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1712. Ways to Split Array Into Three Subarrays

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Description

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

Solutions

Python3

class Solution:
    def waysToSplit(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        s = list(accumulate(nums))
        ans, n = 0, len(nums)
        for i in range(n - 2):
            j = bisect_left(s, s[i] << 1, i + 1, n - 1)
            k = bisect_right(s, (s[-1] + s[i]) >> 1, j, n - 1)
            ans += k - j
        return ans % mod

Java

class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int waysToSplit(int[] nums) {
        int n = nums.length;
        int[] s = new int[n];
        s[0] = nums[0];
        for (int i = 1; i < n; ++i) {
            s[i] = s[i - 1] + nums[i];
        }
        int ans = 0;
        for (int i = 0; i < n - 2; ++i) {
            int j = search(s, s[i] << 1, i + 1, n - 1);
            int k = search(s, ((s[n - 1] + s[i]) >> 1) + 1, j, n - 1);
            ans = (ans + k - j) % MOD;
        }
        return ans;
    }

    private int search(int[] s, int x, int left, int right) {
        while (left < right) {
            int mid = (left + right) >> 1;
            if (s[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    const int mod = 1e9 + 7;

    int waysToSplit(vector<int>& nums) {
        int n = nums.size();
        vector<int> s(n, nums[0]);
        for (int i = 1; i < n; ++i) s[i] = s[i - 1] + nums[i];
        int ans = 0;
        for (int i = 0; i < n - 2; ++i) {
            int j = lower_bound(s.begin() + i + 1, s.begin() + n - 1, s[i] << 1) - s.begin();
            int k = upper_bound(s.begin() + j, s.begin() + n - 1, (s[n - 1] + s[i]) >> 1) - s.begin();
            ans = (ans + k - j) % mod;
        }
        return ans;
    }
};

Go

func waysToSplit(nums []int) (ans int) {
	const mod int = 1e9 + 7
	n := len(nums)
	s := make([]int, n)
	s[0] = nums[0]
	for i := 1; i < n; i++ {
		s[i] = s[i-1] + nums[i]
	}
	for i := 0; i < n-2; i++ {
		j := sort.Search(n-1, func(h int) bool { return h > i && s[h] >= (s[i]<<1) })
		k := sort.Search(n-1, func(h int) bool { return h >= j && s[h] > (s[n-1]+s[i])>>1 })
		ans = (ans + k - j) % mod
	}
	return
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var waysToSplit = function (nums) {
    const mod = 1e9 + 7;
    const n = nums.length;
    const s = new Array(n).fill(nums[0]);
    for (let i = 1; i < n; ++i) {
        s[i] = s[i - 1] + nums[i];
    }
    function search(s, x, left, right) {
        while (left < right) {
            const mid = (left + right) >> 1;
            if (s[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
    let ans = 0;
    for (let i = 0; i < n - 2; ++i) {
        const j = search(s, s[i] << 1, i + 1, n - 1);
        const k = search(s, ((s[n - 1] + s[i]) >> 1) + 1, j, n - 1);
        ans = (ans + k - j) % mod;
    }
    return ans;
};

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