LeetCode #2145 — MEDIUM

Count the Hidden Sequences

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,-3,4]
1
6

Example 2

[3,-4,5,1,-2]
-4
5

Example 3

[4,-7,2]
3
6

Step 02

Core Insight

What unlocks the optimal approach

Fix the first element of the hidden sequence to any value x and ignore the given bounds. Notice that we can then determine all the other elements of the sequence by using the differences array.
We will also be able to determine the difference between the minimum and maximum elements of the sequence. Notice that the value of x does not affect this.
We now have the ‘range’ of the sequence (difference between min and max element), we can then calculate how many ways there are to fit this range into the given range of lower to upper.
Answer is (upper - lower + 1) - (range of sequence)

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2145: Count the Hidden Sequences
class Solution {
    public int numberOfArrays(int[] differences, int lower, int upper) {
        long x = 0, mi = 0, mx = 0;
        for (int d : differences) {
            x += d;
            mi = Math.min(mi, x);
            mx = Math.max(mx, x);
        }
        return (int) Math.max(upper - lower - (mx - mi) + 1, 0);
    }
}

// Accepted solution for LeetCode #2145: Count the Hidden Sequences
func numberOfArrays(differences []int, lower int, upper int) int {
	x, mi, mx := 0, 0, 0
	for _, d := range differences {
		x += d
		mi = min(mi, x)
		mx = max(mx, x)
	}
	return max(0, upper-lower-(mx-mi)+1)
}

# Accepted solution for LeetCode #2145: Count the Hidden Sequences
class Solution:
    def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int:
        x = mi = mx = 0
        for d in differences:
            x += d
            mi = min(mi, x)
            mx = max(mx, x)
        return max(upper - lower - (mx - mi) + 1, 0)

// Accepted solution for LeetCode #2145: Count the Hidden Sequences
/**
 * [2145] Count the Hidden Sequences
 *
 * 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 <= 10^5
 * 	-10^5 <= differences[i] <= 10^5
 * 	-10^5 <= lower <= upper <= 10^5
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-the-hidden-sequences/
// discuss: https://leetcode.com/problems/count-the-hidden-sequences/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn number_of_arrays(differences: Vec<i32>, lower: i32, upper: i32) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2145_example_1() {
        let differences = vec![1, -3, 4];
        let lower = 1;
        let upper = 6;

        let result = 2;

        assert_eq!(
            Solution::number_of_arrays(differences, lower, upper),
            result
        );
    }

    #[test]
    #[ignore]
    fn test_2145_example_2() {
        let differences = vec![3, -4, 5, 1, -2];
        let lower = -4;
        let upper = 5;

        let result = 4;

        assert_eq!(
            Solution::number_of_arrays(differences, lower, upper),
            result
        );
    }

    #[test]
    #[ignore]
    fn test_2145_example_3() {
        let differences = vec![4, -7, 2];
        let lower = 3;
        let upper = 6;

        let result = 0;

        assert_eq!(
            Solution::number_of_arrays(differences, lower, upper),
            result
        );
    }
}

// Accepted solution for LeetCode #2145: Count the Hidden Sequences
function numberOfArrays(differences: number[], lower: number, upper: number): number {
    let [x, mi, mx] = [0, 0, 0];
    for (const d of differences) {
        x += d;
        mi = Math.min(mi, x);
        mx = Math.max(mx, x);
    }
    return Math.max(0, upper - lower - (mx - mi) + 1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.