Count the Hidden Sequences - Practice Coding Problems

Count the Hidden Sequences - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: differences = [1,-3,4], lower = 1, upper = 6

› Output: 2

💡 Note: Hidden sequences of length 4. Starting with 3: [3,4,1,5] ✓. Starting with 4: [4,5,2,6] ✓. Starting with 5: [5,6,3,7] ✗ (7 > 6). Valid sequences: 2

Example 2 — No Valid Sequences

$ Input: differences = [4,-7,2], lower = 3, upper = 6

› Output: 0

💡 Note: Any sequence will go outside bounds. For example, start=3: [3,7,0,2] has 7>6 and 0<3. No valid starting position exists.

Example 3 — Single Valid Start

$ Input: differences = [1,-1,1], lower = 1, upper = 3

› Output: 1

💡 Note: Only starting with 2 works: [2,3,2,3]. All values stay within [1,3]. Other starts go out of bounds.

Constraints

1 ≤ differences.length ≤ 10⁵
-10⁵ ≤ differences[i] ≤ 10⁵
-10⁸ ≤ lower ≤ upper ≤ 10⁸

Visualization

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Understanding the Visualization

Input

Differences [1,-3,4] and bounds [1,6]

Process

Generate sequences for each starting position

Output

Count sequences that stay within bounds

Key Takeaway

🎯 Key Insight: Use prefix sums to efficiently determine the valid range of starting positions without simulating every sequence

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that we can use prefix sums to determine the valid range of starting positions without simulating every sequence. Calculate prefix sums to find min/max relative positions, then compute how many starting values keep all sequence elements within bounds. Best approach uses prefix sum optimization: Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O((upper - lower) × n)	O(1)	Try every possible starting value and simulate the sequence
Prefix Sum Optimization	O(n)	O(1)	Use prefix sums to find the range bounds efficiently

Brute Force — Algorithm Steps

Iterate through all possible starting values from lower to upper
For each start, build the sequence by adding differences
Check if all sequence values are within [lower, upper]
Count valid sequences

Visualization

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Step-by-Step Walkthrough

Try Start=1

Build sequence [1,2,-1,3] - valid

Try Start=2

Build sequence [2,3,0,4] - valid

Count Valid

Count all sequences that stay in bounds

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* differences, int n, int lower, int upper) {
    int count = 0;
    for (int start = lower; start <= upper; start++) {
        int current = start;
        int valid = 1;
        for (int i = 0; i < n; i++) {
            current += differences[i];
            if (current < lower || current > upper) {
                valid = 0;
                break;
            }
        }
        if (valid) count++;
    }
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int differences[1000];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] != '\n' && token[0] != ' ' && token[0] != '\0') {
            differences[n++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int lower, upper;
    scanf("%d", &lower);
    scanf("%d", &upper);
    
    printf("%d\n", solution(differences, n, lower, upper));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O((upper - lower) × n)

For each of (upper-lower+1) starting values, we simulate n steps

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track current position and count

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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