grid47 Exploring patterns and algorithms
Sep 11, 2024
5 min read
Solution to LeetCode 565: Array Nesting Problem
You are given an array nums, a permutation of numbers from [0, n-1], and for each index k, you need to build a set s[k] by repeatedly selecting elements based on the rule nums[k] -> nums[nums[k]] -> nums[nums[nums[k]]], stopping when a duplicate is found. Return the longest length of any such set s[k].
Problem
Approach
Steps
Complexity
Input: The input consists of an array nums containing a permutation of numbers from [0, n-1].
Example: Input: nums = [3, 1, 4, 0, 2]
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] < nums.length
• All values of nums are unique.
Output: The output should be a single integer representing the longest length of any set s[k].
Example: Output: 3
Constraints:
• The result should be an integer.
Goal: The goal is to find the longest set s[k] for any index k in the array nums.
Steps:
• Initialize a variable to store the maximum size of the set found.
• Iterate over the array, for each index k, follow the rule of building the set until a duplicate is found.
• Mark the visited elements as processed (e.g., by setting nums[k] to -1).
• Update the maximum set size during the iteration.
Goal: The problem constraints ensure that the array will contain unique values and its length will be within the specified limits.
Steps:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] < nums.length
• All values of nums are unique.
Assumptions:
• The input array is non-empty and contains unique values.
• The size of the array is within the given constraints.
• Input: Input: nums = [3, 1, 4, 0, 2]
• Explanation: Starting from nums[0] = 3, we build the set {3, 0}. The length of the longest set is 3.
• Input: Input: nums = [0, 1, 2]
• Explanation: Each element points to itself, so the longest set has a length of 1.
Approach: The problem can be solved by iterating through the array and, for each element, following the chain of elements until a repeat is found. The challenge is to avoid revisiting elements unnecessarily.
Observations:
• The process of following the chain of elements from any starting point is deterministic and involves tracking visited elements to avoid infinite loops.
• The solution will involve visiting each element once and marking it as visited to ensure no duplicates are counted in the chain.
Steps:
• Initialize a variable to store the maximum set size.
• Iterate over the array to process each element as the starting point for building the set.
• For each element, follow the chain of elements while marking visited elements to prevent revisits.
• Update the maximum set size found.
Empty Inputs:
• The input array will never be empty as per the problem constraints.
Large Inputs:
• The solution should efficiently handle input sizes up to 100,000 elements.
Special Values:
• If all elements point to themselves (like in a sorted array), the solution should return 1.
Constraints:
• The solution should handle large inputs efficiently and avoid unnecessary recomputations.
intarrayNesting(vector<int>& nums) {
int mxsize =0;
for(int i =0; i < nums.size(); i++) {
int size =0;
for(int k = i; nums[k] >=0; size++) {
int ak = nums[k];
nums[k] =-1;
k = ak;
}
mxsize = max(size, mxsize);
}
return mxsize;
}
1 : Function Definition
intarrayNesting(vector<int>& nums) {
Defines the function `arrayNesting` that takes a vector of integers `nums` and returns the size of the largest cycle (set of elements that point to each other).
2 : Variable Initialization
int mxsize =0;
Initializes the variable `mxsize` to store the size of the largest set found. This will be updated as the function iterates through the array.
3 : Outer Loop - Iterate Through Array
for(int i =0; i < nums.size(); i++) {
Starts an outer loop to iterate through each element in the array `nums`.
4 : Size Initialization
int size =0;
Initializes the variable `size` to 0 for each new cycle. This variable will store the number of elements in the current cycle.
5 : Inner Loop - Cycle Traversal
for(int k = i; nums[k] >=0; size++) {
Starts an inner loop to traverse the cycle starting from index `i`. It continues as long as the element at index `k` has not been visited (i.e., it's non-negative).
6 : Cycle Element Access
int ak = nums[k];
Stores the current value of the element at index `k` in `ak`. This is used to track the next element in the cycle.
7 : Mark Element as Visited
nums[k] =-1;
Marks the element at index `k` as visited by setting it to `-1`. This prevents revisiting the same element in future iterations.
8 : Move to Next Element in Cycle
k = ak;
Sets `k` to the value stored in `ak`, which represents the index of the next element in the cycle.
9 : Update Maximum Size
mxsize = max(size, mxsize);
Updates `mxsize` to the larger value between the current cycle size (`size`) and the largest cycle size found so far (`mxsize`).
10 : Return Maximum Size
return mxsize;
Returns the largest cycle size found during the traversal of the array.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in the size of the input array.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the need to store the visited elements and the intermediate sets.
