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2145. Count the Hidden Sequences

Description

You are given a 0-indexed array of n integers differences, which describes the differences between each pair of consecutive integers of a hidden sequence of length (n + 1). More formally, call the hidden sequence hidden, then we have that differences[i] = hidden[i + 1] - hidden[i].

You are further given two integers lower and upper that describe the inclusive range of values [lower, upper] that the hidden sequence can contain.

  • For example, given differences = [1, -3, 4], lower = 1, upper = 6, the hidden sequence is a sequence of length 4 whose elements are in between 1 and 6 (inclusive).
    • [3, 4, 1, 5] and [4, 5, 2, 6] are possible hidden sequences.
    • [5, 6, 3, 7] is not possible since it contains an element greater than 6.
    • [1, 2, 3, 4] is not possible since the differences are not correct.

Return the number of possible hidden sequences there are. If there are no possible sequences, return 0.

 

Example 1:

Input: differences = [1,-3,4], lower = 1, upper = 6
Output: 2
Explanation: The possible hidden sequences are:
- [3, 4, 1, 5]
- [4, 5, 2, 6]
Thus, we return 2.

Example 2:

Input: differences = [3,-4,5,1,-2], lower = -4, upper = 5
Output: 4
Explanation: The possible hidden sequences are:
- [-3, 0, -4, 1, 2, 0]
- [-2, 1, -3, 2, 3, 1]
- [-1, 2, -2, 3, 4, 2]
- [0, 3, -1, 4, 5, 3]
Thus, we return 4.

Example 3:

Input: differences = [4,-7,2], lower = 3, upper = 6
Output: 0
Explanation: There are no possible hidden sequences. Thus, we return 0.

 

Constraints:

  • n == differences.length
  • 1 <= n <= 105
  • -105 <= differences[i] <= 105
  • -105 <= lower <= upper <= 105

Solutions

Solution 1: Prefix Sum

Since the array $\textit{differences}$ is already determined, the difference between the maximum and minimum values of the elements in the array $\textit{hidden}$ is also fixed. We just need to ensure that this difference does not exceed $\textit{upper} - \textit{lower}$.

Let's assume the first element of the array $\textit{hidden}$ is $0$. Then, $\textit{hidden}[i] = \textit{hidden}[i - 1] + \textit{differences}[i - 1]$, where $1 \leq i \leq n$. Let the maximum value of the array $\textit{hidden}$ be $mx$ and the minimum value be $mi$. If $mx - mi \leq \textit{upper} - \textit{lower}$, then we can construct a valid $\textit{hidden}$ array. The number of possible constructions is $\textit{upper} - \textit{lower} - (mx - mi) + 1$. Otherwise, it is impossible to construct a valid $\textit{hidden}$ array, and we return $0$.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{differences}$. The space complexity is $O(1)$.

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class Solution:
    def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int:
        x = mi = mx = 0
        for d in differences:
            x += d
            mi = min(mi, x)
            mx = max(mx, x)
        return max(upper - lower - (mx - mi) + 1, 0)

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class Solution {
    public int numberOfArrays(int[] differences, int lower, int upper) {
        long x = 0, mi = 0, mx = 0;
        for (int d : differences) {
            x += d;
            mi = Math.min(mi, x);
            mx = Math.max(mx, x);
        }
        return (int) Math.max(upper - lower - (mx - mi) + 1, 0);
    }
}

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class Solution {
public:
    int numberOfArrays(vector<int>& differences, int lower, int upper) {
        long long x = 0, mi = 0, mx = 0;
        for (int d : differences) {
            x += d;
            mi = min(mi, x);
            mx = max(mx, x);
        }
        return max(upper - lower - (mx - mi) + 1, 0LL);
    }
};

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func numberOfArrays(differences []int, lower int, upper int) int {
    x, mi, mx := 0, 0, 0
    for _, d := range differences {
        x += d
        mi = min(mi, x)
        mx = max(mx, x)
    }
    return max(0, upper-lower-(mx-mi)+1)
}

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function numberOfArrays(differences: number[], lower: number, upper: number): number {
    let [x, mi, mx] = [0, 0, 0];
    for (const d of differences) {
        x += d;
        mi = Math.min(mi, x);
        mx = Math.max(mx, x);
    }
    return Math.max(0, upper - lower - (mx - mi) + 1);
}

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