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2750. Ways to Split Array Into Good Subarrays

Description

You are given a binary array nums.

A subarray of an array is good if it contains exactly one element with the value 1.

Return an integer denoting the number of ways to split the array nums into good subarrays. As the number may be too large, return it modulo 109 + 7.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [0,1,0,0,1]
Output: 3
Explanation: There are 3 ways to split nums into good subarrays:
- [0,1] [0,0,1]
- [0,1,0] [0,1]
- [0,1,0,0] [1]

Example 2:

Input: nums = [0,1,0]
Output: 1
Explanation: There is 1 way to split nums into good subarrays:
- [0,1,0]

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 1

Solutions

Solution 1: Multiplication Principle

Thinking

A good subarray contains exactly one \(1\); we count ways to cut the array into such pieces. The cuts are completely determined by the positions of the ones.

Between adjacent ones at \(j\) and \(i\) there are \(i-j\) places for a cut, and the segments are independent, so the answer is the product of those gaps. With no one at all the answer is \(0\).

Based on the problem description, we can draw a dividing line between two \(1\)s. Assuming the indices of the two \(1\)s are \(j\) and \(i\) respectively, then the number of different dividing lines that can be drawn is \(i - j\). We find all the pairs of \(j\) and \(i\) that meet the condition, and then multiply all the \(i - j\) together. If no dividing line can be found between two \(1\)s, it means there are no \(1\)s in the array, and the answer is \(0\).

The time complexity is \(O(n)\), where \(n\) is the length of the array. The space complexity is \(O(1)\).

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class Solution:
    def numberOfGoodSubarraySplits(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        ans, j = 1, -1
        for i, x in enumerate(nums):
            if x == 0:
                continue
            if j > -1:
                ans = ans * (i - j) % mod
            j = i
        return 0 if j == -1 else ans

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class Solution {
    public int numberOfGoodSubarraySplits(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int ans = 1, j = -1;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == 0) {
                continue;
            }
            if (j > -1) {
                ans = (int) ((long) ans * (i - j) % mod);
            }
            j = i;
        }
        return j == -1 ? 0 : ans;
    }
}

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class Solution {
public:
    int numberOfGoodSubarraySplits(vector<int>& nums) {
        const int mod = 1e9 + 7;
        int ans = 1, j = -1;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] == 0) {
                continue;
            }
            if (j > -1) {
                ans = 1LL * ans * (i - j) % mod;
            }
            j = i;
        }
        return j == -1 ? 0 : ans;
    }
};

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func numberOfGoodSubarraySplits(nums []int) int {
    const mod int = 1e9 + 7
    ans, j := 1, -1
    for i, x := range nums {
        if x == 0 {
            continue
        }
        if j > -1 {
            ans = ans * (i - j) % mod
        }
        j = i
    }
    if j == -1 {
        return 0
    }
    return ans
}

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function numberOfGoodSubarraySplits(nums: number[]): number {
    let ans = 1;
    let j = -1;
    const mod = 10 ** 9 + 7;
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        if (nums[i] === 0) {
            continue;
        }
        if (j > -1) {
            ans = (ans * (i - j)) % mod;
        }
        j = i;
    }
    return j === -1 ? 0 : ans;
}

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public class Solution {
    public int NumberOfGoodSubarraySplits(int[] nums) {
        long ans = 1, j = -1;
        int mod = 1000000007;
        int n = nums.Length;
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                continue;
            }
            if (j > -1) {
                ans = ans * (i - j) % mod;
            }
            j = i;
        }
        return j == -1 ? 0 : (int) ans;
    }
}

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