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