Count the Hidden Sequences

MEDIUM

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

Approaches

Checkout 2 different approaches to solve Count the Hidden Sequences. Click on different approaches to view the approach and algorithm in detail.

Brute Force on the First Element

This approach recognizes that the entire hidden sequence is determined by its first element, hidden[0]. We can iterate through all possible values for hidden[0] within the given [lower, upper] range. For each potential starting value, we construct the complete sequence and check if all its elements fall within the [lower, upper] bounds. If they do, we count it as a valid sequence.

Algorithm

Create a relative_seq array of size n+1 to store prefix sums. Initialize relative_seq[0] = 0.
For i from 1 to n, calculate relative_seq[i] = relative_seq[i-1] + differences[i-1].
Initialize a counter count = 0.
Iterate through each possible integer value for the first element, h_0, from lower to upper.
For each h_0:
- Assume the generated sequence is valid by setting a flag isValid = true.
- Iterate from k = 0 to n to check each element of the potential hidden sequence.
- Calculate the current element: current_element = h_0 + relative_seq[k].
- If current_element is outside the [lower, upper] range, set isValid = false and break the inner loop.
If isValid remains true after checking all elements, increment count.
After checking all possible values for h_0, return count.

The core idea is that if we fix the value of hidden[0], say to h_0, every other element hidden[k] is uniquely determined by the formula hidden[k] = h_0 + sum(differences[0]...differences[k-1]).

To implement this, we can first pre-calculate the prefix sums of the differences array. Let's call this relative_seq, where relative_seq[k] is the sum of the first k differences, and relative_seq[0] is 0.

Then, we loop through each possible integer value for h_0 from lower to upper. In each iteration, we generate the full candidate sequence by adding h_0 to each element of relative_seq. We then validate this candidate sequence by checking if all its elements are between lower and upper. A counter is incremented for each valid sequence found.

class Solution {
    public int numberOfArrays(int[] differences, int lower, int upper) {
        int n = differences.length;
        long[] relativeSeq = new long[n + 1];
        relativeSeq[0] = 0;
        for (int i = 0; i < n; i++) {
            relativeSeq[i + 1] = relativeSeq[i] + differences[i];
        }

        int count = 0;
        // Iterate through all possible starting values
        for (long h0 = lower; h0 <= upper; h0++) {
            boolean isValid = true;
            // Check if this starting value creates a valid sequence
            for (int i = 0; i <= n; i++) {
                long element = h0 + relativeSeq[i];
                if (element < lower || element > upper) {
                    isValid = false;
                    break;
                }
            }
            if (isValid) {
                count++;
            }
        }
        return count;
    }
}

Complexity Analysis

Time Complexity: O(N * (U - L)), where N is the length of `differences`, U is `upper`, and L is `lower`. We pre-calculate prefix sums in O(N). Then, we iterate up to `U - L + 1` times for the first element. For each choice, we validate the sequence of length N+1, which takes O(N) time. This leads to a total time complexity dominated by the nested loops.Space Complexity: O(N), where N is the length of `differences`. This space is used to store the `relative_seq` (prefix sums).

Pros and Cons

Pros:

Simple to understand and implement.
Correctly solves the problem for small ranges of [lower, upper].

Cons:

Extremely inefficient and will result in a 'Time Limit Exceeded' error for the given constraints.
The runtime depends on the magnitude of lower and upper, not just the size of the input array, making it unsuitable for large value ranges.

Code Solutions

Checking out 3 solutions in different languages for Count the Hidden Sequences. Click on different languages to view the code.

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));
  }
}

Video Solution

Watch the video walkthrough for Count the Hidden Sequences

Patterns:

Prefix Sum

Data Structures:

Array

Companies:

Zomato |