2145. Count the Hidden Sequences Solution

Q: Is Count the Hidden Sequences easy or hard?

Count the Hidden Sequences is generally classified as a medium difficulty problem. The implementation is simple once you recognize that the differences array defines prefix displacement and that only the minimum and maximum offsets are needed.

Q: Count the Hidden Sequences Python/Java solution

Python and Java implementations both follow the same logic: iterate through the differences array, update the running prefix sum, track min and max prefix values, then compute the count of valid starting numbers using the derived bounds. The algorithm runs in O(n) time with constant extra space.

Q: How to solve Count the Hidden Sequences in O(n)?

Iterate through the differences array and maintain a running prefix sum. Track the smallest and largest prefix value encountered. These represent how far the sequence can move relative to the starting value. The valid starting range becomes [lower - minPrefix, upper - maxPrefix], and the number of integers in that interval is the answer.

Q: What is the best approach for Count the Hidden Sequences?

The prefix sum and range validation approach is the most efficient and commonly used solution. Track the minimum and maximum prefix sums produced by the differences array, then compute the valid starting value range that keeps the sequence within [lower, upper]. This runs in O(n) time and O(1) space.

Q: Is Count the Hidden Sequences asked at Google/Amazon/Meta?

Problems involving prefix sums, difference arrays, and range constraints appear frequently in interviews at companies like Google, Amazon, and Meta. This problem specifically tests the ability to convert relative differences into absolute bounds using prefix sums.

Q: What data structure is used in Count the Hidden Sequences?

The solution mainly relies on arrays and prefix sum calculations. No complex data structures are required. A few variables track cumulative sums and range boundaries while iterating through the differences array.

Q: What is the time complexity of Count the Hidden Sequences?

The optimal solution runs in O(n) time because the differences array is scanned once while computing prefix movement. Only constant extra variables are used to track the minimum and maximum prefix sums, giving O(1) space complexity.

The 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.

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

Approach overview

Problem Overview: You are given a differences array where differences[i] = hidden[i+1] - hidden[i]. The hidden sequence must stay within the range [lower, upper]. The task is to count how many valid starting values produce a full sequence that never goes outside these bounds.

Approach 1: Prefix Sum and Range Validation (O(n) time, O(1) space)

The differences array describes relative movement between consecutive elements. Treat the first hidden value as x and build prefix sums of the differences to represent offsets from that start. While iterating through the array, track the minimum and maximum prefix value reached. These offsets determine the smallest and largest value the sequence could reach relative to x. For the sequence to stay inside [lower, upper], the start value must satisfy x + minPrefix ≥ lower and x + maxPrefix ≤ upper. This produces a valid range for x. The count of integers in that range is the answer. This approach relies on cumulative movement using prefix sum concepts and processes the array once.

Approach 2: Dynamic Range Computation (O(n) time, O(1) space)

Instead of computing prefix extremes first, you can maintain the allowable range of the current hidden value while scanning the array. Start with [lower, upper] as the valid range for the first element. When processing differences[i], shift the allowable range to represent the next value in the sequence: every possible value becomes current + differences[i]. Intersect this shifted range with [lower, upper] to ensure the sequence remains valid. Continue this update for all elements. At the end, the remaining width of the starting range represents the number of valid hidden sequences. This approach models constraints dynamically and works well when reasoning about value intervals across an array.

Recommended for interviews: The prefix sum range validation approach is the one most interviewers expect. It reduces the problem to tracking the minimum and maximum displacement produced by the differences array, then computing the number of valid starting values with simple arithmetic. Showing the dynamic range idea first demonstrates understanding of constraints, but the prefix-sum formulation shows stronger pattern recognition with prefix sums and leads directly to a clean O(n) solution.

Approach 1: Prefix Sum and Range Validation

This approach uses the concept of a prefix sum to efficiently monitor the cumulative effects of the differences array. By calculating the prefix sum, we can track how the potential hidden sequence evolves from a starting point. Subsequently, determining the feasible range for the starting point of the hidden sequence allows us to count how many valid hidden sequences exist within the given bounds.

