DAY 89 - Kadane's Algorithm

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

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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.

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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;

            //adding up the sum
            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.

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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;
    }
};

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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;
    }
};

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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)