2145. Count the Hidden Sequences

Description

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

 

Solutions

Solution: Prefix Sum

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number[]} differences
 * @param {number} lower
 * @param {number} upper
 * @return {number}
 */
const numberOfArrays = function (differences, lower, upper) {
  let maxSum = 0;
  let minSum = 0;
  let sum = 0;

  for (const difference of differences) {
    sum += difference;
    maxSum = Math.max(sum, maxSum);
    minSum = Math.min(sum, minSum);
  }

  return Math.max(0, upper - maxSum - (lower - minSum) + 1);
};