Problem

Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions:

0 < i, i + 1 < j, j + 1 < k < n - 1
Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be equal.
where we define that subarray (L, R) represents a slice of the original array starting from the element indexed L to the element indexed R.
Example:
Input: [1,2,1,2,1,2,1]
Output: True
Explanation:
i = 1, j = 3, k = 5.
sum(0, i - 1) = sum(0, 0) = 1
sum(i + 1, j - 1) = sum(2, 2) = 1
sum(j + 1, k - 1) = sum(4, 4) = 1
sum(k + 1, n - 1) = sum(6, 6) = 1
Note:
1 <= n <= 2000.
Elements in the given array will be in range [-1,000,000, 1,000,000].

Solution

class Solution {
    public boolean splitArray(int[] nums) {
        if (nums == null || nums.length < 7) return false;
        int len = nums.length;
        int[] sum = new int[len];
        sum[0] = nums[0];
        for (int i = 1; i < len; i++) {
            sum[i] = sum[i-1]+nums[i];
        }
        // 0 ~ i-1  |  i+1 ~ mid-1  |  mid+1 ~ k-1  |  k+1 ~ len-1
        for (int mid = 3; mid < len-3; mid++) {
            Set<Integer> set = new HashSet<>();
            for (int i = 1; i <= mid-2; i++) {
                //save quarter sum into hashset
                if (sum[i-1] == sum[mid-1]-sum[i]) set.add(sum[i-1]);
            }
            for (int k = mid+2; k <= len-2; k++) {
                if (sum[len-1]-sum[k] == sum[k-1]-sum[mid]) {
                    int quarterSum = sum[len-1]-sum[k];
                    if (set.contains(quarterSum)) return true;
                }
            }
        }
        return false;
    }
}

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