Problem Description

You are given an array differences that describes the differences between consecutive elements in a hidden sequence. Specifically, if the hidden sequence is [a, b, c, d], then differences = [b-a, c-b, d-c].

The hidden sequence must satisfy two constraints:

1. It has length n + 1 (where n is the length of differences)
2. All elements must be within the range [lower, upper] (inclusive)

Your task is to find how many different starting values for the hidden sequence would result in a valid sequence.

For example, with differences = [1, -3, 4], lower = 1, and upper = 6:

• If we start with 3, we get the sequence [3, 4, 1, 5] (since 3+1=4, 4-3=1, 1+4=5)
• If we start with 4, we get the sequence [4, 5, 2, 6]
• Both are valid since all elements are between 1 and 6

However:

• Starting with 5 gives [5, 6, 3, 7], which is invalid because 7 > 6
• The sequence [1, 2, 3, 4] doesn't match the given differences

The solution uses prefix sums to track the relative positions of all elements. By assuming the first element is 0 and calculating all subsequent elements using the differences, we find the minimum (mi) and maximum (mx) values in this relative sequence. The spread mx - mi represents the range that any valid sequence must span.

If this spread is larger than upper - lower, no valid sequence exists. Otherwise, we can shift the sequence by any amount from lower - mi to upper - mx, giving us upper - lower - (mx - mi) + 1 possible starting positions.

Intuition

The key insight is that once we fix the differences array, the shape of the hidden sequence is completely determined - only its position on the number line can vary.

Think of it like a rigid stick with fixed bends. The differences array tells us exactly how the stick bends, but we can slide this entire stick up or down along the number line. The question becomes: how many positions can we place this stick such that it stays within the boundaries [lower, upper]?

To find this, we can start by placing the first element at position 0 and build the entire sequence using the differences. This gives us a "template" sequence. For instance, with differences = [1, -3, 4], starting at 0 gives us [0, 1, -2, 2].

Now we observe that this sequence spans from -2 to 2, a range of 4 units. No matter where we shift this sequence, it will always span exactly 4 units. If our allowed range [lower, upper] is smaller than 4 units, it's impossible to fit the sequence.

If the allowed range is large enough, we need to count valid starting positions. The lowest we can place our sequence is when its minimum value touches lower (shifting everything up by lower - mi). The highest is when its maximum value touches upper (shifting everything up by upper - mx).

The number of integer positions between these two extremes is (upper - mx) - (lower - mi) + 1, which simplifies to upper - lower - (mx - mi) + 1.

This approach elegantly avoids checking every possible starting value by recognizing that the relative positions are fixed, and we only need to determine the valid range for shifting the entire sequence.

Learn more about Prefix Sum patterns.

Solution Approach

The implementation follows a prefix sum pattern to track the relative positions of all elements in the hidden sequence.

Step 1: Initialize tracking variables

• Set x = 0 to represent the current cumulative sum (position relative to the first element)
• Set mi = 0 and mx = 0 to track the minimum and maximum positions reached

Step 2: Calculate the range of the sequence

for d in differences:
    x += d
    mi = min(mi, x)
    mx = max(mx, x)

We iterate through each difference and accumulate it in x. This simulates building the hidden sequence starting from position 0:

• If differences = [1, -3, 4], then:
  • After first difference: x = 0 + 1 = 1, sequence so far: [0, 1]
  • After second difference: x = 1 + (-3) = -2, sequence so far: [0, 1, -2]
  • After third difference: x = -2 + 4 = 2, sequence so far: [0, 1, -2, 2]

During this process, we track:

• mi = -2 (the lowest point reached)
• mx = 2 (the highest point reached, considering the initial 0)

Step 3: Calculate valid starting positions

return max(upper - lower - (mx - mi) + 1, 0)

The expression mx - mi gives us the total range that the sequence spans. For a valid sequence to exist within [lower, upper]:

• The available space is upper - lower
• The required space is mx - mi
• The number of valid starting positions is the difference plus 1: upper - lower - (mx - mi) + 1

We use max(..., 0) to handle the case where the required range exceeds the available range, returning 0 when no valid sequences exist.

This solution runs in O(n) time with O(1) space, efficiently determining the count without explicitly constructing any sequences.

