Posts tagged as “easy”

花花酱 LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

By zxi on March 2, 2020

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation: 
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1). 
For nums[3]=2 there exist one smaller number than it (1). 
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]

Constraints:

2 <= nums.length <= 500
0 <= nums[i] <= 100

Solution 1: Brute Force

Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

vector<int> smallerNumbersThanCurrent(vector<int>& nums) {

const int n = nums.size();

vector<int> ans(n);

for (int i = 0; i < n; ++i)

for (int j = 0; j < n; ++j)

if (i != j && nums[j] < nums[i]) ++ans[i];

return ans;

}

};

Solution 2: Sort + Binary Search

Time complexity: O(nlogn)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

vector<int> smallerNumbersThanCurrent(vector<int>& nums) {

const int n = nums.size();

vector<int> s(nums);

sort(begin(s), end(s));

vector<int> ans(n);

for (int i = 0; i < n; ++i)

ans[i] = distance(begin(s), lower_bound(begin(s), end(s), nums[i]));

return ans;

}

};

Solution 3: Cumulative frequency

Time complexity: O(n)
Space complexity: O(101)

C++

// Author: Huahua

class Solution {

public:

vector<int> smallerNumbersThanCurrent(vector<int>& nums) {

vector<int> f(101);

for (int x : nums) ++f[x];

partial_sum(begin(f), end(f), begin(f));

for (int& x : nums)

x = x == 0 ? 0 : f[x - 1];

return nums;

}

};

花花酱 LeetCode 1360. Number of Days Between Two Dates

By zxi on February 23, 2020

Write a program to count the number of days between two dates.

The two dates are given as strings, their format is YYYY-MM-DD as shown in the examples.

Example 1:

Input: date1 = "2019-06-29", date2 = "2019-06-30"
Output: 1

Example 2:

Input: date1 = "2020-01-15", date2 = "2019-12-31"
Output: 15

Constraints:

The given dates are valid dates between the years 1971 and 2100.

Solution: Convert to days since epoch

Time complexity: O(1)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int daysBetweenDates(string date1, string date2) {

array<int, 12> m{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

auto isLeap = [](int year) {

return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);

};

auto daysFromEpoch = [&](const string& date) {

int year = stoi(date.substr(0, 4));

int month = stoi(date.substr(5, 2));

int days = stoi(date.substr(8, 2));

for (int i = 1970; i < year; ++i)

days += 365 + isLeap(i);

for (int i = 1; i < month; ++i)

days += m[i - 1];

days += month > 2 && isLeap(year);

return days;

};

return abs(daysFromEpoch(date1) - daysFromEpoch(date2));

}

};

花花酱 LeetCode 1356. Sort Integers by The Number of 1 Bits

By zxi on February 22, 2020

Given an integer array arr. You have to sort the integers in the array in ascending order by the number of 1’s in their binary representation and in case of two or more integers have the same number of 1’s you have to sort them in ascending order.

Return the sorted array.

Example 1:

Input: arr = [0,1,2,3,4,5,6,7,8]
Output: [0,1,2,4,8,3,5,6,7]
Explantion: [0] is the only integer with 0 bits.
[1,2,4,8] all have 1 bit.
[3,5,6] have 2 bits.
[7] has 3 bits.
The sorted array by bits is [0,1,2,4,8,3,5,6,7]

Example 2:

Input: arr = [1024,512,256,128,64,32,16,8,4,2,1]
Output: [1,2,4,8,16,32,64,128,256,512,1024]
Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.

Example 3:

Input: arr = [10000,10000]
Output: [10000,10000]

Example 4:

Input: arr = [2,3,5,7,11,13,17,19]
Output: [2,3,5,17,7,11,13,19]

Example 5:

Input: arr = [10,100,1000,10000]
Output: [10,100,10000,1000]

Constraints:

1 <= arr.length <= 500
0 <= arr[i] <= 10^4

Solution: Sorting

Time complexity: O(nlogn)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

vector<int> sortByBits(vector<int>& arr) {

sort(begin(arr), end(arr), [](const int a, const int b){

int key_a = __builtin_popcount(a);

int key_b = __builtin_popcount(b);

if (key_a != key_b) return key_a < key_b;

return a < b;

});

return arr;

}

};

Python3

# Author: Huahua

class Solution:

def sortByBits(self, arr: List[int]) -> List[int]:

return sorted(arr, key=lambda x: (bin(x).count('1') << 16) + x)

花花酱 LeetCode 1351. Count Negative Numbers in a Sorted Matrix

By zxi on February 16, 2020

Given a m * n matrix grid which is sorted in non-increasing order both row-wise and column-wise.

Return the number of negative numbers in grid.

Example 1:

Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
Output: 8
Explanation: There are 8 negatives number in the matrix.

Example 2:

Input: grid = [[3,2],[1,0]]
Output: 0

Example 3:

Input: grid = [[1,-1],[-1,-1]]
Output: 3

Example 4:

Input: grid = [[-1]]
Output: 1

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 100
-100 <= grid[i][j] <= 100

Solution 1: Brute Force

Time complexity: O(m*n)
Space complexity: O(1)

C++

// Author: Huahua, 16 ms, 10.4 MB

class Solution {

public:

int countNegatives(vector<vector<int>>& grid) {

int ans = 0;

for (const auto& row : grid)

for (const auto x : row)

ans += x < 0;

return ans;

}

};

Solution 2: Find the frontier

Time complexity: O(m+n)
Space complexity: O(1)

C++

// Author: Huahua, 16 ms, 10.4 MB

class Solution {

public:

int countNegatives(vector<vector<int>>& grid) {

int ans = 0;

int m = grid.size();

int n = grid[0].size();

int r = m - 1;

int c = 0;

while (r >= 0 && c < n) {

if (grid[r][c] < 0) {

ans += n - c;

--r;

} else {

++c;

}

return ans;

}

};

花花酱 LeetCode 1342. Number of Steps to Reduce a Number to Zero

By zxi on February 13, 2020

Given a non-negative integer num, return the number of steps to reduce it to zero. If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

Example 1:

Input: num = 14
Output: 6
Explanation: 
Step 1) 14 is even; divide by 2 and obtain 7. 
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3. 
Step 4) 3 is odd; subtract 1 and obtain 2. 
Step 5) 2 is even; divide by 2 and obtain 1. 
Step 6) 1 is odd; subtract 1 and obtain 0.

Example 2:

Input: num = 8
Output: 4
Explanation: 
Step 1) 8 is even; divide by 2 and obtain 4. 
Step 2) 4 is even; divide by 2 and obtain 2. 
Step 3) 2 is even; divide by 2 and obtain 1. 
Step 4) 1 is odd; subtract 1 and obtain 0.

Example 3:

Input: num = 123
Output: 12

Constraints:

0 <= num <= 10^6

Solution: Simulation

Time complexity: O(logn)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int numberOfSteps (int num) {

int steps = 0;

while (num) {

if (num & 1)

--num;

else

num >>= 1;

++steps;

}

return steps;

}

};