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Maximum Subarray Sum (Kadane’s Algorithm)

Difficulty: Medium, Asked-in: Facebook, Microsoft, Amazon, Morgan Stanley, Walmart.

Key takeaways

  • One of the best problems to learn time and space complexity optimisation using various approaches.
  • The idea behind Kadane's algorithm is worth exploring. We can use similar ideas to solve several coding problems.

Let’s understand the problem

Given an array X[] of n integers, write a program to find the maximum sum of a subarray among all subarrays. A subarray is a contiguous segment of elements from X[i] to X[j], where 0 <= i <= j <= n - 1.

  • If array contains all non-negative numbers, the max subarray sum will be the sum of the entire array.
  • We need to to return the value of the maximum subarray sum.

Example 1

Input: X[] = [-4, 5, 7, -6, 10, -15, 3], Output: 16.

Explanation: The subarray [5, 7, -6, 10] has the maximum sum.

Example 2

Input: X[] = [-3, 1, -2, 6, -4, 2], Output: 6.

Explanation: Here single element subarray [6] has the maximum sum.

Example 3

Input: X[] = [5, 7, 6, 10, 3], Output: 31.

Explanation: All array elements are non-negative. So the maximum subarray sum would be the sum of the entire array.

Discussed solution approaches

  • Brute force approach  using three nested loops
  • Using two nested loops
  • Using divide and conquer idea similar to the merge sort
  • Using dynamic programming : Using single loop and O(n) space
  • Kadane's algorithm: Using single loop and variables

Brute force approach  using three nested loops

Solution idea

The most basic solution is to explore all possible subarrays (for all i and j, where i ≤ j), calculate the sum of each subarray and track the maximum among them.

Solution steps

  1. We declare a variable maxSubarraySum to store the maximum subarray sum found so far.
  2. We explore all subarrays (i, j) using a nested loop: the outer loop runs from i = 0 to n - 1, and the inner loop runs from j = i to n - 1.
  3. For each subarray (i, j), we run another loop from k = i to j and calculate the subarray sum between them. We store this value in a variable subarraySum.
  4. If (subarraySum > maxSubarraySum), we update maxSubarraySum with subarraySum.

At the end of the nested loop, we return maxSubarraySum.

Solution pseudocode

findMaxSubarraySum(X, n)
    maxSubarraySum = INT_MIN
    // Consider every possible subarray [i, j]
    for i from 0 to n - 1
        for j from i to n - 1
            subarraySum = 0
            // Calculate the sum of the current subarray
            for k from i to j
                subarraySum = subarraySum + X[k]
            // Keep the largest subarray sum found so far
            if subarraySum > maxSubarraySum
                maxSubarraySum = subarraySum

    return maxSubarraySum

Time and space complexity analysis

We are using three nested loops i.e. exploring each subarray using two outer loops and calculating their sum using the innermost loop. So time complexity = O(n^3). We are using constant extra space, so space complexity = O(1).

Using two nested loops

Solution idea

Now, the critical questions are: Can we optimize the above approach further? Is it necessary to run the innermost loop from k = i to j? If we observe closely, we can easily calculate the subarray sum from i to j + 1 in O(1), if we know the sub-array sum from i to j. The formula is: Subarray sum from i to j + 1 = Subarray sum from i to j + X[j + 1].

So instead of using three nested loops, we can use only two nested loops: the outer loop to pick the starting index, and the inner loop to calculate the sum of all sub-arrays starting from that index.

Solution pseudocode

findMaxSubarraySum(X, n)
    maxSubarraySum = INT_MIN
    // Try each index as the subarray's starting point
    for i from 0 to n - 1
        subarraySum = 0
        // Extend the subarray one element at a time
        for j from i to n - 1
            subarraySum = subarraySum + X[j]
            // Update the maximum sum found so far
            if subarraySum > maxSubarraySum
                maxSubarraySum = subarraySum

    return maxSubarraySum

Time and space complexity analysis

We are using two nested loops to explore each subarray and performing O(1) operation at each iteration. Total count of loop iterations = Total count of different subarrays = n + n - 1 + n - 2 + ... + 2 + 1 = n(n + 1)/2 = O(n²). So time complexity = Total count of loop iterations * O(1) = O(n²). We are using constant extra space, so space complexity = O(1).

