Maximum Sum Circular Subarray LT918

Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.

Here, a circular array means the end of the array connects to the beginning of the array.  (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)

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

Example 1:

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

Example 2:

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

Example 3:

Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4

Example 4:

Input: [3,-2,2,-3]
Output: 3
Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3

Example 5:

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

Idea 1. Similar to Maximum Subarray LT53, the difference is the circular array part, the subarray could be the tail(A[j..n-1]) plus the head(A[0..i]) (i+1 < j), if the subarray of tail + head is the maximum, the subarray in the middle is A[i+1, j-1] has the minSum, once the minSum is found, the maxSum is Sum - minSum, except maxSum = 0 if all values in the array are negative, as minSum == Sum.

Note. negative arrays

Time complexity: O(n)

Space complexity: O(1)

 class Solution {
     public int maxSubarraySumCircular(int[] A) {
         int maxSoFar = 0;
         int minSoFar = 0;

         int sum = 0;
         int minSum = Integer.MAX_VALUE;
         int maxSum = Integer.MIN_VALUE;

         for(int a: A) {
             maxSoFar = a + Math.max(maxSoFar, 0);
             maxSum = Math.max(maxSum, maxSoFar);

             minSoFar = a + Math.min(minSoFar, 0);
             minSum = Math.min(minSum, minSoFar);

             sum += a;
         }

         if(sum == minSum) {
             return maxSum;
         }
         return Math.max(maxSum, sum - minSum);
     }
 }

Idea 2. 另外一种方法求2个intervals (head + tail)组成的，如果直接从左到右+从右到左的和，可能中间有重合，固定一边，求另外一边的和的最大值，保证j >= i+2.

Time complexity: O(n)

Space complexity: O(n)

 class Solution {
     public int maxSubarraySumCircular(int[] A) {
         int curr = 0;
         int n = A.length;
         int maxSum = Integer.MIN_VALUE;

         for(int a: A) {
             curr = a + Math.max(curr, 0);
             maxSum = Math.max(maxSum, curr);
         }

         int[] rightMaxSum = new int[n];
         rightMaxSum[n-1] = A[n-1];
         curr = A[n-1];
         for(int i = n-2; i >= 0; --i) {
             curr += A[i];
             rightMaxSum[i] = Math.max(curr, rightMaxSum[i+1]);
         }

         curr = 0;
         int result = maxSum;
         for(int i = 0; i+2 < n; ++i) {
             curr += A[i];
             result = Math.max(result, curr + rightMaxSum[i+2]);
         }

         return result;
     }
 }

Idea 3. Similar to sliding window minValue, build a min deque for prefix sum, for each prefix sum, find the minimum of previous prefixSum so that the subarray sum ending here is the maximu, since it's circular array, j - i <= n, remove the invalid previous index. Note: store the index

Time complexity: O(n)

Space complexity: O(n)

 class Solution {
     public int maxSubarraySumCircular(int[] A) {
         Deque<Integer> minPrefix = new LinkedList<>();
         minPrefix.addLast(0);

         int n = A.length;

         int[] prefix = new int[2*n + 1];
         for(int i = 0; i < 2*n; ++i) {
             prefix[i+1] = prefix[i] + A[i%n];
         }
         int result = Integer.MIN_VALUE;
         for(int i = 1; i <= 2*n; ++i) {

             if(i - minPrefix.getFirst() > n) {
                 minPrefix.removeFirst();
             }

             result = Math.max(result, prefix[i] - prefix[minPrefix.getFirst()]);

             while(!minPrefix.isEmpty() && prefix[minPrefix.getLast()] >prefix[i]) {
                 minPrefix.removeLast();
             }
             minPrefix.add(i);

         }

         return result;
     }
 }

Note:

1. -30000 <= A[i] <= 30000
2. 1 <= A.length <= 30000