Today I have done Design Browser History and leetcode's July LeetCoding Challenge with cpp.

Design Browser History

Description

You have a browser of one tab where you start on the homepage and you can visit another url, get back in the history number of steps or move forward in the history number of steps.

Implement the BrowserHistory class:

• BrowserHistory(string homepage) Initializes the object with the homepage of the browser.
• void visit(string url) Visits url from the current page. It clears up all the forward history.
• string back(int steps) Move steps back in history. If you can only return x steps in the history and steps > x, you will return only x steps. Return the current url after moving back in history at most steps.
• string forward(int steps) Move steps forward in history. If you can only forward x steps in the history and steps > x, you will forward only x steps. Return the current url after forwarding in history at most steps.

Example:

Input:
["BrowserHistory","visit","visit","visit","back","back","forward","visit","forward","back","back"]
[["leetcode.com"],["google.com"],["facebook.com"],["youtube.com"],[1],[1],[1],["linkedin.com"],[2],[2],[7]]
Output:
[null,null,null,null,"facebook.com","google.com","facebook.com",null,"linkedin.com","google.com","leetcode.com"]

Explanation:
BrowserHistory browserHistory = new BrowserHistory("leetcode.com");
browserHistory.visit("google.com");       // You are in "leetcode.com". Visit "google.com"
browserHistory.visit("facebook.com");     // You are in "google.com". Visit "facebook.com"
browserHistory.visit("youtube.com");      // You are in "facebook.com". Visit "youtube.com"
browserHistory.back(1);                   // You are in "youtube.com", move back to "facebook.com" return "facebook.com"
browserHistory.back(1);                   // You are in "facebook.com", move back to "google.com" return "google.com"
browserHistory.forward(1);                // You are in "google.com", move forward to "facebook.com" return "facebook.com"
browserHistory.visit("linkedin.com");     // You are in "facebook.com". Visit "linkedin.com"
browserHistory.forward(2);                // You are in "linkedin.com", you cannot move forward any steps.
browserHistory.back(2);                   // You are in "linkedin.com", move back two steps to "facebook.com" then to "google.com". return "google.com"
browserHistory.back(7);                   // You are in "google.com", you can move back only one step to "leetcode.com". return "leetcode.com"

Constraints:

• 1 <= homepage.length <= 20
• 1 <= url.length <= 20
• 1 <= steps <= 100
• homepage and url consist of '.' or lower case English letters.
• At most 5000 calls will be made to visit, back, and forward.

Solution

class BrowserHistory {
  string homepage;
  vector<string> history;
  int index = 1;
public:
  BrowserHistory(string _homepage): homepage(_homepage) {
    history.push_back(homepage);
  }
  
  void visit(string url) {
    while(history.size() > index) history.pop_back();
    history.push_back(url);
    index += 1;
  }
  
  string back(int steps) {
    if(index > steps) index = index - steps;
    else index = 1;
    return history[index - 1];
  }
  
  string forward(int steps) {
    if(index + steps > history.size()) index = history.size();
    else index += steps;
    return history[index - 1];
  }
};

// Accepted
// 71/71 cases passed (112 ms)
// Your runtime beats 94.89 % of cpp submissions
// Your memory usage beats 61.84 % of cpp submissions (57.6 MB)

July LeetCoding Challenge 22

Description

Partition Array into Disjoint Intervals

Given an array nums, partition it into two (contiguous) subarrays left and right so that:

• Every element in left is less than or equal to every element in right.
• left and right are non-empty.
• left has the smallest possible size.

Return the length of left after such a partitioning. It is guaranteed that such a partitioning exists.

Example 1:

Input: nums = [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: nums = [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

Note:

1. 2 <= nums.length <= 30000
2. 0 <= nums[i] <= 106
3. It is guaranteed there is at least one way to partition nums as described.

Solution

int mmax[30000];
int mmin[30000];
class Solution {
public:
  int partitionDisjoint(vector<int>& nums) {
    int len = nums.size();
    mmax[0] = nums[0];
    mmin[len - 1] = nums[len - 1];
    for(int i = 1; i < len; ++i) {
      mmax[i] = max(mmax[i - 1], nums[i]);
      mmin[len - 1 - i] = min(mmin[len - i], nums[len - 1 - i]);
    }
    for(int i = 1; i < len; ++i) {
      if(mmax[i - 1] <= mmin[i]) return i;
    }
    return -1;
  }
};