Python

def count_hidden_sequences(differences, lower, upper):
    min_prefix, max_prefix, current = 0, 0, 0
    for diff in differences:
        current += diff
        min_prefix = min(min_prefix, current)
        max_prefix = max(max_prefix, current)
    min_start = lower - min_prefix
    max_start = upper - max_prefix
    return max(0, max_start - min_start + 1)

print(count_hidden_sequences([1, -3, 4], 1, 6))

Java

public class HiddenSequences {
    public int countHiddenSequences(int[] differences, int lower, int upper) {
        long minPrefix = 0, maxPrefix = 0, currPrefix = 0;
        for (int diff : differences) {
            currPrefix += diff;
            minPrefix = Math.min(minPrefix, currPrefix);
            maxPrefix = Math.max(maxPrefix, currPrefix);
        }
        long minStart = lower - minPrefix;
        long maxStart = upper - maxPrefix;
        return maxStart >= minStart ? (int)(maxStart - minStart + 1) : 0;
    }

    public static void main(String[] args) {
        HiddenSequences hs = new HiddenSequences();
        int[] differences = {1, -3, 4};
        System.out.println(hs.countHiddenSequences(differences, 1, 6));
    }
}

C++

#include <vector>
#include <algorithm>
using namespace std;

int countHiddenSequences(vector<int>& differences, int lower, int upper) {
    long long minPrefix = 0, maxPrefix = 0, currentPrefix = 0;
    for (int diff : differences) {
        currentPrefix += diff;
        minPrefix = min(minPrefix, currentPrefix);
        maxPrefix = max(maxPrefix, currentPrefix);
    }
    long long minStart = lower - minPrefix;
    long long maxStart = upper - maxPrefix;
    return max(0LL, maxStart - minStart + 1);
}

JavaScript

function countHiddenSequences(differences, lower, upper) {
    let minPrefix = 0, maxPrefix = 0, current = 0;
    for (let diff of differences) {
        current += diff;
        minPrefix = Math.min(minPrefix, current);
        maxPrefix = Math.max(maxPrefix, current);
    }
    const minStart = lower - minPrefix;
    const maxStart = upper - maxPrefix;
    return Math.max(0, maxStart - minStart + 1);
}

console.log(countHiddenSequences([1, -3, 4], 1, 6));

#include <stdio.h>

int count_hidden_sequences(int* differences, int n, int lower, int upper) {
    long long min_prefix = 0;
    long long max_prefix = 0;
    long long current_prefix = 0;

    for (int i = 0; i < n; i++) {
        current_prefix += differences[i];
        if (current_prefix < min_prefix) min_prefix = current_prefix;
        if (current_prefix > max_prefix) max_prefix = current_prefix;
    }

    long long min_start = lower - min_prefix;
    long long max_start = upper - max_prefix;

    if (min_start > max_start) return 0;
    return (int)(max_start - min_start + 1);
}

int main() {
    int differences[] = {1, -3, 4};
    int n = sizeof(differences) / sizeof(differences[0]);
    int lower = 1, upper = 6;
    printf("%d\n", count_hidden_sequences(differences, n, lower, upper));
    return 0;
}

using System;

public class HiddenSequences {
    public int CountHiddenSequences(int[] differences, int lower, int upper) {
        long minPrefix = 0, maxPrefix = 0, currentPrefix = 0;
        foreach (int diff in differences) {
            currentPrefix += diff;
            minPrefix = Math.Min(minPrefix, currentPrefix);
            maxPrefix = Math.Max(maxPrefix, currentPrefix);
        }
        long minStart = lower - minPrefix;
        long maxStart = upper - maxPrefix;
        return (int)Math.Max(0, maxStart - minStart + 1);
    }

    public static void Main() {
        HiddenSequences hs = new HiddenSequences();
        Console.WriteLine(hs.CountHiddenSequences(new int[]{1, -3, 4}, 1, 6));
    }
}

In this Python implementation, we iterate over the differences list to determine the effect on a running prefix sum. The goal is to find the smallest and largest possible values of the prefix sum, calculate the possible starting points for the hidden sequence, and determine how many such starting points are valid according to the given constraints.

Complexity

Time Complexity: O(n), where n is the number of elements in the differences list. Space Complexity: O(1), with only a fixed number of variables used.

