아래 링크에서 크롬 확장 설치

https://github.com/williambelle/github-contribution-color-graph

1.  우측 상단 모서리에 프로필 사진 클릭한 다음 내 프로필 클릭

2. Contribution settings → Private contributions & Activity overview check

https://docs.github.com/en/enterprise-server@3.1/account-and-profile/setting-up-and-managing-your-github-profile/managing-contribution-graphs-on-your-profile/showing-an-overview-of-your-activity-on-your-profile

1일 1커밋 습관화합시다 !!

https://github.com/jasonlong/isometric-contributions

Consecutive Numbers

 문제 설명

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| num         | varchar |
+-------------+---------+
id is the primary key for this table.

 입출력 예

Example 1:

Input: 
Logs table:
+----+-----+
| id | num |
+----+-----+
| 1  | 1   |
| 2  | 1   |
| 3  | 1   |
| 4  | 2   |
| 5  | 1   |
| 6  | 2   |
| 7  | 2   |
+----+-----+
Output: 
+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+
Explanation: 1 is the only number that appears consecutively for at least three times.

 Oracle Query

SELECT DISTINCT 
	l1.Num As ConsecutiveNums
FROM Logs l1
JOIN Logs l2 ON l1.Id = l2.Id-1
JOIN Logs l3 ON l2.Id = l3.Id-1
WHERE l1.Num = l2.Num
	AND l2.Num = l3.Num;

* 참고 링크 : https://leetcode.com/problems/consecutive-numbers/discuss/185886/Self-Join-Twice-(681-ms-faster-than-100.00-of-Oracle-online-submissions-for-Consecutive-Numbers.)

 출처

https://leetcode.com/problems/consecutive-numbers/

Valid Triangle Number

 문제 설명

Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.

 제한 사항

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 1000

 입출력 예

Example 1:

Input: nums = [2,2,3,4]
Output: 3
Explanation: Valid combinations are: 
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3

Example 2:

Input: nums = [4,2,3,4]
Output: 4

 Python 코드

Python code 

class Solution(object):
    def triangleNumber(self, nums):

        nums, count, n = sorted(nums, reverse=1), 0, len(nums)
        for i in range(n):
            j, k = i + 1, n - 1
            while j < k:
                if nums[j] + nums[k] > nums[i]:
                    count += k - j
                    j += 1
                else:
                    k -= 1
        return count

* 참고 링크 : https://leetcode.com/problems/valid-triangle-number/discuss/104177/O(N2)-solution-for-C%2B%2B-and-Python

 C++ 코드

C ++ code

class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        if(nums.size() < 3) return 0;
        sort(nums.begin(), nums.end(), greater<int>());
        int res = 0;
        int n = nums.size();
        for(int a = 0; a < n-2; ++a){
            int b = a+1;
            int c = n-1;
            while(b < c){
                if(nums[b] + nums[c] > nums[a]){
                    res = res + c - b;
                    ++b;
                }
                else
                    --c;
            }
        }
        return res;
    }
};

* 참고 링크 : https://www.programmerall.com/article/1415760413/

 출처

https://leetcode.com/problems/valid-triangle-number/

Find Peak Element

 문제 설명

A peak element is an element that is strictly greater than its neighbors.

Given an integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

You may imagine that nums[-1] = nums[n] = -∞.

You must write an algorithm that runs in O(log n) time.

 제한 사항

• 1 <= nums.length <= 1000
• -231 <= nums[i] <= 231 - 1
• nums[i] != nums[i + 1] for all valid i.

 입출력 예

Example 1:

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 5
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

 Python 코드

Python code 

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left = 0
        right = len(nums)-1
				# left가 커지고 right가 작아지는 경우 left와 right가 같아질 수 있다.
        # 그 때 원한느 값을 찾은 경우이므로 left와 right가 같아지면 반복문 종료
        while left < right:
            mid = (left + right) // 2      # 가운데 지점
            if nums[mid] < nums[mid+1]:    # 현재 값이 오른쪽 값보다 작다면
                left = mid+1
            else:                          # 현재 값이 오른쪽 값보다 크다면
                right = mid
        return left                        # 반복문이 종료되면 left가 정답

* 참고 링크 : https://velog.io/@hrpp1300/LeetCode-162.-Find-Peak-Element

 C++ 코드

C ++ code

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int left = 0;
        int right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > nums[mid + 1]) {
                right = mid;
            }
            else {
                left = mid + 1;
            }
        }
        return left; 
    }
};

* 참고 링크 : https://myeongcs.tistory.com/80

 출처

https://leetcode.com/problems/find-peak-element/submissions/