1712. Ways to Split Array Into Three Subarrays

1712. Ways to Split Array Into Three Subarrays (Medium)

A split of an integer array is good if:

• The array is split into three non-empty contiguous subarrays - named left, mid, right respectively from left to right.
• The sum of the elements in left is less than or equal to the sum of the elements in mid, and the sum of the elements in mid is less than or equal to the sum of the elements in right.

Given nums, an array of non-negative integers, return the number of good ways to split nums. As the number may be too large, return it modulo 109 + 7.

 

Example 1:

Input: nums = [1,1,1]
Output: 1
Explanation: The only good way to split nums is [1] [1] [1].

Example 2:

Input: nums = [1,2,2,2,5,0]
Output: 3
Explanation: There are three good ways of splitting nums:
[1] [2] [2,2,5,0]
[1] [2,2] [2,5,0]
[1,2] [2,2] [5,0]

Example 3:

Input: nums = [3,2,1]
Output: 0
Explanation: There is no good way to split nums.

 

Constraints:

• 3 <= nums.length <= 105
• 0 <= nums[i] <= 104

Related Topics:
Binary Search

Solution 1. Three Pointers

Turn array A into its prefix sum array.

Let i be the last index of the left part. So A[i] is the sum of the left part.

Given i, the last index of the mid part is a range. Let it be [j, k).

When we increment i, j and k must be monotonically increasing.

To find j, we can increment j from i + 1 until mid >= left i.e. A[j] - A[i] >= A[i].

To find k, we can increment k from j until mid < right, i.e. A[N - 1] - A[k] < A[k] - A[i].

// OJ: https://leetcode.com/problems/ways-to-split-array-into-three-subarrays/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int waysToSplit(vector<int>& A) {
        long mod = 1e9+7, ans = 0;
        for (int i = 1; i < A.size(); ++i) A[i] += A[i - 1];
        long N = A.size(), i = 0, j = 0, k = 0;
        for (; i < N; ++i) {
            long left = A[i];
            j = max(i + 1, j); // `j` is at least one greater than `i`.
            while (j < N && A[j] - left < left) ++j; // find the smallest `j` that satisfies `mid >= left`
            if (j >= N) break; // No room for `k`. Break
            k = max(k, j);
            while (k < N - 1 && A.back() - A[k] >= A[k] - A[i]) ++k;
            ans = (ans + k - j) % mod;
        }
        return ans;
    }
};

Solution 2. Binary Search

// OJ: https://leetcode.com/problems/ways-to-split-array-into-three-subarrays/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int waysToSplit(vector<int>& A) {
        long mod = 1e9 + 7, ans = 0, N = A.size();
        for (int i = 1; i < N; ++i) A[i] += A[i - 1];
        for (int i = 0; i < N; ++i) {
            long left = A[i], other = A.back() - left;
            int j = lower_bound(begin(A) + i + 1, end(A), 2 * left) - begin(A);
            int k = upper_bound(begin(A) + j, end(A) - 1, left + other / 2) - begin(A);
            ans = (ans + k - j) % mod;
        }
        return ans;
    }
};