1712. Ways to Split Array Into Three Subarrays

Problem Link

Description

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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

Solution

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Python3

class Solution:
    def waysToSplit(self, nums: List[int]) -> int:
        prefix = [0]
        for num in nums:
            prefix.append(prefix[-1] + num)
            
        n = len(nums)
        M = 10 ** 9 + 7
        res = 0
 
        for i in range(1, n-1):
            mid = bisect_left(prefix, 2 * prefix[i])
            right = bisect_right(prefix, (prefix[-1] + prefix[i]) // 2)
            res += max(0, min(n,right) - max(i+1, mid))
        
        return res % M

C++

class Solution {
public:
    int waysToSplit(vector<int>& nums) {
        int M = 1e9 + 7;
        int n = nums.size();
        
        for (int i = 1; i < n; i++)
            nums[i] += nums[i-1];
        
        long long res = 0;
        for (int i = 0; i < n - 1; i++){
            int left = nums[i];
            int remaining = nums[n-1] - left;
            int max_mid = remaining / 2;
            
            int mid_start = lower_bound(nums.begin()+i+1, nums.end()-1, left * 2) - nums.begin();
            int right_start = upper_bound(nums.begin()+i+1, nums.end()-1, left + max_mid) - nums.begin();
            
            res += max(0, right_start - mid_start);
            res %= M;
        }
        
        return res % M;
    }
};

Java

class Solution {
    public int waysToSplit(int[] nums) {
 
        int MOD = (int) (1e9 + 7);
 
        int N = nums.length;
 
        // prefix array
        int[] A = new int[N];
        A[0] = nums[0];
        for (int i = 1; i < N; ++i) A[i] = A[i - 1] + nums[i];
 
        int res = 0;
        for (int i = 1; i < N - 1; ++i) {
 
            int left = helper(A, A[i - 1], i, true);
            int right = helper(A, A[i - 1], i, false);
 
            if (left == -1 || right == -1) continue;  // none is satisfied
 
            res = (res + (right - left + 1) % MOD) % MOD;
        }
 
        return res;
    }
 
    private int helper(int[] A, int leftSum, int index, boolean searchLeft) {
 
        int N = A.length;
        int l = index, r = N - 2;
        int res = -1;
 
        while (l <= r) {
 
            int m = (r - l) / 2 + l;
 
            int midSum = A[m] - A[index - 1];
            int rightSum = A[N - 1] - A[m];
 
            if (leftSum <= midSum && midSum <= rightSum) {
                res = m;
                if (searchLeft) r = m - 1;
                else l = m + 1;
            } else if (leftSum > midSum) {  // shrink left
                l = m + 1;
            } else {  // shrink right
                r = m - 1;
            }
        }
 
        return res;
    }
}