Using divide and conquer idea similar to merge sort

Solution idea

Now critical questions are: Can we improve time complexity further? Can we solve this problem using recursion or dividing the problem into smaller subproblems? Let's think!

Suppose we want to calculate the maximum sub-array sum of the array X[l, r], where l and r are the left and right ends. Now we divide the array into two equal subarrays by calculating the mid: X[l, mid] and X[mid + 1, r]. If we observe, the subarray with the maximum sum must lie in one of the following places:

  • Entirely in the left subarray X[l, mid].
  • Entirely in the right subarray X[mid + 1, r].
  • A subarray crossing the mid-point, where one end of the subarray is present in the left half and another end is present on the right half. So here is the relation for any subarray X[i, j] crossing the mid point: l <= i <= mid <= j <= n -1.

We can recursively find the maximum sub-array sum of X[l, mid] and X[mid + 1, r] because these two are smaller instances of the problem of finding the max sub-array sum. Now we need to calculate the maximum sub-array that crosses the mid-point and take the maximum of all three possibilities to get overall max subarray sum.

Base case: When l == r, there is only one element in the array, and we can directly return the maximum subarray sum as X[l] or X[r]. This is the smallest version of the problem, where the recursion will return the value directly.

Finding max subarray sum using divide and conquer approach

Solution pseudocode

getMaxSubarraySum(X, l, r)
    if l == r
        return X[l]

    mid = l + (r - l) / 2
    // Find the max subarray sum entirely in the left half
    leftMaxSum = getMaxSubarraySum(X, l, mid)
    // Find the max subarray sum entirely in the right half
    rightMaxSum = getMaxSubarraySum(X, mid + 1, r)
    // Find the max subarray sum that crosses the midpoint
    crossingMaxSum = maxCrossingSum(X, l, mid, r)
    return max(leftMaxSum, rightMaxSum, crossingMaxSum)

Implementation  maxCrossingSum(X, l, mid, r)

Suppose X[i, j] is the subarray with the maximum sum crossing the midpoint. So we can break this subarray into two parts: 1) Subarray X[i, mid] with the max sum in the left part, 2) Subarray X[mid + 1, j] with the max sum in the right part. So, to get the overall max sum crossing the midpoint, we find these two sums by separately traversing the left and right parts and adding their values. Here are the steps:

Step 1: We find the max contiguous sum of the left half, X[i, mid].

  • We initialize a variable sum to store the continuous sum from mid to some index i and a variable maxLeftSum to store the max sum found so far in the left part.
  • We run a loop starting from index i = mid to l, calculate the contiguous sum, and store it in the variable sum. Whenever sum > maxLeftSum, we update maxLeftSum with sum.
sum = 0
maxLeftSum = INT_MIN
// Find the maximum sum ending at mid
for i from mid down to l
    sum = sum + X[i]
    if sum > maxLeftSum
        maxLeftSum = sum

Step 2: Now we find the maximum contiguous sum of the right half, X[mid + 1, j].

  • We use the same variable sum and initialize it with 0.
  • We initialize a variable maxRightSum to store the max sum found so far in the right part. 
  • Now we run a loop from j = mid + 1 to r, calculate the contiguous sum, and store it in the variable sum. Whenever sum > maxRightSum, we update maxRightSum with sum.
sum = 0
maxRightSum = INT_MIN
// Find the maximum sum starting at mid + 1
for i from mid + 1 to r
    sum = sum + X[i]
    if sum > maxRightSum
        maxRightSum = sum

Step 3: Finally, to get the maximum contiguous subarray sum crossing the mid, we return the sum of variables maxLeftSum and maxRightSum.

