Huahua’s Tech Road

花花酱 LeetCode 1995. Count Special Quadruplets

By zxi on December 31, 2021

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

nums[a] + nums[b] + nums[c] == nums[d], and
a < b < c < d

Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

Constraints:

4 <= nums.length <= 50
1 <= nums[i] <= 100

Solution 1: Brute force (224ms)

Enumerate a, b, c, d.

Time complexity: O(C(n, 4)) = O(n⁴/24)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

const int n = nums.size();

int ans = 0;

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

for (int b = a + 1; b < n; ++b)

for (int c = b + 1; c < n; ++c)

for (int d = c + 1; d < n; ++d)

ans += nums[a] + nums[b] + nums[c] == nums[d];

return ans;

}

};

Solution 2: Static frequency table + binary search (39ms)

For each element, we store its indices (sorted).

Given a, b, c, target t = nums[a] + nums[b] + nums[c], we check the hashtable and use binary search to find how many times it occurred after index c.

Time complexity: O(n³/6*logn)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

constexpr int kMax = 100;

const int n = nums.size();

int ans = 0;

vector<vector<int>> m(kMax + 1);

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

m[nums[i]].push_back(i);

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

for (int b = a + 1; b < n; ++b)

for (int c = b + 1; c < n; ++c) {

const int t = nums[a] + nums[b] + nums[c];

if (t > kMax) continue;

ans += end(m[t]) - lower_bound(begin(m[t]), end(m[t]), c + 1);

}

return ans;

}

};

Solution 3: Dynamic frequency table (29ms)

Similar to 花花酱 LeetCode 1. Two Sum, we dynamically add elements (from right to left) into the hashtable.

Time complexity: O(n³/6)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

constexpr int kMax = 100;

const int n = nums.size();

int ans = 0;

vector<int> m(kMax + 1);

// for every c we had seen, we can use it as target (nums[d]).

for (int c = n - 1; c >= 0; ++m[nums[c--]])

for (int a = 0; a < c; ++a)

for (int b = a + 1; b < c; ++b) {

const int t = nums[a] + nums[b] + nums[c];

if (t > kMax) continue;

ans += m[t];

}

return ans;

}

};

花花酱 LeetCode 1991. Find the Middle Index in Array

By zxi on December 31, 2021

Given a 0-indexed integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones).

A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1].

If middleIndex == 0, the left side sum is considered to be 0. Similarly, if middleIndex == nums.length - 1, the right side sum is considered to be 0.

Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index.

Example 1:

Input: nums = [2,3,-1,8,4]
Output: 3
Explanation: The sum of the numbers before index 3 is: 2 + 3 + -1 = 4
The sum of the numbers after index 3 is: 4 = 4

Example 2:

Input: nums = [1,-1,4]
Output: 2
Explanation: The sum of the numbers before index 2 is: 1 + -1 = 0
The sum of the numbers after index 2 is: 0

Example 3:

Input: nums = [2,5]
Output: -1
Explanation: There is no valid middleIndex.

Constraints:

1 <= nums.length <= 100
-1000 <= nums[i] <= 1000

Solution: Pre-compute + prefix sum

Pre-compute the sum of entire array. We scan the array and accumulate prefix sum and we can compute the sum of the rest of array.

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

C++

// Author: Huahua

class Solution {

public:

int findMiddleIndex(vector<int>& nums) {

int sum = accumulate(begin(nums), end(nums), 0);

for (int i = 0, cur = 0; i < nums.size(); ++i) {

if (cur == sum - cur - nums[i]) return i;

cur += nums[i];

}

return -1;

}

};

花花酱 LeetCode 1985. Find the Kth Largest Integer in the Array

By zxi on December 31, 2021

You are given an array of strings nums and an integer k. Each string in nums represents an integer without leading zeros.

Return the string that represents the k^th largest integer in nums.

Note: Duplicate numbers should be counted distinctly. For example, if nums is ["1","2","2"], "2" is the first largest integer, "2" is the second-largest integer, and "1" is the third-largest integer.

Example 1:

Input: nums = ["3","6","7","10"], k = 4
Output: "3"
Explanation:
The numbers in nums sorted in non-decreasing order are ["3","6","7","10"].
The 4^th largest integer in nums is "3".

Example 2:

Input: nums = ["2","21","12","1"], k = 3
Output: "2"
Explanation:
The numbers in nums sorted in non-decreasing order are ["1","2","12","21"].
The 3^rd largest integer in nums is "2".

Example 3:

Input: nums = ["0","0"], k = 2
Output: "0"
Explanation:
The numbers in nums sorted in non-decreasing order are ["0","0"].
The 2^nd largest integer in nums is "0".

Constraints:

1 <= k <= nums.length <= 10⁴
1 <= nums[i].length <= 100
nums[i] consists of only digits.
nums[i] will not have any leading zeros.

