53. Maximum Subarray

53. Maximum Subarray
Photo by Takashi Miyazaki / Unsplash


Given an integer array nums, find the

subarraywith the largest sum, and returnits sum.

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.


  • 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.


Bruteforce Solution


  1. Initialize a variable maxSubarray = -infinity to keep track of the best subarray. We need to use negative infinity, not 0, because it is possible that there are only negative numbers in the array.
  2. Use a for loop that considers each index of the array as a starting point.
  3. For each starting point, create a variable currentSubarray = 0. Then, loop through the array from the starting index, adding each element to currentSubarray. Every time we add an element it represents a possible subarray - so continuously update maxSubarray to contain the maximum out of the currentSubarray and itself.
  4. Return maxSubarray.
class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        max_subarray = -math.inf
        for i in range(len(nums)):
            current_subarray = 0
            for j in range(i, len(nums)):
                current_subarray += nums[j]
                max_subarray = max(max_subarray, current_subarray)
        return max_subarray

Dynamic Programming


Whenever you see a question that asks for the maximum or minimum of something, consider Dynamic Programming as a possibility. The difficult part of this problem is figuring out when a negative number is "worth" keeping in a subarray. This question in particular is a popular problem that can be solved using an algorithm called Kadane's Algorithm. If you're good at problem solving though, it's quite likely you'll be able to come up with the algorithm on your own. This algorithm also has a very greedy-like intuition behind it.

Let's focus on one important part: where the optimal subarray begins. We'll use the following example.

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

We can see that the optimal subarray couldn't possibly involve the first 3 values - the overall sum of those numbers would always subtract from the total. Therefore, the subarray either starts at the first 4, or somewhere further to the right.

What if we had this example though?

nums = [-2,1000000000,-3,4,-1,2,1,-5,4]

We need a general way to figure out when a part of the array is worth keeping.

As expected, any subarray whose sum is positive is worth keeping. Let's start with an empty array, and iterate through the input, adding numbers to our array as we go along. Whenever the sum of the array is negative, we know the entire array is not worth keeping, so we'll reset it back to an empty array.

However, we don't actually need to build the subarray, we can just keep an integer variable current_subarray and add the values of each element there. When it becomes negative, we reset it to 0 (an empty array).


Initialize 2 integer variables. Set both of them equal to the first value in the array.

  • currentSubarray will keep the running count of the current subarray we are focusing on.
  • maxSubarray will be our final return value. Continuously update it whenever we find a bigger subarray.
  1. Iterate through the array, starting with the 2nd element (as we used the first element to initialize our variables). For each number, add it to the currentSubarray we are building. If currentSubarray becomes negative, we know it isn't worth keeping, so throw it away. Remember to update maxSubarray every time we find a new maximum.
  2. Return maxSubarray.


A clever way to update currentSubarray is using currentSubarray = max(num, currentSubarray + num). If currentSubarray is negative, then num > currentSubarray + num.

class Solution(object):
    def maxSubArray(self, nums):
        :type nums: List[int]
        :rtype: int
        current_subarray = nums[0]
        max_subarray = nums[0]

        # since current_subarray uses our first element
        for i in nums[1:]:
            # Work out whether or not we reset our current subarray 
            # so if we go 1 6, -10
            # we do not want to include -10, so we find the max here
            # equally if we go -10, -5, 6
            # we want to start at 6 instead of -10 as it's higher
            current_subarray = max(i, current_subarray + i)

            # Our max subarray is the maximum between what we currently have vs the max we've found
            # since we do not need to remember which elements make up the max we can just store the value
            max_subarray = max(max_subarray, current_subarray)
        return max_subarray