1955. Count Number of Special Subsequences

Description

A sequence is special if it consists of a positive number of 0s, followed by a positive number of 1s, then a positive number of 2s.

  • For example, [0,1,2] and [0,0,1,1,1,2] are special.
  • In contrast, [2,1,0], [1], and [0,1,2,0] are not special.

Given an array nums (consisting of only integers 0, 1, and 2), return the number of different subsequences that are special. Since the answer may be very large, return it modulo 109 + 7.

A subsequence of an array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements. Two subsequences are different if the set of indices chosen are different.

 

Example 1:

Input: nums = [0,1,2,2]
Output: 3
Explanation: The special subsequences are bolded [0,1,2,2], [0,1,2,2], and [0,1,2,2].

Example 2:

Input: nums = [2,2,0,0]
Output: 0
Explanation: There are no special subsequences in [2,2,0,0].

Example 3:

Input: nums = [0,1,2,0,1,2]
Output: 7
Explanation: The special subsequences are bolded:
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 2

Solutions

Solution 1

Python Code
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class Solution:
    def countSpecialSubsequences(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        n = len(nums)
        f = [[0] * 3 for _ in range(n)]
        f[0][0] = nums[0] == 0
        for i in range(1, n):
            if nums[i] == 0:
                f[i][0] = (2 * f[i - 1][0] + 1) % mod
                f[i][1] = f[i - 1][1]
                f[i][2] = f[i - 1][2]
            elif nums[i] == 1:
                f[i][0] = f[i - 1][0]
                f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1]) % mod
                f[i][2] = f[i - 1][2]
            else:
                f[i][0] = f[i - 1][0]
                f[i][1] = f[i - 1][1]
                f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2]) % mod
        return f[n - 1][2]

Java Code
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class Solution {
    public int countSpecialSubsequences(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int n = nums.length;
        int[][] f = new int[n][3];
        f[0][0] = nums[0] == 0 ? 1 : 0;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == 0) {
                f[i][0] = (2 * f[i - 1][0] % mod + 1) % mod;
                f[i][1] = f[i - 1][1];
                f[i][2] = f[i - 1][2];
            } else if (nums[i] == 1) {
                f[i][0] = f[i - 1][0];
                f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1] % mod) % mod;
                f[i][2] = f[i - 1][2];
            } else {
                f[i][0] = f[i - 1][0];
                f[i][1] = f[i - 1][1];
                f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2] % mod) % mod;
            }
        }
        return f[n - 1][2];
    }
}

C++ Code
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class Solution {
public:
    int countSpecialSubsequences(vector<int>& nums) {
        const int mod = 1e9 + 7;
        int n = nums.size();
        int f[n][3];
        memset(f, 0, sizeof(f));
        f[0][0] = nums[0] == 0;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == 0) {
                f[i][0] = (2 * f[i - 1][0] % mod + 1) % mod;
                f[i][1] = f[i - 1][1];
                f[i][2] = f[i - 1][2];
            } else if (nums[i] == 1) {
                f[i][0] = f[i - 1][0];
                f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1] % mod) % mod;
                f[i][2] = f[i - 1][2];
            } else {
                f[i][0] = f[i - 1][0];
                f[i][1] = f[i - 1][1];
                f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2] % mod) % mod;
            }
        }
        return f[n - 1][2];
    }
};

Go Code
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func countSpecialSubsequences(nums []int) int {
	const mod = 1e9 + 7
	n := len(nums)
	f := make([][3]int, n)
	if nums[0] == 0 {
		f[0][0] = 1
	}
	for i := 1; i < n; i++ {
		if nums[i] == 0 {
			f[i][0] = (2*f[i-1][0] + 1) % mod
			f[i][1] = f[i-1][1]
			f[i][2] = f[i-1][2]
		} else if nums[i] == 1 {
			f[i][0] = f[i-1][0]
			f[i][1] = (f[i-1][0] + 2*f[i-1][1]) % mod
			f[i][2] = f[i-1][2]
		} else {
			f[i][0] = f[i-1][0]
			f[i][1] = f[i-1][1]
			f[i][2] = (f[i-1][1] + 2*f[i-1][2]) % mod
		}
	}
	return f[n-1][2]
}

