花花酱 LeetCode 2006. Count Number of Pairs With Absolute Difference K

By zxi on November 25, 2021

Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.

The value of |x| is defined as:

x if x >= 0.
-x if x < 0.

Example 1:

Input: nums = [1,2,2,1], k = 1
Output: 4
Explanation: The pairs with an absolute difference of 1 are:
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]

Example 2:

Input: nums = [1,3], k = 3
Output: 0
Explanation: There are no pairs with an absolute difference of 3.

Example 3:

Input: nums = [3,2,1,5,4], k = 2
Output: 3
Explanation: The pairs with an absolute difference of 2 are:
- [3,2,1,5,4]
- [3,2,1,5,4]
- [3,2,1,5,4]

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= 100
1 <= k <= 99

Solution: Hashtable
|y – x| = k
y = x + k or y = x – k

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

C++

// Author: Huahua

class Solution {

public:

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

unordered_map<int, int> m;

int ans = 0;

for (int x : nums) {

ans += m[x - k];

ans += m[x + k];

++m[x];

}

return ans;

}

};

花花酱 LeetCode 1711. Count Good Meals

By zxi on January 2, 2021

A good meal is a meal that contains exactly two different food items with a sum of deliciousness equal to a power of two.

You can pick any two different foods to make a good meal.

Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the i^th item of food, return the number of different good meals you can make from this list modulo 10⁹ + 7.

Note that items with different indices are considered different even if they have the same deliciousness value.

Example 1:

Input: deliciousness = [1,3,5,7,9]
Output: 4
Explanation: The good meals are (1,3), (1,7), (3,5) and, (7,9).
Their respective sums are 4, 8, 8, and 16, all of which are powers of 2.

Example 2:

Input: deliciousness = [1,1,1,3,3,3,7]
Output: 15
Explanation: The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.

Constraints:

1 <= deliciousness.length <= 10⁵
0 <= deliciousness[i] <= 2²⁰

Solution: Hashtable

Same idea as LeetCode 1: Two Sum

Use a hashtable to store the occurrences of all the numbers added so far. For a new number x, check all possible 2^i – x. ans += freq[2^i – x] 0 <= i <= 21

Time complexity: O(22n)
Space complexity: O(n)

C++

class Solution {

public:

int countPairs(vector<int>& A) {

constexpr int kMod = 1e9 + 7;

unordered_map<int, int> m;

long ans = 0;

for (int x : A) {

for (int t = 1; t <= 1 << 21; t *= 2) {

auto it = m.find(t - x);

if (it != end(m)) ans += it->second;

}

++m[x];

}

return ans % kMod;

}

};

Python3

class Solution:

def countPairs(self, deliciousness: List[int]) -> int:

sums = [1<<i for i in range(22)]

m = defaultdict(int)

ans = 0

for x in deliciousness:

for t in sums:

if t - x in m: ans += m[t - x]

m[x] += 1

return ans % (10**9 + 7)

花花酱 LeetCode 447. Number of Boomerangs

By zxi on March 24, 2018

Problem

Given n points in the plane that are all pairwise distinct, a “boomerang” is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters).

Find the number of boomerangs. You may assume that n will be at most 500 and coordinates of points are all in the range [-10000, 10000] (inclusive).

Example:

Input:
[[0,0],[1,0],[2,0]]

Output:
2

Explanation:
The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

Solution: HashTable

For each point, compute the distance to the rest of the points and count.

if there are k points that have the same distance to current point, then there are P(k,2) = k*k-1 Boomerangs.

for example, p1, p2, p3 have the same distance to p0, then there are P(3,2) = 3 * (3-1) = 6 Boomerangs

(p1, p0, p2), (p1, p0, p3)

(p2, p0, p1), (p2, p0, p3)

(p3, p0, p1), (p3, p0, p2)

C++

// Author: Huahua

// Running time: 210 ms (beats 83.33%)

class Solution {

public:

int numberOfBoomerangs(vector<pair<int, int>>& points) {

int ans = 0;

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

unordered_map<int, int> dist(points.size());

for (int j = 0; j < points.size(); ++j) {

const int dx = points[i].first - points[j].first;

const int dy = points[i].second - points[j].second;

++dist[dx * dx + dy * dy];

}

for (const auto& kv : dist)

ans += kv.second * (kv.second - 1);

}

return ans;

}

};

Solution 2: Sorting

dist = [1, 2, 1, 2, 1, 5]

sorted_dist = [1, 1, 1, 2, 2, 5], 1*3, 2*2, 5*1

ans = 3*(3-1) + 2 * (2 – 1) * 1 * (1 – 1) = 8

Time complexity: O(n*nlogn)

Space complexity: O(n)

// Author: Huahua

// Running time: 150 ms (beats 98.52%)

class Solution {

public:

int numberOfBoomerangs(vector<pair<int, int>>& points) {

const int n = points.size();

int ans = 0;

vector<int> dist(points.size());

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

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

const int dx = points[i].first - points[j].first;

const int dy = points[i].second - points[j].second;

dist[j] = dx * dx + dy * dy;

}

std::sort(dist.begin(), dist.end());

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

int k = 1;

while (j < n && dist[j] == dist[j - 1]) { ++j; ++k; }

ans += k * (k - 1);

}

return ans;

}

};

Posts tagged as “pairs”

花花酱 LeetCode 2006. Count Number of Pairs With Absolute Difference K

Solution: Hashtable|y – x| = ky = x + k or y = x – k

C++

花花酱 LeetCode 1711. Count Good Meals

Solution: Hashtable

C++

Python3

花花酱 LeetCode 447. Number of Boomerangs

Problem

Solution: HashTable

Solution 2: Sorting

Solution: Hashtable
|y – x| = k
y = x + k or y = x – k