Pseudocode to find max subarray sum crossing

maxCrossingSum(X, l, mid, r)
    sum = 0
    maxLeftSum = INT_MIN
    // Find the maximum sum from the midpoint back to l
    for i from mid down to l
        sum = sum + X[i]
        maxLeftSum = max(maxLeftSum, sum)

    sum = 0
    maxRightSum = INT_MIN
    // Find the maximum sum from mid + 1 through r
    for i from mid + 1 to r
        sum = sum + X[i]
        maxRightSum = max(maxRightSum, sum)

    return maxLeftSum + maxRightSum

Time and space complexity analysis

Suppose T(n) is the time complexity of finding the maximum subarray sum using divide and conquer approach. To calculate the overall time complexity, we need to add the time complexities of the divide, conquer, and combine steps.

  • For the base case (n = 1), the algorithm takes constant time => T(1) = O(1).
  • We only need to calculate mid to divide the array. So the time complexity of divide part = O(1). 
  • We are solving two subproblems of size n/2. So the time complexity of conquer part = 2T(n/2).
  • To find the maximum subarray sum crossing the mid, we run two single loops in the opposite direction. In the worst case, each loop will run n/2 times. So the time complexity of combine part = O(n).

Final recurrence relation

  • T(n) = O(1), if n = 1.
  • T(n) = 2T(n/2) + O(n), if n > 1.

This recurrence is the same as a recurrence relation of the merge sort. So overall time complexity = O(nlogn). For a better understanding of this analysis, you can explore analysis of recursion blog.

Space complexity is equal to the size of the recursion call stack, which depends on the height of the recursion tree. At each stage, the input size decreases by 2, so the height of the recursion tree will be O(logn). Space complexity = O(logn). Think!

Using dynamic programming

Solution idea

If we observe, the subarray with maximum sum must be ending at some index in the array. So one idea would be to find the maximum subarray sum ending at all indexes and store their values in an extra memory. Now we can get the max subarray sum of the whole array by finding the maximum of the values stored in the extra memory. The critical question is: How can we implement this? Let's think!

Suppose we take an array maxSumEnding[] of size n, where maxSumEnding[i] is the maximum subarray sum ending at index i. Now how can we fill values in this array? Here is an insight: If we know the max subarray sum ending at i - 1 index, we can easily calculate the max subarray sum ending at index i.

  • maxSumEnding[i] always includes the value X[i].
  • If (maxSumEnding[i - 1] > 0), we need to include maxSumEnding[i - 1] to calculate maxSumEnding[i]. The idea is simple: maxSumEnding[i - 1] + X[i] will be always greater than X[i]. So in this case, we update maxSumEnding[i] = maxSumEnding[i - 1] + X[i].
  • If (maxSumEnding[i -1] < 0), we need to ignore maxSumEnding[i -1], because maxSumEnding[i - 1] + X[i] will be always less than X[i]. So we just update maxSumEnding[i] = X[i].
  • In simpler terms, we can write: maxSumEnding[i] = max(maxSumEnding[i - 1] + X[i], X[i]).

Finally, to get the maximum subarray sum of the whole array, we return the maximum value stored in the array maxSumEnding[]. This problem falls under the dynamic programming category because we are storing the solution of sub-problems and solving the larger problem using the solution of smaller sub-problems.

Solution steps

  1. We create an extra array maxSumEnding[] of size n.
  2. We initialize the array with the first value, i.e., maxSumEnding[0] = X[0].

  3. Now we traverse the array from i = 1 to n - 1. At ith iteration, we calculate the max subarray sum ending at index i and store it at maxSumEnding[i].

    • If (maxSumEnding[i - 1] < 0), maxSumEnding[i] = X[i].
    • Otherwise, maxSumEnding[i] = X[i] + maxSumEnding[i - 1].
  4. We return the maximum value stored in maxSumEnding[].