Example Walkthrough

Let's walk through the solution with differences = [2, -5, 3], lower = -1, and upper = 4.

Step 1: Build the relative sequence starting from 0

Starting with x = 0, we process each difference:

• Initial position: x = 0, sequence: [0]
• After difference 2: x = 0 + 2 = 2, sequence: [0, 2]
• After difference -5: x = 2 + (-5) = -3, sequence: [0, 2, -3]
• After difference 3: x = -3 + 3 = 0, sequence: [0, 2, -3, 0]

Track minimum and maximum:

• mi = -3 (lowest point in the sequence)
• mx = 2 (highest point in the sequence)

Step 2: Calculate the sequence range

The sequence spans from -3 to 2, requiring a range of mx - mi = 2 - (-3) = 5 units.

Step 3: Determine valid starting positions

Our allowed range is [lower, upper] = [-1, 4], which spans 4 - (-1) = 5 units.

To find valid starting positions:

• If we shift the sequence so its minimum -3 aligns with lower = -1, we add 2 to each element: [2, 4, -1, 2] ✓
• If we shift the sequence so its maximum 2 aligns with upper = 4, we add 2 to each element: [2, 4, -1, 2] ✓

In this case, both extremes give us the same sequence! This happens because the required range (5) exactly matches the available range (5).

Valid starting positions = upper - lower - (mx - mi) + 1 = 4 - (-1) - 5 + 1 = 1

Only one starting value (2) produces a valid sequence. Any other starting value would push either the minimum below -1 or the maximum above 4.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array differences. This is because the algorithm iterates through the differences array exactly once, performing constant-time operations (addition, min/max comparisons) for each element.

The space complexity is O(1). The algorithm only uses a fixed number of variables (x, mi, mx) regardless of the input size, requiring constant extra space.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Consider the Initial Element (0th position)

A common mistake is only tracking the cumulative sum values without considering that the initial element (before any differences are applied) also counts toward the minimum and maximum values.

Incorrect approach:

# Wrong: Only considers positions after applying differences
min_cumulative = float('inf')
max_cumulative = float('-inf')
cumulative_sum = 0

for difference in differences:
    cumulative_sum += difference
    min_cumulative = min(min_cumulative, cumulative_sum)
    max_cumulative = max(max_cumulative, cumulative_sum)

Why it fails: If all differences are positive (e.g., differences = [2, 3, 1]), the minimum position would incorrectly be set to 2 instead of 0, leading to an incorrect range calculation.

Correct approach:

# Initialize with 0 to include the starting position
min_cumulative = 0
max_cumulative = 0
cumulative_sum = 0

for difference in differences:
    cumulative_sum += difference
    min_cumulative = min(min_cumulative, cumulative_sum)
    max_cumulative = max(max_cumulative, cumulative_sum)

2. Off-by-One Error in Counting Valid Positions

Another frequent error is miscalculating the number of valid starting positions by forgetting to add 1 to account for inclusive boundaries.

Incorrect formula:

# Wrong: Missing the +1
return max(upper - lower - (max_cumulative - min_cumulative), 0)

Why it fails: When calculating the number of integers in a range [a, b], the count is b - a + 1, not b - a. The same principle applies here when counting valid starting positions.

Correct formula:

# Correct: Include the +1 for inclusive counting
return max(upper - lower - (max_cumulative - min_cumulative) + 1, 0)

Example to illustrate: If lower = 1, upper = 3, and the sequence span is 1, the valid starting values could be 1, 2, or 3 (three positions), but 3 - 1 - 1 = 1 would incorrectly give only one position.

3. Integer Overflow in Languages with Fixed Integer Size

While Python handles arbitrary precision integers automatically, in languages like Java or C++, accumulating differences could cause overflow.

Solution for other languages:

// Use long instead of int in Java
long cumulativeSum = 0;
long minCumulative = 0;
long maxCumulative = 0;

4. Not Handling Edge Cases Properly

Failing to handle cases where the required range exceeds the available range, resulting in negative counts.

Incorrect:

# Wrong: Could return negative values
return upper - lower - (max_cumulative - min_cumulative) + 1

Correct:

# Use max to ensure non-negative result
return max(upper - lower - (max_cumulative - min_cumulative) + 1, 0)