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Approach 2: Dynamic Range Computation

An alternative method involves dynamically computing the range of possible values for each element in the hidden sequence. Starting with a base value, each element's value is recursively determined based on previous elements and the differences array. Finally, the overall range of valid starting points is deduced by verifying constraints imposed by lower and upper bounds.

Python

def count_hidden_sequences(differences, lower, upper):
    prefix_sums = [0]
    for diff in differences:
        prefix_sums.append(prefix_sums[-1] + diff)
    min_val, max_val = min(prefix_sums), max(prefix_sums)
    min_start = lower - min_val
    max_start = upper - max_val
    return max(0, max_start - min_start + 1)

print(count_hidden_sequences([3, -4, 5, 1, -2], -4, 5))

Java

import java.util.*;

public class HiddenSequences {
    public int countHiddenSequences(int[] differences, int lower, int upper) {
        long[] prefixSums = new long[differences.length + 1];
        for (int i = 0; i < differences.length; i++) {
            prefixSums[i + 1] = prefixSums[i] + differences[i];
        }
        long minVal = Arrays.stream(prefixSums).min().getAsLong();
        long maxVal = Arrays.stream(prefixSums).max().getAsLong();
        long minStart = lower - minVal;
        long maxStart = upper - maxVal;
        return maxStart >= minStart ? (int)(maxStart - minStart + 1) : 0;
    }

    public static void main(String[] args) {
        HiddenSequences hs = new HiddenSequences();
        int[] differences = {3, -4, 5, 1, -2};
        System.out.println(hs.countHiddenSequences(differences, -4, 5));
    }
}

C++

#include <vector>
#include <algorithm>
using namespace std;

int countHiddenSequences(vector<int>& differences, int lower, int upper) {
    vector<long long> prefix_sums(differences.size() + 1, 0);
    for (int i = 0; i < differences.size(); ++i) {
        prefix_sums[i + 1] = prefix_sums[i] + differences[i];
    }
    auto minmax = minmax_element(prefix_sums.begin(), prefix_sums.end());
    long long minVal = *minmax.first;
    long long maxVal = *minmax.second;
    long long minStart = lower - minVal;
    long long maxStart = upper - maxVal;
    return max(0LL, maxStart - minStart + 1);
}

JavaScript

function countHiddenSequences(differences, lower, upper) {
    let prefixSums = [0];
    for (let i = 0; i < differences.length; i++) {
        prefixSums.push(prefixSums[i] + differences[i]);
    }
    const minVal = Math.min(...prefixSums);
    const maxVal = Math.max(...prefixSums);
    const minStart = lower - minVal;
    const maxStart = upper - maxVal;
    return Math.max(0, maxStart - minStart + 1);
}

console.log(countHiddenSequences([3, -4, 5, 1, -2], -4, 5));

#include <stdio.h>

int count_hidden_sequences(int* differences, int n, int lower, int upper) {
    long long* range_starts = (long long*)malloc((n + 1) * sizeof(long long));
    range_starts[0] = 0;

    for (int i = 1; i <= n; i++) {
        range_starts[i] = range_starts[i - 1] + differences[i - 1];
    }

    long long minVal = 0, maxVal = 0;
    for (int i = 0; i <= n; i++) {
        if (range_starts[i] < minVal) minVal = range_starts[i];
        if (range_starts[i] > maxVal) maxVal = range_starts[i];
    }

    free(range_starts);
    long long min_start = lower - minVal;
    long long max_start = upper - maxVal;

    return min_start > max_start ? 0 : (int)(max_start - min_start + 1);
}

int main() {
    int differences[] = {3, -4, 5, 1, -2};
    int n = sizeof(differences) / sizeof(differences[0]);
    int lower = -4, upper = 5;
    printf("%d\n", count_hidden_sequences(differences, n, lower, upper));
    return 0;
}

using System;
using System.Linq;

public class HiddenSequences {
    public int CountHiddenSequences(int[] differences, int lower, int upper) {
        long[] prefixSums = new long[differences.Length + 1];
        for (int i = 0; i < differences.Length; ++i) {
            prefixSums[i + 1] = prefixSums[i] + differences[i];
        }
        long minVal = prefixSums.Min();
        long maxVal = prefixSums.Max();
        long minStart = lower - minVal;
        long maxStart = upper - maxVal;
        return Math.Max(0, (int)(maxStart - minStart + 1));
    }

    public static void Main() {
        HiddenSequences hs = new HiddenSequences();
        Console.WriteLine(hs.CountHiddenSequences(new int[]{3, -4, 5, 1, -2}, -4, 5));
    }
}

This Python example constructs prefix sums, representing logical ranges shaped by differences. By establishing the smallest and largest of these calculated sums, the algorithm computes how many possible starting values exist within the specified bounds.