Solution pseudocode

findMaxSubarraySum(X, n)
    maxSumEnding = array of size n
    maxSumEnding[0] = X[0]
    // Find the maximum subarray sum ending at each index
    for i from 1 to n - 1
        if maxSumEnding[i - 1] > 0
            maxSumEnding[i] = X[i] + maxSumEnding[i - 1]
        else
            maxSumEnding[i] = X[i]

    maxSubarraySum = INT_MIN
    // Find the largest value among the computed sums
    for i from 0 to n - 1
        maxSubarraySum = max(maxSubarraySum, maxSumEnding[i])

    return maxSubarraySum

Here is another implementation: Instead of calculating the max of the maxSumEnding[] in a separate loop, we can track the maximum on the go in the same loop where we are updating maxSumEnding[].

findMaxSubarraySum(X, n)
    maxSumEnding = array of size n
    maxSumEnding[0] = X[0]
    maxSubarraySum = X[0]
    for i from 1 to n - 1
        if maxSumEnding[i - 1] > 0
            maxSumEnding[i] = X[i] + maxSumEnding[i - 1]
        else
            maxSumEnding[i] = X[i]

        // Update the largest subarray sum found so far
        maxSubarraySum = max(maxSubarraySum, maxSumEnding[i])

    return maxSubarraySum

Time and space complexity analysis

We are traversing the array and performing O(1) operation at each iteration. Time complexity = O(n). Space complexity = O(n), for extra array maxSumEnd[n].

Efficient solution using Kadane's algorithm

Solution idea

Can we solve this problem in O(1) space? Can we solve this problem in a single traversal? Here is the thought process for optimising the solution using Kadane's algorithm.

If we observe the filling pattern of array in the previous approach, we only need one previous value at i - 1 index to calculate the max subarray sum ending at index i. So instead of using an extra array of size n, we can track the max subarray sum ending at the index i using some variable maxSumEndingHere.

  • Before the ith iteration, maxSumEndingHere will store the max subarray sum ending at the i - 1 index. So to calculate the max subarray sum ending at index i, we can use the idea of the previous approach i.e. maxSumEndingHere = max(maxSumEndingHere + X[i], X[i]).
  • During this process, we track the max subarray sum so far using another variable maxSumSoFar i.e. if (maxSumEndingHere > maxSumSoFar), maxSumSoFar = maxSumEndingHere.

So in this approach, we can use a single loop and two extra variables. This means there is no need for O(n) space. This is the Kadane's algorithm idea! We can use similar ideas to solve various questions.

Solution steps

  1. We initialize variables maxSumSoFar and maxSumEndingHere with X[0].

  2. Now we run a loop from i = 1 to n - 1. At each ith iteration:

    • We calculate the max subarray sum ending at the ith index, i.e., maxSumEndingHere = max (maxSumEndingHere + X[i], X[i]).
    • We also update maxSumSoFar, i.e., if (maxSumSoFar < maxSumEndingHere), maxSumSoFar = maxSumEndingHere.
  3. By the end of the loop, the maximum subarray sum value gets stored in the variable maxSumSoFar. We return this value as the output.

Solution pseudocode

findMaxSubarraySum(X, n)
    maxSumSoFar = X[0]
    maxSumEndingHere = X[0]
    for i from 1 to n - 1
        maxSumEndingHere = max(maxSumEndingHere + X[i], X[i])
        // Track the largest subarray sum found so far
        if maxSumEndingHere > maxSumSoFar
            maxSumSoFar = maxSumEndingHere

    return maxSumSoFar

Time and space complexity analysis

We are traversing each element using a single loop and performing a constant operation. So time complexity = O(n). Since we are using a constant number of variables, space complexity = O(1).

Critical ideas to think!

  • How can we modify the above approaches to find the start and end indices of the subarray with the maximum sum?
  • Do the above algorithms handle the case of negative numbers?
  • Can we think of solving this problem using other approaches?
  • How can we design a solution to calculate the maximum sum submatrix of size k x k in the m x n matrix?
  • How do we solve the problem if a sorted array is provided as an input?

Comparison of time and space complexities

  • Three nested loops: Time = O(n^3), Space = O(1).
  • Two nested loops: Time = O(n^2), Space = O(1).
  • Divide and conquer: Time = O(nlogn), Space = O(logn).
  • Dynamic programming: Time = O(n), Space = O(n).
  • Kadane's algorithm: Time = O(n), Space = O(1).

Suggested coding questions to practice

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