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Formatted question description: https://leetcode.ca/all/2145.html
2145. 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 length4
whose elements are in between1
and6
(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 than6
.[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 1.
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/ // 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); } };
-
class Solution: def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int: num = mi = mx = 0 for d in differences: num += d mi = min(mi, num) mx = max(mx, num) return max(0, upper - lower - (mx - mi) + 1) ############ # 2145. Count the Hidden Sequences # https://leetcode.com/problems/count-the-hidden-sequences/ class Solution: def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int: mmax = mmin = s = 0 for x in differences: s += x mmax = max(mmax, s) mmin = min(mmin, s) return max(0, (upper - lower) - (mmax - mmin) + 1)
-
class Solution { public int numberOfArrays(int[] differences, int lower, int upper) { long num = 0, mi = 0, mx = 0; for (int d : differences) { num += d; mi = Math.min(mi, num); mx = Math.max(mx, num); } return Math.max(0, (int) (upper - lower - (mx - mi) + 1)); } }
-
func numberOfArrays(differences []int, lower int, upper int) int { num, mi, mx := 0, 0, 0 for _, d := range differences { num += d mi = min(mi, num) mx = max(mx, num) } return max(0, upper-lower-(mx-mi)+1) } func max(a, b int) int { if a > b { return a } return b } func min(a, b int) int { if a < b { return a } return b }
Discuss
https://leetcode.com/problems/count-the-hidden-sequences/discuss/1709710/C%2B%2B-One-pass-O(N)-time