Complexity

Time Complexity: O(n), due to prefix sum computation and iterative processing. Space Complexity: O(n), using space for calculated prefix sums.

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Approach 3: Prefix Sum

Since the array differences is already determined, the difference between the maximum and minimum values of the elements in the array hidden is also fixed. We just need to ensure that this difference does not exceed upper - lower.

Let's assume the first element of the array hidden is 0. Then, hidden[i] = hidden[i - 1] + differences[i - 1], where 1 leq i leq n. Let the maximum value of the array hidden be mx and the minimum value be mi. If mx - mi leq upper - lower, then we can construct a valid hidden array. The number of possible constructions is upper - lower - (mx - mi) + 1. Otherwise, it is impossible to construct a valid hidden array, and we return 0.

The time complexity is O(n), where n is the length of the array differences. The space complexity is O(1).

TypeScript

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);
}

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)
}

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Complexity comparison

Approach	Complexity
Prefix Sum and Range Validation	Time Complexity: O(n), where n is the number of elements in the differences list. Space Complexity: O(1), with only a fixed number of variables used.
Dynamic Range Computation	Time Complexity: O(n), due to prefix sum computation and iterative processing. Space Complexity: O(n), using space for calculated prefix sums.
Prefix Sum	–

Complexity analysis

Approach	Time	Space	When to Use
Prefix Sum and Range Validation	O(n)	O(1)	Best general solution. Quickly computes valid starting values using prefix displacement tracking.
Dynamic Range Computation	O(n)	O(1)	Useful when reasoning about interval constraints step-by-step across the sequence.

Frequently asked questions

Is Count the Hidden Sequences easy or hard?

Count the Hidden Sequences is generally classified as a medium difficulty problem. The implementation is simple once you recognize that the differences array defines prefix displacement and that only the minimum and maximum offsets are needed.

Count the Hidden Sequences Python/Java solution

Python and Java implementations both follow the same logic: iterate through the differences array, update the running prefix sum, track min and max prefix values, then compute the count of valid starting numbers using the derived bounds. The algorithm runs in O(n) time with constant extra space.

How to solve Count the Hidden Sequences in O(n)?

Iterate through the differences array and maintain a running prefix sum. Track the smallest and largest prefix value encountered. These represent how far the sequence can move relative to the starting value. The valid starting range becomes [lower - minPrefix, upper - maxPrefix], and the number of integers in that interval is the answer.

What is the best approach for Count the Hidden Sequences?

The prefix sum and range validation approach is the most efficient and commonly used solution. Track the minimum and maximum prefix sums produced by the differences array, then compute the valid starting value range that keeps the sequence within [lower, upper]. This runs in O(n) time and O(1) space.

Is Count the Hidden Sequences asked at Google/Amazon/Meta?

Problems involving prefix sums, difference arrays, and range constraints appear frequently in interviews at companies like Google, Amazon, and Meta. This problem specifically tests the ability to convert relative differences into absolute bounds using prefix sums.

What data structure is used in Count the Hidden Sequences?

The solution mainly relies on arrays and prefix sum calculations. No complex data structures are required. A few variables track cumulative sums and range boundaries while iterating through the differences array.

What is the time complexity of Count the Hidden Sequences?

The optimal solution runs in O(n) time because the differences array is scanned once while computing prefix movement. Only constant extra variables are used to track the minimum and maximum prefix sums, giving O(1) space complexity.

Count the Hidden Sequences solution and explanation

The problem

Approach overview

Approach 1: Prefix Sum and Range Validation

Approach 2: Dynamic Range Computation

Approach 3: Prefix Sum

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

Count the Hidden Sequences solution and explanation

The problem

Approach overview

Approach 1: Prefix Sum and Range Validation

Approach 2: Dynamic Range Computation

Approach 3: Prefix Sum

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

Related problems