# Max Consecutive Ones (Kotlin) Solutions

I am currently learning Data structure and Algorithms deeper in kotlin and I decided to start with Leetcode.

The problem

Given a binary array `nums`

, return *the maximum number of consecutive *`1`

*'s in the array*.

**Example 1:**

**Input:** nums = [1,1,0,1,1,1]

**Output:** 3

**Explanation:** The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s. The maximum number of consecutive 1s is 3.

**Solution:**

The first step is to understand the problem and think of the best possible way to solve it. what Leetcode wants you to do is get the highest number of consecutive **(following each other continuously) **1`s in the given array

**Examples:**

nums = [1,1,0,**1,1,1**] the correct answer is 3

nums = [1,1,0,**1,1,1,1**] the correct answer is 4

nums = [1,0,0,**1,1**,0] the correct answer is 2

**Kotlin Code**

**Code Explanation**

there are so many ways to solve a problem, but this is how I solved mine,

Firstly, you have to create two variables that will hold the largest number and get the count for the current value.

which I called current and largest and assign 0 to them.

`var current = 0`

var largest = 0

Secondly, you will loop through the array of 0 and 1 array, get the elements of the array and check for the value 0.

`for (u in ty) {`

if (u == 0) {

if you find 0, then set the current to 0, and if not zero(0) then increment the current by 1.

`current = 0`

} else {

current++

}

Thirdly, Before ending the current iteration of the loop, you will check if the `current`

is greater than the `largest`

.

If yes, set `current`

to `largest`

.

`if (current > largest) largest = current`

Then return the `largest`

Thanks for reading.