TypeScript Code
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function countSpecialSubsequences(nums: number[]): number {
    const mod = 1e9 + 7;
    const n = nums.length;
    const f: number[][] = Array(n)
        .fill(0)
        .map(() => Array(3).fill(0));
    f[0][0] = nums[0] === 0 ? 1 : 0;
    for (let i = 1; i < n; ++i) {
        if (nums[i] === 0) {
            f[i][0] = (((2 * f[i - 1][0]) % mod) + 1) % mod;
            f[i][1] = f[i - 1][1];
            f[i][2] = f[i - 1][2];
        } else if (nums[i] === 1) {
            f[i][0] = f[i - 1][0];
            f[i][1] = (f[i - 1][0] + ((2 * f[i - 1][1]) % mod)) % mod;
            f[i][2] = f[i - 1][2];
        } else {
            f[i][0] = f[i - 1][0];
            f[i][1] = f[i - 1][1];
            f[i][2] = (f[i - 1][1] + ((2 * f[i - 1][2]) % mod)) % mod;
        }
    }
    return f[n - 1][2];
}

Solution 2

Python Code
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class Solution:
    def countSpecialSubsequences(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        n = len(nums)
        f = [0] * 3
        f[0] = nums[0] == 0
        for i in range(1, n):
            if nums[i] == 0:
                f[0] = (2 * f[0] + 1) % mod
            elif nums[i] == 1:
                f[1] = (f[0] + 2 * f[1]) % mod
            else:
                f[2] = (f[1] + 2 * f[2]) % mod
        return f[2]

Java Code
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class Solution {
    public int countSpecialSubsequences(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int n = nums.length;
        int[] f = new int[3];
        f[0] = nums[0] == 0 ? 1 : 0;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == 0) {
                f[0] = (2 * f[0] % mod + 1) % mod;
            } else if (nums[i] == 1) {
                f[1] = (f[0] + 2 * f[1] % mod) % mod;
            } else {
                f[2] = (f[1] + 2 * f[2] % mod) % mod;
            }
        }
        return f[2];
    }
}

C++ Code
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class Solution {
public:
    int countSpecialSubsequences(vector<int>& nums) {
        const int mod = 1e9 + 7;
        int n = nums.size();
        int f[3]{0};
        f[0] = nums[0] == 0;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == 0) {
                f[0] = (2 * f[0] % mod + 1) % mod;
            } else if (nums[i] == 1) {
                f[1] = (f[0] + 2 * f[1] % mod) % mod;
            } else {
                f[2] = (f[1] + 2 * f[2] % mod) % mod;
            }
        }
        return f[2];
    }
};

Go Code
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func countSpecialSubsequences(nums []int) int {
	const mod = 1e9 + 7
	n := len(nums)
	f := [3]int{}
	if nums[0] == 0 {
		f[0] = 1
	}
	for i := 1; i < n; i++ {
		if nums[i] == 0 {
			f[0] = (2*f[0] + 1) % mod
		} else if nums[i] == 1 {
			f[1] = (f[0] + 2*f[1]) % mod
		} else {
			f[2] = (f[1] + 2*f[2]) % mod
		}
	}
	return f[2]
}

TypeScript Code
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function countSpecialSubsequences(nums: number[]): number {
    const mod = 1e9 + 7;
    const n = nums.length;
    const f: number[] = [0, 0, 0];
    f[0] = nums[0] === 0 ? 1 : 0;
    for (let i = 1; i < n; ++i) {
        if (nums[i] === 0) {
            f[0] = (((2 * f[0]) % mod) + 1) % mod;
            f[1] = f[1];
            f[2] = f[2];
        } else if (nums[i] === 1) {
            f[0] = f[0];
            f[1] = (f[0] + ((2 * f[1]) % mod)) % mod;
            f[2] = f[2];
        } else {
            f[0] = f[0];
            f[1] = f[1];
            f[2] = (f[1] + ((2 * f[2]) % mod)) % mod;
        }
    }
    return f[2];
}