011100010000

POTD Solutions | 4 Novâ€™ 23 | Find Transition Point

View all POTD Solutions Welcome to the daily solutions of our PROBLEM OF THE DAY (POTD). We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Binary Search algorithm but will also help you build up

Count Subarrays of 1 in Binary Array

Given an array arr[] of size N, the array contains only 1s and 0s, and the task is to return the count of the total number of subarrays where all the elements of the subarrays are 1. Examples: Input: N = 4, arr[] = {1, 1, 1, 0}Output: 6Explanation: Subarrays of 1 will look like the following: [1], [

Count 1's in a sorted binary array

Given a binary array arr[] of size n, which is sorted in non-increasing order, count the number of 1's in it. Examples: Input: arr[] = [1, 1, 0, 0, 0, 0, 0]Output: 2Explanation: Count of the 1's in the given array is 2. Input: arr[] = [1, 1, 1, 1, 1, 1, 1]Output: 7 Input: arr[] = [0, 0, 0, 0, 0, 0,

Minimum number of 1's to be replaced in a binary array

Given a binary array arr[] of zero's and one's only. The task is to find the minimum number of one's to be changed to zero such if there exist any index [Tex]i [/Tex]in the array such that arr[i] = 0 then arr[i-1] and arr[i+1] both should not be equals to [Tex]1 [/Tex]at the same time. That is, for