Leetcode 2145: Count the Hidden Sequences
You are given an array representing the differences between each pair of consecutive integers of a hidden sequence. You are also given two integers, lower and upper, which define an inclusive range that the elements of the hidden sequence can take. Your task is to determine the number of possible hidden sequences that fit the given differences and lie within the specified range.
Problem
Approach
Steps
Complexity
Input: The input consists of the following elements:
- An integer array differences of size n (1 ≤ n ≤ 10^5), where differences[i] = hidden[i + 1] - hidden[i].
- Two integers lower and upper (−10^5 ≤ lower ≤ upper ≤ 10^5) which specify the inclusive range of values the hidden sequence can contain.
Example: For example, given differences = [2, -5, 3], lower = -4, upper = 6, we are tasked with finding the number of possible hidden sequences.
Constraints:
• 1 <= n <= 10^5
• -10^5 <= differences[i] <= 10^5
• -10^5 <= lower <= upper <= 10^5
Output: The output should be a single integer that represents the number of valid hidden sequences that lie within the specified range, or 0 if no valid sequences exist.
Example: For differences = [2, -5, 3], lower = -4, upper = 6, the output would be 3.
Constraints:
Goal: To calculate the number of possible hidden sequences that satisfy the given differences and lie within the specified range.
Steps:
• 1. Initialize two variables, mn and mx, to track the minimum and maximum sum encountered during the process of generating the sequence from the given differences.
• 2. Iterate through the differences array and calculate the cumulative sum, updating mn and mx as necessary.
• 3. Check if the range between lower and upper is large enough to accommodate the difference between mx and mn. If not, return 0.
• 4. Otherwise, calculate the number of valid sequences by determining how many positions in the range [lower, upper] can fit the sequence's difference.
Goal: The input values should fall within the following constraints.
Steps:
• The length of differences (n) must be between 1 and 100,000.
• The values in the differences array must be between -100,000 and 100,000.
• The values of lower and upper must be between -100,000 and 100,000, and lower must be less than or equal to upper.
Assumptions:
• It is assumed that the input array differences is not empty and contains at least one element.
• It is assumed that the given range [lower, upper] is valid, with lower ≤ upper.
• Input: Example 1: differences = [2, -5, 3], lower = -4, upper = 6.
• Explanation: In this case, the possible hidden sequences are: [-3, 0, -5, -2], [0, 5, 0, 3], and [1, 6, 1, 4]. Thus, there are 3 valid sequences, so the output is 3.
• Input: Example 2: differences = [1, -3, 4], lower = 1, upper = 6.
• Explanation: Here, the two possible sequences are [3, 4, 1, 5] and [4, 5, 2, 6]. The output is 2.
Approach: To solve this problem, we can use a greedy approach by calculating the cumulative sum of the differences and checking if the resulting sequence can fit within the given range [lower, upper].
Observations:
• The problem is essentially about adjusting the sequence to fit the given range after applying the cumulative differences.
• The sequence will need to be adjusted so that the difference between the maximum and minimum values is within the bounds defined by lower and upper.
Steps:
• 1. Initialize two variables to track the cumulative minimum and maximum sums.
• 2. Iterate through the differences array to accumulate the values and update the min and max values.
• 3. Check if the difference between the max and min exceeds the allowed range. If so, return 0.
• 4. Otherwise, return the number of valid sequences by calculating how many positions in the range [lower, upper] can accommodate the sequence.
Empty Inputs:
• Handle the case where the input differences array is empty.
Large Inputs:
• For large input sizes (n up to 100,000), ensure the algorithm runs efficiently with O(n) time complexity.
Special Values:
• Handle the edge cases where differences contain large negative or positive numbers.
Constraints:
• Ensure that the solution handles the case where the hidden sequence cannot fit within the specified range.
int numberOfArrays(vector<int>& diff, long lower, long upper) {
// last - fist of n + 1 seq
long mn = 0, mx = 0, x = 0;
for(int k : diff) {
x += k;
mx = max(mx, x);
mn = min(mn, x);
}
int sum = mx - mn;
if (upper - lower < sum)
return 0;
else {
return (int)(upper - lower) - sum + 1;
}
}
1 : Function Declaration
int numberOfArrays(vector<int>& diff, long lower, long upper) {
This line defines the function `numberOfArrays`, which takes three parameters: a reference to a vector `diff`, and two long integers `lower` and `upper` representing the bounds of the valid range.
2 : Variable Initialization
long mn = 0, mx = 0, x = 0;
This initializes three long integers: `mn` for the minimum prefix sum, `mx` for the maximum prefix sum, and `x` for the current running sum of differences.
3 : Loop
for(int k : diff) {
This starts a loop that iterates over each element `k` in the vector `diff`, representing the difference array.
4 : Accumulate Sum
x += k;
This adds the current element `k` to the running sum `x`.
5 : Update Maximum
mx = max(mx, x);
This updates the maximum prefix sum `mx` by comparing it with the current running sum `x`.
6 : Update Minimum
mn = min(mn, x);
This updates the minimum prefix sum `mn` by comparing it with the current running sum `x`.
7 : Calculate Range
int sum = mx - mn;
This calculates the difference between the maximum and minimum prefix sums, storing the result in `sum`.
8 : Condition Check
if (upper - lower < sum)
This condition checks if the difference between the `upper` and `lower` bounds is smaller than the calculated sum `sum`.
9 : Return Zero
return 0;
If the condition is true, it means the range is too small to accommodate the required sum, so the function returns 0.
10 : Else Block
else {
If the condition is false (the range is large enough), the function proceeds to the next step.
11 : Return Result
return (int)(upper - lower) - sum + 1;
This line calculates and returns the number of valid arrays that can be formed by subtracting the sum from the range and adding 1.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear, as we need to process each element of the differences array once.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant, as we only need a few extra variables for tracking the cumulative min and max sums.
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