# Find Transition Point Problem in Java

Given a sorted array that only contains 0s and 1s, find the transition point or the first index where 1 was observed and only 0 was observed before.

Our task is to finish the transitionPoint() function, which accepts arrays a and N as input parameters and outputs the point where 0 stops and 1 begins on a 0-based index. Return -1 in the event that the array contains no 1s. Return 0 if there are no 0 values in the array.

Example 1:

Input:

`int N = 6int a[] = {0, 0, 0, 1, 1, 1}`

Output:

`The transition point is 3`

Explanation:

At index 3, the occurrence of 1 begins. Therefore, the transition point is 3.

Example 2:

Input:

`int N = 5int a[] = {0, 0, 0, 0, 0}`

Output:

`The transition point is -1`

Explanation:

Since no -1 exists in the array, the transition point is -1.

Example 3:

Input:

`int N = 5int a[] = { 0, 1, 0, 1, 1}`

Output:

`The transition point is 1`

Explanation:

At index 1, the occurrence of 1 begins; therefore, the transition point is 1.

## Approach: Using if-Else

More specifically, the function identifies the index where the array transitions from 0 to 1 and defines a class called FindTransitionPoint with a method called transitionPoint. As it iterates across the array, it looks for pairs of nearby elements to determine where the transition point is. This method returns the index of the first instance of a 1 after a 0, assuming the array consists of alternating 0s and 1s. Its linear time complexity and inability to handle situations in which the array deviates from this pattern make it potentially inefficient for large arrays.

## Implementation:

FileName: FindTransitionPoint.java

`import java.io.*;import java.util.*;public class FindTransitionPoint{    int transitionPoint(int arr[], int n)    {        //If the first element of the array        // s 1, then return 0 as an index        if(arr[0]==1)            return 0;        for(int i=1;i<arr.length;i=i+2)        {            // if the array's current element is 1            // and the array's previous element is 0            // return current element index            if(arr[i]==1 && arr[i-1]==0)            {                return i;            }            // if the array's current element is 1            // and the array's previous element is 1            // return previous element index            else if(arr[i]==1 && arr[i-1]==1)            {                return i-1;            }        }        //If the array does not contain 1 return -1        return -1;    }    public static void main(String[] args)    {        FindTransitionPoint obj = new FindTransitionPoint();        int[] arr = {0, 1, 0, 1, 1};        int n = arr.length;        System.out.println("The transition point of the given array is: " + obj.transitionPoint(arr, n));    }}`

Output:

`The transition point of the given array is: 1`

Complexity Analysis:

The time complexity is O(N), and the space complexity is O(1), where N represents the size of the given array.