How do you solve Find the Number of Copy Arrays on LeetCode?

Find the Number of Copy Arrays (LeetCode #3468) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The first element determines the rest of the array.

What is the time complexity of Find the Number of Copy Arrays?

The optimal solution for Find the Number of Copy Arrays runs in O(n) time and uses O(1) space. We only make a single pass through the array, updating ranges, resulting in O(n) time complexity.

What is the brute force solution for Find the Number of Copy Arrays?

The brute force approach for Find the Number of Copy Arrays is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible arrays and check if they satisfy the conditions. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Number of Copy Arrays?

Find the Number of Copy Arrays uses the following concepts: Array, Math. The recommended patterns to study are: Range Queries, Dynamic Programming.

What are the interview tips for Find the Number of Copy Arrays?

Explain your thought process clearly. Use examples to illustrate your approach. Practice similar problems for better understanding.

What are common mistakes when solving Find the Number of Copy Arrays?

Not considering the bounds correctly. Forgetting to check if low exceeds high.

#3468

Find the Number of Copy Arrays

Medium

ArrayMathRange QueriesDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can derive the range of valid values for each copy[i] based on the previous value and the bounds, reducing the problem to a single pass.

⚙️

Algorithm

3 steps

1Step 1: Initialize the range for copy[0] as [u[0], v[0]].
2Step 2: For each index i, update the range for copy[i] based on the previous range and the difference from original.
3Step 3: Calculate the number of valid integers in the final range.

solution.py8 lines

1def countCopyArrays(original, bounds):
2    low, high = bounds[0][0], bounds[0][1]
3    for i in range(1, len(original)):
4        diff = original[i] - original[i - 1]
5        low = max(low + diff, bounds[i][0])
6        high = min(high + diff, bounds[i][1])
7        if low > high: return 0
8    return high - low + 1

ℹ

Complexity note: We only make a single pass through the array, updating ranges, resulting in O(n) time complexity.

1The first element determines the rest of the array.
2The difference between consecutive elements must be consistent.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.