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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 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 1.

Assume hidden = 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, 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