
Problem Statement
Given a circular integer array nums
, which connects at the end back to its own start, the task is to determine the maximum possible sum attainable by any non-empty subarray of this array. This means that for any index i
of nums
, the next index is (i + 1) % n
, and the index before it is (i - 1 + n) % n
, reflecting its circular structure. A subarray, by definition here, ensures that there are no overlapping elements directly due to the array's cyclical nature. Therefore, a subarray nums[i] ... nums[j]
includes elements such that they are not visited more than once in the subarray within one loop from start to end of the circular construction of nums
.
Examples
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
Approach and Intuition
To tackle this problem of finding the maximum subarray sum in a circular integer array, one effective approach follows these reasoning strategies:
- In a typical linear array, subarray sums can be efficiently calculated using Kadane's Algorithm. However, the circular nature of this problem adds complexity.
- The maximum circular subarray sum could essentially be formed by either:
- A subarray that doesn't cross the boundaries of the array.
- A subarray that crosses the boundaries (wraps around).
- For subarrays not crossing boundaries:
- Directly apply Kadane's Algorithm to find the maximum subarray sum.
- For subarrays that cross the boundaries, max crossing subarray sum can be formulated by:
- Identifying the minimum subarray sum using Kadane's Algorithm.
- Subtracting this sum from the total sum of the array gives an indirect value of the wrapping subarray.
- The solution is the maximum value between the largest sum obtained by Kadane’s Algorithm without wrapping and the sum derived from the total array sum minus the smallest subarray sum. This considers both internal and wrapping subarrays.
- Implementing this involves iterating over the array to compute both the total sum and the potential maximum/minimum subarray sums, taking into account various edge cases represented by constraints and input sizes.
Examples Illustrated:
Example 1:
nums = [1, -2, 3, -2]
— Direct application of Kadane's provides a maximum subarray sum of3
using just[3]
.
Example 2:
nums = [5, -3, 5]
— Due to the wrapping potential, one can take subarray[5, 5]
from the ends of the array which sum up to10
, whereas traditional Kadane would have opted for[5]
at either instance providing suboptimal5
.
Example 3:
nums = [-3, -2, -3]
— All negative numbers, but Kadane's still picks the least negative[-2]
as the maximal sum possible.
These examples show that careful consideration between simple linear subarrays and those that take advantage of the circular nature of the array is necessary to correctly solve this problem.
Solutions
- C++
- Java
class Solution {
public:
int maxSumCircularSubarray(vector<int>& arr) {
int currentMax = 0;
int currentMin = 0;
int highestSum = arr[0];
int lowestSum = arr[0];
int sumTotal = 0;
for (int value: arr) {
// Standard Kadane's to find maximum subarray sum
currentMax = max(currentMax, 0) + value;
highestSum = max(highestSum, currentMax);
// Inverse Kadane's to find the minimum subarray sum
currentMin = min(currentMin, 0) + value;
lowestSum = min(lowestSum, currentMin);
sumTotal += value;
}
if (sumTotal == lowestSum) {
return highestSum;
}
return max(highestSum, sumTotal - lowestSum);
}
};
The given C++ code implementation addresses the problem of finding the maximum sum of a circular subarray from a given list of integers. Here's how the algorithm effectively tackles this problem:
- Initialize variables to keep track of the maximum and minimum subarray sums (
currentMax
,currentMin
), the highest and lowest sums found (highestSum
,lowestSum
), and the total sum of the array (sumTotal
). - Iterate through each element in the array:
- Update
currentMax
using a variation of Kadane’s algorithm to capture the maximum subarray sum without wrapping. - Update
currentMin
simultaneously, implementing an inverted form of Kadane’s algorithm to determine the minimum subarray sum. - Accumulate the total sum of the array in
sumTotal
.
- Update
- Establish a check-condition to ensure the entire array sum is not a minimum subarray sum, which could happen if all elements are negative.
- Return the maximum value between
highestSum
and the difference ofsumTotal
andlowestSum
. This step takes into consideration both the non-wrapped and wrapped subarrays.
This approach ensures the solution handles both scenarios where the maximum sum subarray is either contiguous in the regular order or spans the circular boundary of the array. Make sure the input vector is non-empty to avoid undefined behavior.
class Solution {
public int maximumCircularSubarraySum(int[] array) {
int currentMax = 0;
int currentMin = 0;
int highestSum = array[0];
int lowestSum = array[0];
int sum = 0;
for (int value : array) {
// Implementing maximum subarray sum using Kadane's Algorithm
currentMax = Math.max(currentMax, 0) + value;
highestSum = Math.max(highestSum, currentMax);
// Implementing minimum subarray sum to find out the minimum subarray
currentMin = Math.min(currentMin, 0) + value;
lowestSum = Math.min(lowestSum, currentMin);
sum += value;
}
if (sum == lowestSum) {
return highestSum;
}
return Math.max(highestSum, sum - lowestSum);
}
}
The problem at hand aims to solve the Maximum Sum Circular Subarray using Java. The provided solution implements a variation of Kadane's Algorithm to find both the maximum subarray sum in linear runtime and handle the circular nature of the problem. Below are the breakdowns and pertinent details of the solution:
Initialization:
- Initialize variables to track maximum and minimum sums (
highestSum
andlowestSum
) as well as the current maximum and minimum during iteration (currentMax
andcurrentMin
). - The starting values for
highestSum
andlowestSum
are the first element of the array to ensure proper comparison across the single-element array case.
- Initialize variables to track maximum and minimum sums (
Iteration over the Array:
- For each element in the array, update the
currentMax
sum, which represents the sum of the subarray ending at the current index. IfcurrentMax
drops below zero, reset it to zero (Kadane's reset step). - Simultaneously, track the
currentMin
sum for finding the minimum subarray sum to compute the maximum circular sum later. - Maintain an overall
sum
of the array elements for later calculations.
- For each element in the array, update the
Circular Subarray Handling:
- After iterating through the array, extract the subarray contributing the least to the overall array sum (using
sum - lowestSum
). - This helps in managing the circular aspect where the maximum sum subarray might span from the end of the array back to the beginning.
- After iterating through the array, extract the subarray contributing the least to the overall array sum (using
Edge Case:
- If the total sum of the array and the minimum subarray sum are the same, this indicates that all elements are either the same, or the only maximum sum subarray option available isn't improved by any circular computation. Thus, return the highest sum from the standard Kadane's result.
Result Calculation:
- Determine the result by comparing
highestSum
(from standard Kadane's algorithm) andsum - lowestSum
(considering circular subarray possibilities).
- Determine the result by comparing
This solution efficiently computes the maximum circular subarray sum in O(n) time by cleverly utilizing the modifications to Kadane's Algorithm and handling unique aspects brought about by the circular array constraint.
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