leetcode Day20----array.easy

Non-decreasing Array

Given an array with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.

We define an array is non-decreasing if array[i] <= array[i + 1] holds for every i (1 <= i < n).

Example 1:
Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
Example 2:
Input: [4,2,1]
Output: False
Explanation: You can’t get a non-decreasing array by modify at most one element.
Note: The n belongs to [1, 10,000].

C语言

bool checkNonDes(int* nums, int numsSize, int* i) {
    for (int j = 0; j+1 < numsSize; j++) {
        if (nums[j] > nums[j+1]) {
            *i = j;
            return false;
        }
    }
    return true;
}


bool checkPossibility(int* nums, int numsSize){
    int i;
    if (numsSize < 3)
        return true;
    if (checkNonDes(nums,numsSize,&i) == false){
        if (i < numsSize-2) {
            if (nums[i] < nums[i+2]) {
                nums[i+1] = nums[i];
                if (checkNonDes(nums,numsSize,&i) == false){
                    return false;
                }
            } else {
                nums[i] = nums[i+1];
                if (checkNonDes(nums,numsSize,&i) == false){
                    return false;
                }
            }
        }
    }
    return true;
}

Success
Details
Runtime: 28 ms, faster than 12.56% of C online submissions for Non-decreasing Array.
Memory Usage: 8.3 MB, less than 100.00% of C online submissions for Non-decreasing Array.

python3

class Solution:
    def checkPossibility(self, nums: List[int]) -> bool:
        count_dec = 0
        for i in range(len(nums) - 1):
            if nums[i] > nums[i + 1]:
                count_dec += 1
                if i == 0:
                    nums[i] = nums[i + 1]
                elif nums[i - 1] <= nums[i + 1]:
                    nums[i] = nums[i - 1]
                else:
                    nums[i + 1] = nums[i]
            if count_dec > 1:
                return False
        return True

Success
Details
Runtime: 48 ms, faster than 99.86% of Python3 online submissions for Non-decreasing Array.
Memory Usage: 14.2 MB, less than 5.95% of Python3 online submissions for Non-decreasing Array.

Longest Continuous Increasing Subsequence

Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).

Example 1:
Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3.
Even though [1,3,5,7] is also an increasing subsequence, it’s not a continuous one where 5 and 7 are separated by 4.
Example 2:
Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1.
Note: Length of the array will not exceed 10,000.

C语言

int findLengthOfLCIS(int* nums, int numsSize){
    int i=0,count=0,res=0;
    if(numsSize<1)return 0;
    while(i<numsSize-1){if(nums[i]<nums[i+1]){count++;if(count>res)res=count;}else count=0;i++;}
    return res+1;
}

Success
Details
Runtime: 8 ms, faster than 42.35% of C online submissions for Longest Continuous Increasing Subsequence.
Memory Usage: 7.5 MB, less than 100.00% of C online submissions for Longest Continuous Increasing Subsequence.

python3

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        if not nums:
            return 0
        res = 1
        c = 1
        for i in range(1, len(nums)):
            c = c + 1 if nums[i] > nums[i-1] else 1
            res = max(res, c)
        return res

Success
Details
Runtime: 48 ms, faster than 48.89% of Python3 online submissions for Longest Continuous Increasing Subsequence.
Memory Usage: 14.1 MB, less than 6.67% of Python3 online submissions for Longest Continuous Increasing Subsequence.

Degree of an Array

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Example 1:
Input: [1, 2, 2, 3, 1]
Output: 2
Explanation:
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.
Example 2:
Input: [1,2,2,3,1,4,2]
Output: 6
Note:

nums.length will be between 1 and 50,000.
nums[i] will be an integer between 0 and 49,999.

C语言

int findShortestSubArray(int* nums, int numsSize){
    int count[50000]={0};
    int first[50000]={0};
    int res=0,degree=0;
    for(int i=0;i<numsSize;i++){
        if(first[nums[i]]==0)
            first[nums[i]]=i+1;
        count[nums[i]]++;
        if(count[nums[i]]>degree){
            degree=count[nums[i]];
            res=i-first[nums[i]]+2;
        }
        else if(count[nums[i]]==degree){
            if((i-first[nums[i]]+2)<res)
                res=i-first[nums[i]]+2;
        }
    }
    return res;
}

Success
Details
Runtime: 28 ms, faster than 40.54% of C online submissions for Degree of an Array.
Memory Usage: 9.3 MB, less than 100.00% of C online submissions for Degree of an Array.

python3

class Solution:
    def findShortestSubArray(self, nums: List[int]) -> int:
        dic = {}
        for i in nums:
            if i in dic:
                dic[i] += 1
            else:
                dic[i] = 1
        degree = max(dic.values())
        if degree == 1:
            return 1
        max_vec = []
        for i in dic:
            if dic[i] == degree:
                max_vec.append(i)
                
        shortest_len = float('inf')
        for i in range(len(max_vec)):
            temp = len(nums) - nums.index(max_vec[i]) - nums[::-1].index(max_vec[i])
            shortest_len = min(shortest_len,temp)
        return shortest_len

Success
Details
Runtime: 60 ms, faster than 96.88% of Python3 online submissions for Degree of an Array.
Memory Usage: 14.5 MB, less than 14.06% of Python3 online submissions for Degree of an Array.