Leetcode 565. Array Nesting

Problem Explanation:

The problem deals with arrays and sets. Given an array, we are supposed to establish sets following a certain indexing rule. The rule requires each subsequent element in the set to be fetched from the array using the previous element as its index. The process continues until we encounter a duplicate element in the set. The task is to determine the size of the longest set we can obtain following this rule.

For example, consider the array A = [5,4,0,3,1,6,2]. Here, we can form the following set for index 0: S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}. We stop at this point as the next element would be 5, which is already present in the set.

In order to solve this problem, the key is to track the elements that have already been seen to avoid unnecessary repetitive operations. For each integer num in the array, if it hasn't been visited (-1 is used to indicate that an element has been visited), we start from it and keep accessing the array by using the current number as the index, until a duplicate is found.

Solution Approach:

The solution is simple in terms of both idea and implementation. We use a loop to iterate over every element of the array and for each element, we begin the operation of building a set following the rule stipulated in the problem statement. If we encounter a number that we have seen before, we immediately stop and move to the next iteration. After each iteration, we compare the size of the current set with the maximum size seen so far and update the maximum size accordingly.

Python Solution:

1
2python
3class Solution:
4    def arrayNesting(self, nums):
5        res = 0
6        for num in nums:
7            if num != -1:
8                start = num
9                count = 0
10                while nums[start] != -1:
11                    temp = start
12                    start = nums[start]
13                    nums[temp] = -1
14                    count += 1
15                res = max(res, count)
16        return res

Java Solution:

1
2java
3class Solution {
4    public int arrayNesting(int[] nums) {
5        int ans = 0;
6        for (int num : nums) {
7            if (num != -1) {
8                int start = num;
9                int count = 0;
10                while (nums[start] != -1) {
11                    int temp = start;
12                    start = nums[start];
13                    nums[temp] = -1;
14                    count++;
15                }
16                ans = Math.max(ans, count);
17            }
18        }
19        return ans;
20    }
21}

JavaScript Solution:

1
2javascript
3class Solution {
4  arrayNesting(nums) {
5    let ans = 0;
6    for (let num of nums) {
7      if (num != -1) {
8        let start = num;
9        let count = 0;
10        while (nums[start] != -1) {
11          let temp = start;
12          start = nums[start];
13          nums[temp] = -1;
14          count++;
15        }
16        ans = Math.max(ans, count);
17      }
18    }
19    return ans;
20  }
21};

C++ Solution:

1
2cpp
3class Solution {
4public:
5    int arrayNesting(vector<int>& nums) {
6        int ans = 0;
7        for  (int num : nums) {
8            if (num != -1) {
9                int start = num;
10                int count = 0;
11                while (nums[start] != -1) {
12                    int temp = start;
13                    start = nums[start];
14                    nums[temp] = -1;
15                    count++;
16                }
17                ans = max(ans, count);
18            }
19        }
20        return ans;
21    }
22};

C# Solution:

1
2csharp
3public class Solution {
4    public int ArrayNesting(int[] nums) {
5        int ans = 0;
6        for  (int num : nums) {
7            if (num != -1) {
8                int start = num;
9                int count = 0;
10                while (nums[start] != -1) {
11                    int temp = start;
12                    start = nums[start];
13                    nums[temp] = -1;
14                    count++;
15                }
16                ans = Math.Max(ans, count);
17            }
18        }
19        return ans;
20    }
21}

In all solution examples, ans is the maximum size of the array we have found so far, start is the start point of our current search, count is the size of the current array. During each iteration, num is the current integer, temp is used to cache the current start point, so that we can mark it as visited afterwards. If an integer has been seen before, we skip it and continue to the next iteration. All the solutions shared above loop through the array and keep track of each set seen so far. They also make a smart note of elements visited by marking them as -1. The solutions effectively illustrate the use of loops, conditional statements, and array manipulation techniques in Python, Java, JavaScript, C++, and C#.

One of the crucial learnings from this problem is the awareness to not reprocess an index which has already been processed. Processing the same index again will lead us to the same cycle thus, wasting computational resources. The elements of the visited cycles can be marked as explained above.

This problem illustrates the concept of array indexing and manipulation. It underlies that the array indices can themselves be the array's contents, leading to a system of deterministic jumps or transitions from one element to another. Array indexing is a fundamental concept used widely in the creation of more complex data structures.

Key points to remember while dealing with similar problems include understanding the problem constraints, keeping an eye out for potential areas of optimization, making proper use of data structures, and avoiding unnecessary computation.

Understanding this problem can be very helpful for those preparing for coding interviews as it touches upon essential programming concepts and demonstrates how understanding of basics can lead to optimization in problem-solving. Moreover, getting the solutions in multiple programming languages helps in building a better understanding of language syntax and their applications.


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