Posts tagged as “easy”

花花酱 LeetCode 2828. Check if a String Is an Acronym of Words

By zxi on August 26, 2023

Given an array of strings words and a string s, determine if s is an acronym of words.

The string s is considered an acronym of words if it can be formed by concatenating the first character of each string in words in order. For example, "ab" can be formed from ["apple", "banana"], but it can’t be formed from ["bear", "aardvark"].

Return true if s is an acronym of words, and false otherwise.

Example 1:

Input: words = ["alice","bob","charlie"], s = "abc"
Output: true
Explanation: The first character in the words "alice", "bob", and "charlie" are 'a', 'b', and 'c', respectively. Hence, s = "abc" is the acronym.

Example 2:

Input: words = ["an","apple"], s = "a"
Output: false
Explanation: The first character in the words "an" and "apple" are 'a' and 'a', respectively. 
The acronym formed by concatenating these characters is "aa". 
Hence, s = "a" is not the acronym.

Example 3:

Input: words = ["never","gonna","give","up","on","you"], s = "ngguoy"
Output: true
Explanation: By concatenating the first character of the words in the array, we get the string "ngguoy". 
Hence, s = "ngguoy" is the acronym.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 10
1 <= s.length <= 100
words[i] and s consist of lowercase English letters.

Solution: Check the first letter of each word

No need to concatenate, just check the first letter of each word.

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

C++

// Author: Huahua

class Solution {

public:

bool isAcronym(vector<string>& words, string_view s) {

if (words.size() != s.size()) return false;

for (int i = 0; i < s.size(); ++i)

if (words[i][0] != s[i]) return false;

return true;

}

};

花花酱 LeetCode 2824. Count Pairs Whose Sum is Less than Target

By zxi on August 23, 2023

Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs(i, j)where0 <= i < j < nandnums[i] + nums[j] < target.

Example 1:

Input: nums = [-1,1,2,3,1], target = 2
Output: 3
Explanation: There are 3 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target
- (0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target 
- (0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target
Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.

Example 2:

Input: nums = [-6,2,5,-2,-7,-1,3], target = -2
Output: 10
Explanation: There are 10 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = -4 < target
- (0, 3) since 0 < 3 and nums[0] + nums[3] = -8 < target
- (0, 4) since 0 < 4 and nums[0] + nums[4] = -13 < target
- (0, 5) since 0 < 5 and nums[0] + nums[5] = -7 < target
- (0, 6) since 0 < 6 and nums[0] + nums[6] = -3 < target
- (1, 4) since 1 < 4 and nums[1] + nums[4] = -5 < target
- (3, 4) since 3 < 4 and nums[3] + nums[4] = -9 < target
- (3, 5) since 3 < 5 and nums[3] + nums[5] = -3 < target
- (4, 5) since 4 < 5 and nums[4] + nums[5] = -8 < target
- (4, 6) since 4 < 6 and nums[4] + nums[6] = -4 < target

Constraints:

1 <= nums.length == n <= 50
-50 <= nums[i], target <= 50

Solution 1: Brute force

Enumerate all pairs.

Time complexity: O(n²)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int countPairs(vector<int>& nums, int target) {

const int n = nums.size();

int ans = 0;

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

for (int j = i + 1; j < n; ++j)

ans += nums[i] + nums[j] < target;

return ans;

}

};

Solution 2: Two Pointers

Sort the numbers.

Use two pointers i, and j.
Set i to 0 and j to n – 1.
while (nums[i] + nums[j] >= target) –j
then we have nums[i] + nums[k] < target (i < k <= j), in total (j – i) pairs.
++i, move to the next starting number.
Time complexity: O(nlogn + n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int countPairs(vector<int>& nums, int target) {

const int n = nums.size();

sort(begin(nums), end(nums));

int ans = 0;

for (int i = 0, j = n - 1; i < n; ++i) {

while (j >= i && nums[i] + nums[j] >= target) --j;

ans += max(0, j - i);

}

return ans;

}

};

花花酱 LeetCode 2815. Max Pair Sum in an Array

By zxi on August 13, 2023

You are given a 0-indexed integer array nums. You have to find the maximum sum of a pair of numbers from nums such that the maximum digit in both numbers are equal.

Return the maximum sum or -1 if no such pair exists.

Example 1:

Input: nums = [51,71,17,24,42]
Output: 88
Explanation: 
For i = 1 and j = 2, nums[i] and nums[j] have equal maximum digits with a pair sum of 71 + 17 = 88. 
For i = 3 and j = 4, nums[i] and nums[j] have equal maximum digits with a pair sum of 24 + 42 = 66.
It can be shown that there are no other pairs with equal maximum digits, so the answer is 88.

Example 2:

Input: nums = [1,2,3,4]
Output: -1
Explanation: No pair exists in nums with equal maximum digits.

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 10⁴

Solution: Brute Force

Enumerate all pairs of nums and check their sum and max digit.

Time complexity: O(n²)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int maxSum(vector<int>& nums) {

auto maxDigit = [](int x) {

int ans = 0;

while (x) {

ans = max(ans, x % 10);

x /= 10;

}

return ans;

};

int ans = -1;

const int n = nums.size();

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

for (int j = i + 1; j < n; ++j)

if (nums[i] + nums[j] > ans

&& maxDigit(nums[i]) == maxDigit(nums[j]))

ans = nums[i] + nums[j];

return ans;

}

};

花花酱 LeetCode 2769. Find the Maximum Achievable Number

By zxi on July 11, 2023

You are given two integers, num and t.

An integer x is called achievable if it can become equal to num after applying the following operation no more than t times:

Increase or decrease x by 1, and simultaneously increase or decrease num by 1.

Return the maximum possible achievable number. It can be proven that there exists at least one achievable number.

Example 1:

Input: num = 4, t = 1
Output: 6
Explanation: The maximum achievable number is x = 6; it can become equal to num after performing this operation:
1- Decrease x by 1, and increase num by 1. Now, x = 5 and num = 5. 
It can be proven that there is no achievable number larger than 6.

Example 2:

Input: num = 3, t = 2
Output: 7
Explanation: The maximum achievable number is x = 7; after performing these operations, x will equal num: 
1- Decrease x by 1, and increase num by 1. Now, x = 6 and num = 4.
2- Decrease x by 1, and increase num by 1. Now, x = 5 and num = 5.
It can be proven that there is no achievable number larger than 7.

Constraints:

1 <= num, t <= 50

Solution: Greedy

Always decrease x and always increase num, the max achievable number x = num + t * 2

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

C++

// Author: Huahua

class Solution {

public:

int theMaximumAchievableX(int num, int t) {

return num + t * 2;

}

};

花花酱 LeetCode 2667. Create Hello World Function

By zxi on May 7, 2023

Write a function createHelloWorld. It should return a new function that always returns "Hello World".

Example 1:

Input: args = []
Output: "Hello World"
Explanation:
const f = createHelloWorld();
f(); // "Hello World"
The function returned by createHelloWorld should always return "Hello World".

Example 2:

Input: args = [{},null,42]
Output: "Hello World"
Explanation:
const f = createHelloWorld();
f({}, null, 42); // "Hello World"
Any arguments could be passed to the function but it should still always return "Hello World".

Constraints:

0 <= args.length <= 10

Solution:

JavaScript

// Author: Huahua

/**

* @return {Function}

var createHelloWorld = function() {

return function(...args) {

return "Hello World";

}

};