Nayan Pahuja

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# DAY 89 - Kadane's Algorithm

Hey Folks!. Welcome to day 89 of 100DaysOfCode Challenge. We shall be covering the kadane's algorithm today.

## Question: Maximum Subarray

Given an integer array `nums`, find the subarray with the largest sum, and return its sum.

### Examples:

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: The subarray [4,-1,2,1] has the largest sum 6.

Example 2:

Input: nums = [1]
Output: 1
Explanation: The subarray [1] has the largest sum 1.

Example 3:

Input: nums = [5,4,-1,7,8]
Output: 23
Explanation: The subarray [5,4,-1,7,8] has the largest sum 23.

Constraints:

1 <= nums.length <= 105

-104 <= nums[i] <= 104

Follow up: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

## Question Breakdown:

The question is pretty simple. Let's define what it wants:-

• Subarray

A subarray is a contiguous non-empty sequence of elements within an array.

• We need to find the largest sum of all the subarrays.

### Intuition:

#### Brute Force Approach:

• In a brute force approach or naïve approach, we can run three loops to find it out.
• We will use the first two for loops to form the subarray and another for loop using a third pointer to find the sum inside that subarray.
• We will then update the maximum sum found accordingly.

#### Code: Brute Force Approach:

``````
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int maxi = INT_MIN; //to store our max sum
int n = nums.size();
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
// subarray = arr[i.....j]
int sum = 0;

for (int k = i; k <= j; k++) {
sum += nums[k];
}

maxi = max(maxi, sum);
}
}

return maxi;
}
};

``````

This gives us a time limit exceeded error as it is a solution of third-order (cubic) time complexity.

#### Better Approach:

• If we check our previous approach carefully, we are adding the next element to the previously calculated sum.
• At a moment we don't need all the elements but the previously calculated sum and the next element to add.
• So if we try to optimize it, we can remove our third for loop and calculate the sum while our j pointer is traversing as it can also accumulate the sum for that subarray.

#### Code:

``````
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int maxi = INT_MIN; // maximum sum
int n = nums.size();
for (int i = 0; i < n; i++) {
int sum = 0;
for (int j = i; j < n; j++) {

sum += nums[j];

maxi = max(maxi, sum); // getting the maximum
}
}

return maxi;
}
};

``````

Time Complexity: O(N2), where N = size of the array.
Reason: We are using two nested loops, each running approximately N times.

This also gives us a TLE, so we need to find a better approach.

### Optimal Approach Intuition:

• The intuition of Kadane's Algorithm says that if a subarray is contributing a negative sum, it should not be considered further.
• Instead of calculating the actual subarray Kadane's algorithm asks us to move further than negative contributions and count only when the sum is not negative.
• Here, we will traverse the given array with a single loop and while iterating we will add the elements to our sum. If at any point the sum becomes less than 0, we will set the sum as 0 as we are not going to consider any subarray with a negative sum.

#### Approach:

• Initialize two variables sum and maxi to calculate the temporary sum and our final result.
• Traverse the array using a for loop with iterator i to the last index and keep adding our sum to the sum variable.
• If at any point our sum is negative means we should not consider this subarray as a part of our answer otherwise update the maximum
• Return our final result as maxi.

#### Code:

``````
class Solution {
public:
int maxSubArray(vector<int>& nums) {
long long sum = 0;
long long maxi = LONG_MIN;
for(int i = 0; i < nums.size(); i++){
sum += nums[i];

if(sum > maxi){
maxi = sum;
}

if(sum < 0){
sum = 0;
}
}
return maxi;
}
};

``````

BONUS: Printing the subarray:
If we want to print the subarray we use the following approach in addition to our optimal approach.

``````
class Solution {
public:
int maxSubArray(vector<int>& nums) {
long long sum = 0;
long long maxi = LONG_MIN;
int ansStart = -1;
int ansEnd = -1;
int tempStart = 0;
for(int i = 0; i < nums.size(); i++){

if(sum == 0) tempStart = i; //means our last subarray contributed negative and we have to start new.

sum += nums[i];

if(sum > maxi){
maxi = sum;
ansStart = tempStart;
ansEnd = i;
}

if(sum < 0){
sum = 0;
}
}

for(int i = ansStart; i<=ansEnd; i++){
cout << nums[i] << " ";
}

return maxi;
}
};

``````

## Complexity Analysis:

Algorithm Time Complexity Space Complexity
Binary Search Approach O(N3) O(1)
Better Approach: O(N2) O(1)
Optimal Approach: O(N) O(1)