Solution: nth_element / quick selection

Use std::nth_element to find the k-th largest element. When comparing two strings, compare their lengths first and compare their content if they have the same length.

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

C++

// Author: Huahua

class Solution {

public:

string kthLargestNumber(vector<string>& nums, int k) {

nth_element(begin(nums), begin(nums) + k - 1, end(nums), [](const auto& a, const auto& b) {

return a.length() == b.length() ? a > b : a.length() > b.length();

});

return nums[k - 1];

}

};

花花酱 LeetCode 1984. Minimum Difference Between Highest and Lowest of K Scores

By zxi on December 31, 2021

You are given a 0-indexed integer array nums, where nums[i] represents the score of the i^th student. You are also given an integer k.

Pick the scores of any k students from the array so that the difference between the highest and the lowest of the k scores is minimized.

Return the minimum possible difference.

Example 1:

Input: nums = [90], k = 1
Output: 0
Explanation: There is one way to pick score(s) of one student:
- [90]. The difference between the highest and lowest score is 90 - 90 = 0.
The minimum possible difference is 0.

Example 2:

Input: nums = [9,4,1,7], k = 2
Output: 2
Explanation: There are six ways to pick score(s) of two students:
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 4 = 5.
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 1 = 8.
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 7 = 2.
- [9,4,1,7]. The difference between the highest and lowest score is 4 - 1 = 3.
- [9,4,1,7]. The difference between the highest and lowest score is 7 - 4 = 3.
- [9,4,1,7]. The difference between the highest and lowest score is 7 - 1 = 6.
The minimum possible difference is 2.

Constraints:

1 <= k <= nums.length <= 1000
0 <= nums[i] <= 10⁵

Solution: Sliding Window

Sort the array, to minimize the difference, k numbers must be consecutive (i.e, from a subarray). We use a sliding window size of k and try all possible subarrays.
Ans = min{(nums[k – 1] – nums[0]), (nums[k] – nums[1]), … (nums[n – 1] – nums[n – k])}

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

C++

// Author: Huahua

class Solution {

public:

int minimumDifference(vector<int>& nums, int k) {

int ans = INT_MAX;

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

for (int i = k - 1; i < nums.size(); ++i)

ans = min(ans, nums[i] - nums[i - k + 1]);

return ans;

}

};

花花酱 LeetCode 1980. Find Unique Binary String

By zxi on December 31, 2021

Given an array of strings nums containing n unique binary strings each of length n, return a binary string of length n that does not appear in nums. If there are multiple answers, you may return any of them.

Example 1:

Input: nums = ["01","10"]
Output: "11"
Explanation: "11" does not appear in nums. "00" would also be correct.

Example 2:

Input: nums = ["00","01"]
Output: "11"
Explanation: "11" does not appear in nums. "10" would also be correct.

Example 3:

Input: nums = ["111","011","001"]
Output: "101"
Explanation: "101" does not appear in nums. "000", "010", "100", and "110" would also be correct.

Constraints:

n == nums.length
1 <= n <= 16
nums[i].length == n
nums[i] is either '0' or '1'.
All the strings of nums are unique.

Solution 1: Hashtable

We can use bitset to convert between integer and binary string.

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

C++

// Author: Huahua

class Solution {

public:

string findDifferentBinaryString(vector<string>& nums) {

const int n = nums.size();

unordered_set<int> seen(1 << n);

for (const string& num : nums)

seen.insert(bitset<16>(num).to_ulong());

for (int i = 0; i < 1 << n; ++i)

if (!seen.count(i)) return bitset<16>(i).to_string().substr(16 - n);

return "";

}

};

Solution 2: One bit a time

Let ans[i] = ‘1’ – nums[i][i], s.t. ans is at least one bit different from any strings.

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

C++

// Author: Huahua

class Solution {

public:

string findDifferentBinaryString(vector<string>& nums) {

const int n = nums.size();

string ans(n, '0');

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

ans[i] = '1' - nums[i][i] + '0';

return ans;

}

};

Huahua's Tech Road

花花酱 LeetCode 1995. Count Special Quadruplets

Solution 1: Brute force (224ms)

C++

Solution 2: Static frequency table + binary search (39ms)

C++

Solution 3: Dynamic frequency table (29ms)

C++

花花酱 LeetCode 1991. Find the Middle Index in Array

Solution: Pre-compute + prefix sum

C++

花花酱 LeetCode 1985. Find the Kth Largest Integer in the Array

Solution: nth_element / quick selection

C++

花花酱 LeetCode 1984. Minimum Difference Between Highest and Lowest of K Scores

Solution: Sliding Window

C++

花花酱 LeetCode 1980. Find Unique Binary String

Solution 1: Hashtable

C++

Solution 2: One bit a time

C++