2249. Count The Hidden Sequences¶

Difficulty: Medium

2249. Count the Hidden Sequences

Medium

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

Solution¶

class Solution {
    public int numberOfArrays(int[] arr, int lower, int upper) {
        int n = arr.length;
        int pref[] = new int[n];
        pref[0] = arr[0];
        for (int i = 1; i < n; i++) pref[i] = pref[i - 1] + arr[i];
        int a = 0, b = 0;
        for (int i = 0; i < n; i++) {
            a = Math.min(a, pref[i]);
            b = Math.max(b, pref[i]);
            if (b - a > upper - lower) return 0;
        }
        return (upper - lower) - (b - a) + 1;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here