You are given 3 positive integers zero, one, and limit.
A binary array arr is called stable if:
arr is exactly zero.arr is exactly one.arr with a size greater than limit must contain both 0 and 1.Return the total number of stable binary arrays.
Since the answer may be very large, return it modulo 109 + 7.
Example 1:
Input: zero = 1, one = 1, limit = 2
Output: 2
Explanation:
The two possible stable binary arrays are [1,0] and [0,1].
Example 2:
Input: zero = 1, one = 2, limit = 1
Output: 1
Explanation:
The only possible stable binary array is [1,0,1].
Example 3:
Input: zero = 3, one = 3, limit = 2
Output: 14
Explanation:
All the possible stable binary arrays are [0,0,1,0,1,1], [0,0,1,1,0,1], [0,1,0,0,1,1], [0,1,0,1,0,1], [0,1,0,1,1,0], [0,1,1,0,0,1], [0,1,1,0,1,0], [1,0,0,1,0,1], [1,0,0,1,1,0], [1,0,1,0,0,1], [1,0,1,0,1,0], [1,0,1,1,0,0], [1,1,0,0,1,0], and [1,1,0,1,0,0].
Constraints:
1 <= zero, one, limit <= 1000This approach utilizes a hash map to store elements as keys and their frequencies as values. It helps in quick look-up and insertion, providing an efficient way to solve the problem.
This C solution implements a simple hash map using a hash function to solve the frequency counting problem. We define a hash map with buckets and use a linked list to handle collisions. Each array element is processed by hashing it and incrementing its frequency in the hash map. The solution handles collisions using a chaining method.
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Time Complexity: O(n) for processing n elements and O(1) average look-up time.
Space Complexity: O(n) for storing the elements in the hash map.
This approach involves sorting the array first, which groups identical elements together. After sorting, a single pass through the array is sufficient to count the frequencies of all elements.
This C solution sorts the array using the qsort function, making subsequent traversal straightforward for counting each element's frequency. After sorting, we iterate over the array, counting occurrences of each distinct element.
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Time Complexity: O(n log n) for sorting and O(n) for traversal, leading to O(n log n).
Space Complexity: O(1) if sorting is in-place or O(n) otherwise.
| Approach | Complexity |
|---|---|
| Approach 1: Using a Hash Map | Time Complexity: O(n) for processing n elements and O(1) average look-up time. |
| Approach 2: Using Sorting | Time Complexity: O(n log n) for sorting and O(n) for traversal, leading to O(n log n). |
Median of Two Sorted Arrays - Binary Search - Leetcode 4 • NeetCode • 551,948 views views
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