1712. Ways to Split Array Into Three Subarrays

Problem Description

In this problem, we need to find out how many ways we can split an array of non-negative integers into three non-empty contiguous subarrays wherein each subarray is named as left, mid, and right from left to right. The splitting must satisfy the condition that the sum of the elements in left is less than or equal to the sum of the elements in mid, and in turn, the sum in mid should be less than or equal to the sum of the elements in right. If we find a way to split the array that meets these conditions, we call it a "good" split.

Our goal is to count the total number of these good splits and return that count modulo 10^9 + 7 as the number can be very large.

Intuition

The intuition behind the solution is based on the concept of prefix sums and binary search. The array nums is processed to create another array, say s, which is a cumulative sum array (using accumulate(nums)). The cumulative sum array helps us quickly calculate the sum of any contiguous subarray.

Given the prefix sum array s, for each index i that could potentially be the end of the left subarray, we want to find suitable indices j and k where j marks the end of the mid subarray, and k represents the first index such that the sum of elements in the right subarray is not less than the sum of mid.

We use binary search (with bisect_left and bisect_right) to find the positions j and k based on the sum of left.

1. We start by fixing the end of the left subarray at index i.
2. We search for the smallest index j after i where the sum of mid is at least twice the sum of left. This guarantees that sum(left) ≤ sum(mid).
3. We then search for the largest index k where the sum of mid is still less than or equal to the sum of right. This is ensured by the sum of the entire array minus the sum up to i, divided by 2.
4. The number of positions for k minus the number of positions for j gives us the count of good splits for a fixed i.
5. We loop over all possible ends for the left subarray and sum up the counts to get the answer.
6. Finally, we return the answer modulo 10^9 + 7 to respect the constraints of the problem.

This approach is efficient as it trims down the search space using binary search rather than checking every possible split by brute-force which would be too slow.

Learn more about Two Pointers, Binary Search and Prefix Sum patterns.

Solution Approach

The solution approach is a combination of algorithmic strategies involving prefix sums, binary search, and modular arithmetic. Let's go through the steps of the implementation:

1. Prefix Sum Calculation: The first step involves calculating the prefix sums of the nums array and storing it in the array s. This is done using accumulate(nums). Prefix sums allow us to calculate the sum of any continuous subarray in constant time.

2. Modular Arithmetic Constant: A constant mod = 10**9 + 7 is defined at the beginning of the solution. This is used to perform modular arithmetic to prevent integer overflow and to return the final result modulo 10^9 + 7 as required by the problem statement.

3. Initialization: We initialize ans to store the count of good ways to split the array, and n to store the length of the nums array.

4. Iterating Over Potential Left Subarray Ends: Using a for loop, we iterate over each index i which could be the end of the left subarray. We are careful to stop at n-2 because we need at least two more elements to form the mid and right subarrays as they must be non-empty.

5. Finding the Lower and Upper Bounds for Mid Subarray Ends:

   • To find the lower bound j for mid's end, we use bisect_left. We search the prefix sum array s for the smallest index j such that 2 * s[i] <= s[j]. This ensures that the sum of mid is at least as much as the sum of left.
   • Similarly, to find the upper bound k for mid's end, we use bisect_right. Here we look for the last index k such that s[k] <= (s[-1] + s[i]) / 2. This ensures that sum(mid) ≤ sum(right).

6. Counting Good Splits: For each position of i, the good splits number is given by k - j, since for each index from j to k-1, we can form a valid mid subarray maintaining the condition of good split.

7. Accumulating and Modulo Operation: The number of good splits calculated for each i is added to ans. After the loop is finished, ans may be a very large number, so we return ans % mod to keep the result within the specified numeric range.

This solution is efficient due to binary search, which performs O(log n) comparisons rather than linear scans, reducing the overall complexity to O(n log n) for the problem. Without binary search, a brute-force approach would have a complexity of O(n^2), which would not be practical for large arrays.

Example Walkthrough

Let’s walk through a small example to illustrate the solution approach described above. Suppose we have the following array nums of non-negative integers:

1nums = [1, 2, 2, 2, 5, 0]

Following the steps of the solution:

1. Prefix Sum Calculation: Create a prefix sum array s using accumulate:

1s = [1, 3, 5, 7, 12, 12] // The cumulative sum of `nums`

2. Modular Arithmetic Constant: Define mod = 10**9 + 7 for later use.

3. Initialization: Initialize ans = 0 and n = 6 (length of nums).

4. Iterating Over Potential Left Subarray Ends: We consider the potential ends of the left subarray. Index i can go from 0 to n-3 (inclusive) to leave room for mid and right:

1For i = 0, s[i] = 1
2For i = 1, s[i] = 3
3For i = 2, s[i] = 5
4For i = 3, s[i] = 7

We stop at i = 3 because i = n-2 would not leave enough elements for mid and right.

5. Finding the Lower and Upper Bounds for Mid Subarray Ends:

   • For each i, find j and k using binary search:

   1For i = 0 (s[i] = 1):
   2- Search `j` such that `2 * s[i] <= s[j]`: `j = 2`
   3- Search `k` such that `s[k] <= (s[-1] + s[i]) / 2`: `k = 4`
   4- Good splits for i = 0: `k - j` = `4 - 2` = 2
   5
   6For i = 1 (s[i] = 3):
   7- `j = 3`
   8- `k = 5`
   9- Good splits for i = 1: `k - j` = `5 - 3` = 2
   10
   11For i = 2 (s[i] = 5):
   12- `j = 4`
   13- `k = 5`
   14- Good splits for i = 2: `k - j` = `5 - 4` = 1
   15
   16For i = 3 (s[i] = 7):
   17- `j = 4`
   18- `k = 5`
   19- Good splits for i = 3: `k - j` = `5 - 4` = 1

6. Counting Good Splits: Sum up all the good splits:

1ans = 2 + 2 + 1 + 1 = 6

7. Accumulating and Modulo Operation: Since our ans is small, ans % mod is trivially 6. If ans were large, the modulo operation would ensure we get a result within the numeric limits of 10^9 + 7.

Therefore, there are 6 good splits of the array nums following the rules set out by the problem.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the waysToSplit method can be broken down into the following parts:

1. Calculating the prefix sum of the nums list using accumulate. This takes O(n) time, where n is the length of nums.

2. The outer for-loop runs from 0 to n - 3 inclusive, which gives us O(n) iterations.

3. Inside the for-loop, it uses bisect_left and bisect_right methods to find indices j and k. Both of these binary search operations run in O(log n) time.

Therefore, with every iteration of the for-loop, we perform two O(log n) operations. Since we iterate O(n) times, this leads to a compound time complexity of O(n log n) for this part of the function.

The total time complexity of the function is therefore O(n) + O(n log n) = O(n log n).

Space Complexity

The space complexity is determined by the additional space used by the algorithm, which includes:

1. The space used by the prefix sum array s. This array is the same length as nums, so it requires O(n) space.

2. Variables used for iteration and indexing, such as i, j, k, ans, mod, and n, which use O(1) space.

The dominant term here is the space used by the prefix sum array, so the total space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.