0918 - Maximum Sum Circular Subarray (Medium)
Problem Link
https://leetcode.com/problems/maximum-sum-circular-subarray/
Problem Statement
Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray ofnums.
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.length1 <= n <= 3 * 10^4-3 * 10^4 <= nums[i] <= 3 * 10^4
Approach 1: Kadane's Algorithm
- C++
- Java
- Python
- Rust
class Solution {
public:
// kadane's algo
int kadane(vector<int>& nums) {
int local = nums[0], global = nums[0];
for (int i = 1; i < nums.size(); i++) {
local = max(nums[i], local + nums[i]);
global = max(global, local);
}
return global;
}
// case 1: max subarray sum in [0 .. n - 1]
// i.e. kadane's algo
// case 2. circular subarray in [0 .. | n - 1 .. | .. 2 * n - 1]
// i.e. total sum - min subarray sum in [0 .. n - 1]
int maxSubarraySumCircular(vector<int>& nums) {
// use kadane's algo to find out max sub array sum (case 1)
int maxSubArraySum = kadane(nums);
// handle cases like [-3,-2,-3]
if (maxSubArraySum < 0) return maxSubArraySum;
// calculate the total sum
int totalSum = accumulate(nums.begin(), nums.end(), 0);
// in order to use the same kadane function, we flip the sign
for (auto &x : nums) x *= -1;
// use kadane's algo to find out min sub array sum
int minSubArraySum = kadane(nums) * -1;
// compare case 1 & case 2, take the max
return max(maxSubArraySum, totalSum - minSubArraySum);
}
};
class Solution {
// kadane's algo
public int kadane(int[] nums) {
int local = nums[0];
int global = nums[0];
for (int i = 1; i < nums.length; i++) {
local = Math.max(nums[i], local + nums[i]);
global = Math.max(global, local);
}
return global;
}
// case 1: max subarray sum in [0 .. n - 1]
// i.e. kadane's algo
// case 2. circular subarray in [0 .. | n - 1 .. | .. 2 * n - 1]
// i.e. total sum - min subarray sum in [0 .. n - 1]
public int maxSubarraySumCircular(int[] nums) {
int n = nums.length;
// use kadane's algo to find out max sub array sum (case 1)
int maxSubArraySum = kadane(nums);
// handle cases like [-3,-2,-3]
if (maxSubArraySum < 0) return maxSubArraySum;
// calculate the total sum
int totalSum = 0;
// in order to use the same kadane function, we flip the sign
for (int i = 0; i < n; i++) {
totalSum += nums[i];
nums[i] = -nums[i];
}
// use kadane's algo to find out min sub array sum
int minSubArraySum = kadane(nums);
// compare case 1 & case 2, take the max
return Math.max(maxSubArraySum, totalSum + minSubArraySum);
}
}
class Solution:
def maxSubarraySumCircular(self, nums):
# kadane's algo
def kadane(nums):
local_sum = nums[0]
global_sum = nums[0]
for i in range(1, len(nums)):
local_sum = max(nums[i], local_sum + nums[i])
global_sum = max(global_sum, local_sum)
return global_sum
# case 1: max subarray sum in [0 .. n - 1]
# i.e. kadane's algo
# case 2. circular subarray in [0 .. | n - 1 .. | .. 2 * n - 1]
# i.e. total sum - min subarray sum in [0 .. n - 1]
n = len(nums)
# use kadane's algo to find out max sub array sum (case 1)
max_sub_array_sum = kadane(nums)
# handle cases like [-3,-2,-3]
if max_sub_array_sum < 0:
return max_sub_array_sum
# calculate the total sum
total_sum = sum(nums)
# in order to use the same kadane function, we flip the sign
for i in range(n):
nums[i] *= -1
# use kadane's algo to find out min sub array sum
min_sub_array_sum = kadane(nums) * -1
# compare case 1 & case 2, take the max
return max(max_sub_array_sum, total_sum - min_sub_array_sum)
use std::cmp::max;
impl Solution {
// kadane's algo
fn kadane(nums: &Vec<i32>) -> i32 {
let mut local = nums[0];
let mut global = nums[0];
for i in 1..nums.len() {
local = max(nums[i], local + nums[i]);
global = max(global, local);
}
return global;
}
// case 1: max subarray sum in [0 .. n - 1]
// i.e. kadane's algo
// case 2. circular subarray in [0 .. | n - 1 .. | .. 2 * n - 1]
// i.e. total sum - min subarray sum in [0 .. n - 1]
fn max_subarray_sum_circular(nums: Vec<i32>) -> i32 {
let n = nums.len();
// use kadane's algo to find out max sub array sum (case 1)
let max_sub_array_sum = Solution::kadane(&nums);
// handle cases like [-3,-2,-3]
if max_sub_array_sum < 0 {
return max_sub_array_sum;
}
// calculate the total sum
let total_sum: i32 = nums.iter().sum();
let mut nums = nums.clone();
for i in 0 .. n {
nums[i] = -nums[i];
}
// use kadane's algo to find out min sub array sum
let min_sub_array_sum = Solution::kadane(&nums);
// compare case 1 & case 2, take the max
return max(max_sub_array_sum, total_sum + min_sub_array_sum);
}
}