2963. Count the Number of Good Partitions

Description

You are given a 0-indexed array nums consisting of positive integers.

A partition of an array into one or more contiguous subarrays is called good if no two subarrays contain the same number.

Return the total number of good partitions of nums.

Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,2,3,4]
Output: 8
Explanation: The 8 possible good partitions are: ([1], [2], [3], [4]), ([1], [2], [3,4]), ([1], [2,3], [4]), ([1], [2,3,4]), ([1,2], [3], [4]), ([1,2], [3,4]), ([1,2,3], [4]), and ([1,2,3,4]).

Example 2:

Input: nums = [1,1,1,1]
Output: 1
Explanation: The only possible good partition is: ([1,1,1,1]).

Example 3:

Input: nums = [1,2,1,3]
Output: 2
Explanation: The 2 possible good partitions are: ([1,2,1], [3]) and ([1,2,1,3]).

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Solutions

Solution 1: Hash Table + Grouping + Fast Power

According to the problem description, we know that the same number must be in the same subarray. Therefore, we use a hash table $last$ to record the index of the last occurrence of each number.

Next, we use an index $j$ to mark the index of the last element that has appeared among the elements, and use a variable $k$ to record the number of subarrays that can currently be divided.

Then, we traverse the array $nums$ from left to right. For the current number $nums[i]$ we are traversing, we get its last occurrence index and update $j = \max(j, last[nums[i]])$. If $i = j$, it means that the current position can be the end of a subarray, and we increase $k$ by $1$. Continue to traverse until the entire array is traversed.

Finally, we consider the number of division schemes for $k$ subarrays. The number of subarrays is $k$, and there are $k-1$ positions that can be divided (concatenated), so the number of schemes is $2^{k-1}$. Since the answer may be very large, we need to take the modulo of $10^9 + 7$. Here we can use fast power to speed up the calculation.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

Python3JavaC++GoTypeScript

class Solution:
    def numberOfGoodPartitions(self, nums: List[int]) -> int:
        last = {x: i for i, x in enumerate(nums)}
        mod = 10**9 + 7
        j, k = -1, 0
        for i, x in enumerate(nums):
            j = max(j, last[x])
            k += i == j
        return pow(2, k - 1, mod)

class Solution {
    public int numberOfGoodPartitions(int[] nums) {
        Map<Integer, Integer> last = new HashMap<>();
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            last.put(nums[i], i);
        }
        final int mod = (int) 1e9 + 7;
        int j = -1;
        int k = 0;
        for (int i = 0; i < n; ++i) {
            j = Math.max(j, last.get(nums[i]));
            k += i == j ? 1 : 0;
        }
        return qpow(2, k - 1, mod);
    }

    private int qpow(long a, int n, int mod) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return (int) ans;
    }
}

class Solution {
public:
    int numberOfGoodPartitions(vector<int>& nums) {
        unordered_map<int, int> last;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            last[nums[i]] = i;
        }
        const int mod = 1e9 + 7;
        int j = -1, k = 0;
        for (int i = 0; i < n; ++i) {
            j = max(j, last[nums[i]]);
            k += i == j;
        }
        auto qpow = [&](long long a, int n, int mod) {
            long long ans = 1;
            for (; n; n >>= 1) {
                if (n & 1) {
                    ans = ans * a % mod;
                }
                a = a * a % mod;
            }
            return (int) ans;
        };
        return qpow(2, k - 1, mod);
    }
};

func numberOfGoodPartitions(nums []int) int {
    qpow := func(a, n, mod int) int {
        ans := 1
        for ; n > 0; n >>= 1 {
            if n&1 == 1 {
                ans = ans * a % mod
            }
            a = a * a % mod
        }
        return ans
    }
    last := map[int]int{}
    for i, x := range nums {
        last[x] = i
    }
    const mod int = 1e9 + 7
    j, k := -1, 0
    for i, x := range nums {
        j = max(j, last[x])
        if i == j {
            k++
        }
    }
    return qpow(2, k-1, mod)
}

function numberOfGoodPartitions(nums: number[]): number {
    const qpow = (a: number, n: number, mod: number) => {
        let ans = 1;
        for (; n; n >>= 1) {
            if (n & 1) {
                ans = Number((BigInt(ans) * BigInt(a)) % BigInt(mod));
            }
            a = Number((BigInt(a) * BigInt(a)) % BigInt(mod));
        }
        return ans;
    };
    const last: Map<number, number> = new Map();
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        last.set(nums[i], i);
    }
    const mod = 1e9 + 7;
    let [j, k] = [-1, 0];
    for (let i = 0; i < n; ++i) {
        j = Math.max(j, last.get(nums[i])!);
        if (i === j) {
            ++k;
        }
    }
    return qpow(2, k - 1, mod);
}

2963. Count the Number of Good Partitions

Description

Solutions

Solution 1: Hash Table + Grouping + Fast Power

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