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 length4
whose elements are in between1
and6
(inclusive).<ul> <li><code>[3, 4, 1, 5]</code> and <code>[4, 5, 2, 6]</code> are possible hidden sequences.</li> <li><code>[5, 6, 3, 7]</code> is not possible since it contains an element greater than <code>6</code>.</li> <li><code>[1, 2, 3, 4]</code> is not possible since the differences are not correct.</li> </ul> </li>
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
Assume hidden[0] = 0
.
We can get all hidden[i+1] = hidden[i] + diff[i]
.
The hidden
array forms a polyline. Assume the max/min values are max
/min
.
By changing hidden[0]
, we can shift this range up or down.
If we snap max
to upper
, we move up by upper - max
steps. Then the number of possible of hidden sequences is min + (upper - max) - lower + 1
.
Another way to think about it:
// OJ: https://leetcode.com/problems/count-the-hidden-sequences/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int numberOfArrays(vector<int>& A, int lower, int upper) {
long sum = 0, mn = 0, mx = 0;
for (int n : A) {
sum += n;
mn = min(mn, sum);
mx = max(mx, sum);
}
return max(0L, mn - mx + upper - lower + 1);
}
};
https://leetcode.com/problems/count-the-hidden-sequences/discuss/1709710/C%2B%2B-One-pass-O(N)-time