Kadane's Algorithm for Largest Subarray Sum:

• Kadane's Algorithm is quite a popular algorithm for finding the largest sum of SubArrays in linear time with constant Space Complexity.
• So now first let's see what are SubArrays:

What are SubArrays:

• As the name suggests, SubArrays are Subset of Arrays or SubArrays are Contiguous part of Arrays.
• Let's look at an example for better understanding:

Now that we have an understanding of what SubArrays are, let's look at What is the problem that we have to solve:

Problem:

We have to find the largest SubArray Sum in the given array of integers

Approaches or Solutions:

1. Naive Approach or Brute Force Approach:

Idea:

• We run two nested loops to traverse each element and for each element, we find its sum with other elements. We update the Maximum Sum as soon as we get a larger value.
• After completion, we will get the maximum sum of the subarray.
• Time Complexity: O(N²) and Space Complexity: O(1) where N = Number of elements in the Array

Code:

#include <algorithm>

#include <iostream>

#include <bits/stdc++.h>

using namespace std;

int maximumSubarraySum(vector < int > array) {

  int length = array.size();

  int maxSum = INT_MIN;



  for (int i = 0; i <= length - 1; i++) {

    int currSum = 0;

    for (int j = i; j <= length - 1; j++) {

      currSum += arr[j];

      if (currSum > maxSum) {

        maxSum = currSum;

      }

    }

  }



  return maxSum;

}

2. Efficient Approach:

Idea:

• We maintain two pointers, maxSum (For finding maximum Sum) and currSum(For finding current Sum).
• We traverse through the array once and we do the following:
  • If we get the positive element, we add it to currentSum and check if maxSum < currSum:
    • If yes, we update the maxSum to currSum else we just move on
  • If we get a negative element, we initialize currentSum to zero again
• Time Complexity: O(n), Space Complexity: O(1)

Code:

int maximumSubarraySum(int arr[], int n) {

 int maxSum = 0;

 int currSum = 0;


 for (i = 0; i <= n - 1; i++) {

   currSum += arr[i];

   if (currSum > maxSum) { // If current sum is greater than maxSum, we update it

     maxSum = currSum;

   }

   if (currSum < 0) { // If we get a negative, we update currSum = 0

     currSum = 0;

   }

 }

 return maxSum;

}

3. Modified Kadane's Approach:

• Above approach will return 0 if the entire array is of negative elements, So the only modification will be to initialize maxSum = INT_MIN (INT_MIN is a C++ variable that denotes a huge negative value (-2147483647 - 1)).

Code:

int maximumSubarraySum(int arr[], int n) {

 int maxSum = INT_MIN;

 int currSum = 0;


 for (i = 0; i <= n - 1; i++) {

   currSum += arr[i];

   if (currSum > maxSum) { // If current sum is greater than maxSum, we update it

     maxSum = currSum;

   }

   if (currSum < 0) { // If we get a negative, we update currSum = 0

     currSum = 0;

   }

 }

 return maxSum;

}

Example: