Today I have done Array Nesting and leetcode's June LeetCoding Challenge with cpp.

Array Nesting

Description

You are given an integer array nums of length n where nums is a permutation of the numbers in the range [0, n - 1].

You should build a set s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } subjected to the following rule:

• The first element in s[k] starts with the selection of the element nums[k] of index = k.
• The next element in s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.
• We stop adding right before a duplicate element occurs in s[k].

Return the longest length of a set s[k].

Example 1:

Input: nums = [5,4,0,3,1,6,2]
Output: 4
Explanation: 
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}

Example 2:

Input: nums = [0,1,2]
Output: 1

Constraints:

• 1 <= nums.length <= 10^5
• 0 <= nums[i] < nums.length
• All the values of nums are unique.

Solution

simple simulation

auto speedup = [](){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  return 0;
}();
class Solution {
public:
  int arrayNesting(vector<int>& nums) {
    int answer = 0;
    int len = nums.size();
    vector<bool> vis(len);
    for(int i = 0; i < len; ++i) {
      if(vis[i]) continue;
      int cur = i;
      int chain = 0;
      while(!vis[cur]) {
        chain += 1;
        vis[cur] = true;
        cur = nums[cur];
      }
      answer = max(answer, chain);
    }
    return answer;
  }
};

June LeetCoding Challenge 7

Description

Min Cost Climbing Stairs

You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

Example 1:

Input: cost = [10,15,20]
Output: 15
Explanation: Cheapest is: start on cost[1], pay that cost, and go to the top.

Example 2:

Input: cost = [1,100,1,1,1,100,1,1,100,1]
Output: 6
Explanation: Cheapest is: start on cost[0], and only step on 1s, skipping cost[3].

Constraints:

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

Solution

class Solution {
public:
  int minCostClimbingStairs(vector<int>& cost) {
    int dp[3];
    int len = cost.size();
    for(int i = 2; i <= len; ++i) {
      dp[i % 3] = min(dp[(i + 1) % 3] + cost[i - 2], dp[(i + 2) % 3] + cost[i - 1]);
    }
    return dp[len % 3];
  }
};