Tracking Hidden Values

Number of Valid Hidden Sequences 

Problem Statement

You're given a 0-indexed array differences of n integers, describing how each element in a hidden sequence changes from the previous one. That is:

differences[i] = hidden[i + 1] - hidden[i]

You're also given two integers lower and upper, defining the inclusive range that all elements of the hidden sequence must lie within.

Return the number of possible hidden sequences that satisfy all constraints.

Best Data Structure

This problem doesn't require advanced data structures, but the core technique involves prefix sums. This helps in tracking how the hidden sequence evolves and in determining its valid range of starting points.

Different Approaches

Brute Force Simulation (Naive and Inefficient)

We try every possible value for the first element in the hidden sequence, ranging from lower to upper, then simulate the full sequence and check if it remains within bounds.


class Solution {
    public int numberOfArrays(int[] differences, int lower, int upper) {
        int count = 0;
        for (int start = lower; start <= upper; start++) {
            int val = start;
            boolean isValid = true;
            for (int diff : differences) {
                val += diff;
                if (val < lower || val > upper) {
                    isValid = false;
                    break;
                }
            }
            if (isValid) count++;
        }
        return count;
    }
}

This brute force method is simple but inefficient. 

Optimized Approach – Using Prefix Sum Ranges

Rather than checking every start value, we simulate the sequence assuming it starts at zero, and track the minimum and maximum prefix sums along the way. These values tell us how far the sequence might go above or below its start. We then shift the entire sequence up or down within the [lower, upper] range to find valid starting points.

class Solution {
    public int numberOfArrays(int[] differences, int lower, int upper) {
        long prefixSum = 0;
        long minPrefixSum = 0;
        long maxPrefixSum = 0;
        
        for (int diff : differences) {
            prefixSum += diff;
            minPrefixSum = Math.min(minPrefixSum, prefixSum);
            maxPrefixSum = Math.max(maxPrefixSum, prefixSum);
        }

        long validStartRange = (upper - lower) - (maxPrefixSum - minPrefixSum) + 1;
        return (int)Math.max(0, validStartRange);
    }
}

This approach is highly efficient.

Cleaner Variant – Same Logic, Neater Variables

class Solution {
    public int numberOfArrays(int[] differences, int lower, int upper) {
        long current = 0, min = 0, max = 0;
        for (int diff : differences) {
            current += diff;
            min = Math.min(min, current);
            max = Math.max(max, current);
        }
        return (int)Math.max(0, (upper - lower) - (max - min) + 1);
    }
}

⚠️Common Pitfalls

  • Forgetting to track both min and max of the prefix sums. These bounds are crucial to understand the range of values the sequence can reach.

  • Incorrectly assuming the sequence always starts at lower or 0. The valid start could be anywhere in [lower, upper] that allows the entire sequence to stay within bounds.

  • Overflow risk. Prefix sums may go beyond the int range, especially with large values, so use long to be safe.

📊Time and Space Complexity Analysis

The brute force solution has a time complexity of O((upper - lower + 1) * n) and a space complexity of O(1). This makes it impractical for large ranges of possible start values.

The optimized prefix sum approach has a time complexity of O(n) since we only iterate through the differences array once. The space complexity is O(1) because no extra data structures are used, only a few tracking variables.

🌍Real-World Applications

  • Signal Recovery: Reconstructing a signal from its differences under known amplitude limits.

  • Version Control Systems: Estimating valid states after a sequence of changes or patches.

  • Predictive Modelling: Understanding possible values a parameter can take after a series of incremental updates.

  • Constraint Satisfaction Problems: Finding sequences within allowed bounds after transformations.

I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

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