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# Find the transition point in a binary array

• Difficulty Level : Easy
• Last Updated : 13 Jul, 2022

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```
Recommended Practice

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

1. Traverse the array from the start to the end of the array.
2. 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``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= 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

 ``

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.

1. Create two variables, l and r, initialize l = 0 and r = n-1 and a variable ans = -1 to store the answer.
2. Iterate the steps below till l <= r, the lower-bound is less than the upper-bound.
3. Check if the element at middle index mid = (l+r)/2, is one or not.
4. 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.
5. 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``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);` `    ``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

 ` 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

 ``

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.

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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