Today is Sunday, I gonna review the tasks I've done this week, and finish today's leetcode's November LeetCoding Challenge with cpp.

LeetCode Review

3Sum Closest

loop with binary search

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int diff = INT_MAX;
        int answer = -1;
        int len = nums.size();
        sort(nums.begin(), nums.end());
        for(int i = 0; i < len-2; ++i) {
            for(int j = i+1; j < len-1; ++j) {
                if(j != i+1 && nums[j] == nums[j-1]) continue;
                int tar = target-nums[i]-nums[j], start = j+1, end = len;
                while(start < end) {
                    int mid = (start + end) / 2;
                    if(nums[mid] > tar) {
                        if(nums[mid] - tar < diff) {
                            diff = nums[mid]-tar;
                            answer = nums[mid] + nums[i] + nums[j];
                        }
                        end = mid;
                    } else {
                        if(tar - nums[mid] < diff) {
                            diff = tar - nums[mid];
                            answer = nums[mid] + nums[i] + nums[j];
                        }
                        start = mid + 1;
                    }
                }
            }
        }
        return answer;
    }
};

December LeetCoding Challenge 27

Description

Jump Game IV

Given an array of integers arr, you are initially positioned at the first index of the array.

In one step you can jump from index i to index:

• i + 1 where: i + 1 < arr.length.
• i - 1 where: i - 1 >= 0.
• j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:

Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You don't need to jump.

Example 3:

Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Example 4:

Input: arr = [6,1,9]
Output: 2

Example 5:

Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3

Constraints:

• 1 <= arr.length <= 5 * 10^4
• -10^8 <= arr[i] <= 10^8

Solution

simple BFS, but need some optimization

class Solution {
public:
    int minJumps(vector<int>& arr) {
        int len = arr.size();
        unordered_map<int, vector<int>> positions;
        for(int i = 0; i < len; ++i) {
            int n = arr[i];
            if(positions.count(n)) {
                positions[n].push_back(i);
            } else {
                positions.insert(make_pair(n, vector<int>{i}));
            }
        }
        queue<int> q;
        q.push(0);
        vector<int> step(len);
        // so don't need a `visited` vector
        step[0] = 1;
        while(q.size()) {
            auto pos = q.front();
            q.pop();
            if(pos == len-1) return step[pos]-1;
            if(pos && !step[pos-1]) {
                q.push(pos-1);
                step[pos-1] = step[pos] + 1;
            }
            if(!step[pos+1]) {
                q.push(pos+1);
                step[pos+1] = step[pos] + 1;
            }
            for(auto anotherPos : positions[arr[pos]]) {
                if(anotherPos == pos || step[anotherPos]) continue;
                q.push(anotherPos);
                step[anotherPos] = step[pos] + 1;
            }
            // optimization here
            positions[arr[pos]].clear();
        }
        return -1;
    }
};