HiπDevs,

In the previous post, we have solved and understood the ** Next Permutation** problem and in this post, we will tackle the next one

**Kadaneβs algorithms**.

## #4 Kadane's Algorithms

*in this problem we have given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.*

Example 1:

**Input**: `nums`

= [-2,1,-3,__ 4,-1,2,1__,-5,4]

**Output**:

`6`

*Explanation*: [4,-1,2,1] has the largest sum = 6.

Example 2:

**Input**: `nums`

= [__ 1__]

**Output**:

`1`

##
__Solution__

we can easily solve this problem by using the kadane's algorithms.

lets discuss the Kadane's algorithms step by step.

**step-1** initialize three int variable `sum = 0`

,`max = nums[0]`

, `arrSize = nums.length`

.

**step-2** run a loop from `i = 0`

to `arrSize`

.

1.sum = sum + nums[i].

2.if`sum > max`

then`max = sum`

.

3.if`sum < 0`

then`sum = 0`

.

**step-3** return `max`

**step-4** end

**why reinitialize sum = 0 if sum < 0?**

*negative-sum never contribute to maximum sum, so it is better to reinitialize sum to 0 and start adding from next element*

before coding this algorithm in java, have a look at this image for a better understanding of this algo.

Java

```
class Solution {
public int maxSubArray(int[] nums) {
int sum = 0;
int max = nums[0];
int arrSize = nums.length;
for(int i=0; i < arrSize; i++){
sum = sum + nums[i];
if(sum > max){
max = sum;
}
if(sum < 0) {
sum =0
}
}
return max;
}
}
```

**But how? can we know,**

length of the max subarray

start & end index of the max subarray

elements of the max subarray

so, for these, we can make two more int variables `firstIndex`

& `lastIndex`

that store the start & last index number of the max subarray and then do some modification in the **step-2**.

step-2run a loop from`i = 0`

to`arrSize`

.

1.sum = sum + nums[i].

2.if`sum > max`

then`max = sum`

,.`lastIndex = i`

3.if`sum < 0`

then`sum = 0`

,.`firstIndex = i+1`

by using `firstIndex`

and `lastIndex`

variables, now we can answer the following questions.

**length of the subarray**.

`subArrSize = lastIndex - firstIndex`

**start & end index of subarray**.

by using`firstIndex`

and`lastIndex`

variable**elements of the max subarray**.

by traversing from`firstIndex`

to`lastIndex`

we can find the elements

Java

```
class Solution {
public int maxSubArray(int[] nums) {
int sum = 0;
int max = nums[0];
int arrSize = nums.length;
int firstIndex = 0,lastIndex = 0;
for(int i=0; i < arrSize; i++){
sum = sum + nums[i];
if(sum > max){
max = sum;
lastIndex = i;
}
if(sum < 0) {
sum =0;
firstIndex = i + 1;
}
}
return max;
}
}
```

Time Complexityβ±οΈ

running a for loop from 0 to arrSize.

so, **O(arrSize)** will be it's time complexity.

Space Complexityβ°οΈ

the algo is not using any extra space, then **O(1)** will be its space complexity.

if you like this article please let me know in the comment section.

if you find anything wrong in the post,plz correct me.

Thank you for reading my article.

## Top comments (2)

Wish you good luck to complete the sheet successfully.

thanks bro