Find the transition point in a binary array

Last Updated : 28 Mar, 2025

Given a sorted array, arr[], containing only 0s and 1s, find the transition point, i.e., the first index where 1 was observed, and before that, only 0 was observed. If arr does not have any 1, return -1. If the array does not have any 0, return 0.

Examples :

Input: 0 0 0 1 1
Output: 3
Explanation: Index 3 is the transition point where first 1 was observed.

Input: arr[] = [0, 0, 0, 0]
Output: -1
Explanation: Since, there is no “1”, the answer is -1.

Input: 0 0 0 0 1 1 1 1
Output: 4
Explanation: Index 4 is the transition point where first 1 was observed.

Input: arr[] = [1, 1, 1]
Output: 0
Explanation: There are no 0s in the array, so the transition point is 0.

[Naive Approach] Using a for Loop – O(n) Time and O(1) Space

The idea for this approach is to use a loop to traverse the array and print the index of the first 1.

C++

#include<bits/stdc++.h> using namespace std;  // Function to find the transition point int transitionPoint(vector<int> arr) {     int n = arr.size();          //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; }  int main() {     vector<int> arr = {0, 0, 0, 0, 1, 1};      cout<<transitionPoint(arr);     return 0; }

Java

import java.util.*;  class GfG {  // Function to find the transition point static int transitionPoint(int arr[]) {     int n = arr.length;          // 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; }  public static void main (String[] args)     {         int arr[] = {0, 0, 0, 0, 1, 1};                  System.out.print(transitionPoint(arr));     } }

Python

def transitionPoint(arr):     n = len(arr)          # 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   arr = [0, 0, 0, 0, 1, 1]  print(transitionPoint(arr))

using System;  class GfG  {  // Function to find the transition point static int transitionPoint(int []arr) {     int n = arr.Length;          // 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; }     public static void Main()      {         int []arr = {0, 0, 0, 0, 1, 1};         Console.Write(transitionPoint(arr));     } }

JavaScript

function transitionPoint(arr)     {         let n =  arr.length;                  // 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];     console.log(transitionPoint(arr));

Output

[Expected Approach] – Using Binary Search – O(log n) Time and O(1) Space

The idea is to use Binary Search, and find the smallest index of 1 in the array. As the array is sorted.

Use two variables (l, r) to point left end and right end of the array and 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 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.

C++

#include<bits/stdc++.h> using namespace std;  // Function to find the transition point int findTransitionPoint(vector<int> arr) {     int n = arr.size();          // 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; }  int main() {     vector<int> arr = {0, 0, 0, 0, 1, 1};     cout<<findTransitionPoint(arr);     return 0; }

Java

class GfG {     // Method to find the transition point     static int transitionPoint(int arr[])     {         int n = arr.length;          // 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;     }      public static void main(String args[])     {         int arr[] = { 0, 0, 0, 0, 1, 1 };          System.out.println(transitionPoint(arr));     } }

Python

def transitionPoint(arr):     n = len(arr)     # 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         elif(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  arr = [0, 0, 0, 0, 1, 1] print(transitionPoint(arr));

using System;          class GfG  {     // Method to find the transition point     static int transitionPoint(int []arr)     {         int n = arr.Length;         // 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;     }          public static void Main()      {         int []arr = {0, 0, 0, 0, 1, 1};         Console.Write(transitionPoint(arr));     } }

JavaScript

function transitionPoint(arr)     {         let n = arr.length         // 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];     console.log(transitionPoint(arr));

Output

Find winner in the game of Binary Array

Sahil Chhabra

Improve

Article Tags :

Practice Tags :

Find the transition point in a binary array

[Naive Approach] Using a for Loop – O(n) Time and O(1) Space

[Expected Approach] – Using Binary Search – O(log n) Time and O(1) Space

Similar Reads