LeetCode #918 — MEDIUM

Maximum Sum Circular Subarray

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums.

A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].

A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.

Example 1:

Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.

Example 2:

Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.

Example 3:

Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.

Constraints:

n == nums.length
1 <= n <= 3 * 10⁴
-3 * 10⁴ <= nums[i] <= 3 * 10⁴

Step 01

Brute Force Baseline

Problem summary: Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums. A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n]. A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Monotonic Queue

Example 1

[1,-2,3,-2]

Example 2

[5,-3,5]

Example 3

[-3,-2,-3]

Step 02

Core Insight

What unlocks the optimal approach

For those of you who are familiar with the <b>Kadane's algorithm</b>, think in terms of that. For the newbies, Kadane's algorithm is used to finding the maximum sum subarray from a given array. This problem is a twist on that idea and it is advisable to read up on that algorithm first before starting this problem. Unless you already have a great algorithm brewing up in your mind in which case, go right ahead!
What is an alternate way of representing a circular array so that it appears to be a straight array? Essentially, there are two cases of this problem that we need to take care of. Let's look at the figure below to understand those two cases: <br> <img src="https://assets.leetcode.com/uploads/2019/10/20/circular_subarray_hint_1.png" width="700"/>
The first case can be handled by the good old Kadane's algorithm. However, is there a smarter way of going about handling the second case as well?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #918: Maximum Sum Circular Subarray
class Solution {
    public int maxSubarraySumCircular(int[] nums) {
        final int inf = 1 << 30;
        int pmi = 0, pmx = -inf;
        int ans = -inf, s = 0, smi = inf;
        for (int x : nums) {
            s += x;
            ans = Math.max(ans, s - pmi);
            smi = Math.min(smi, s - pmx);
            pmi = Math.min(pmi, s);
            pmx = Math.max(pmx, s);
        }
        return Math.max(ans, s - smi);
    }
}

// Accepted solution for LeetCode #918: Maximum Sum Circular Subarray
func maxSubarraySumCircular(nums []int) int {
	const inf = 1 << 30
	pmi, pmx := 0, -inf
	ans, s, smi := -inf, 0, inf
	for _, x := range nums {
		s += x
		ans = max(ans, s-pmi)
		smi = min(smi, s-pmx)
		pmi = min(pmi, s)
		pmx = max(pmx, s)
	}
	return max(ans, s-smi)
}

# Accepted solution for LeetCode #918: Maximum Sum Circular Subarray
class Solution:
    def maxSubarraySumCircular(self, nums: List[int]) -> int:
        pmi, pmx = 0, -inf
        ans, s, smi = -inf, 0, inf
        for x in nums:
            s += x
            ans = max(ans, s - pmi)
            smi = min(smi, s - pmx)
            pmi = min(pmi, s)
            pmx = max(pmx, s)
        return max(ans, s - smi)

// Accepted solution for LeetCode #918: Maximum Sum Circular Subarray
impl Solution {
    pub fn max_subarray_sum_circular(nums: Vec<i32>) -> i32 {
        let (mut global_max, mut global_min) = (nums[0], nums[0]);
        let (mut current_max, mut current_min) = (0, 0);
        let mut total = 0;

        for num in nums {
            current_max = i32::max(num, current_max + num);
            current_min = i32::min(num, current_min + num);
            total += num;
            global_max = i32::max(global_max, current_max);
            global_min = i32::min(global_min, current_min);
        }

        if global_max > 0 {
            return i32::max(global_max, total - global_min);
        } else {
            return global_max;
        }
    }
}

// Accepted solution for LeetCode #918: Maximum Sum Circular Subarray
function maxSubarraySumCircular(nums: number[]): number {
    let [pmi, pmx] = [0, -Infinity];
    let [ans, s, smi] = [-Infinity, 0, Infinity];
    for (const x of nums) {
        s += x;
        ans = Math.max(ans, s - pmi);
        smi = Math.min(smi, s - pmx);
        pmi = Math.min(pmi, s);
        pmx = Math.max(pmx, s);
    }
    return Math.max(ans, s - smi);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.