README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2700-2799/2748.Number%20of%20Beautiful%20Pairs/README_EN.md	1301	Weekly Contest 351 Q1	Array Hash Table Math Counting Number Theory

2748. Number of Beautiful Pairs

中文文档

Description

You are given a 0-indexed integer array nums. A pair of indices i, j where 0 <= i < j < nums.length is called beautiful if the first digit of nums[i] and the last digit of nums[j] are coprime.

Return the total number of beautiful pairs in nums.

Two integers x and y are coprime if there is no integer greater than 1 that divides both of them. In other words, x and y are coprime if gcd(x, y) == 1, where gcd(x, y) is the greatest common divisor of x and y.

 

Example 1:

Input: nums = [2,5,1,4]
Output: 5
Explanation: There are 5 beautiful pairs in nums:
When i = 0 and j = 1: the first digit of nums[0] is 2, and the last digit of nums[1] is 5. We can confirm that 2 and 5 are coprime, since gcd(2,5) == 1.
When i = 0 and j = 2: the first digit of nums[0] is 2, and the last digit of nums[2] is 1. Indeed, gcd(2,1) == 1.
When i = 1 and j = 2: the first digit of nums[1] is 5, and the last digit of nums[2] is 1. Indeed, gcd(5,1) == 1.
When i = 1 and j = 3: the first digit of nums[1] is 5, and the last digit of nums[3] is 4. Indeed, gcd(5,4) == 1.
When i = 2 and j = 3: the first digit of nums[2] is 1, and the last digit of nums[3] is 4. Indeed, gcd(1,4) == 1.
Thus, we return 5.

Example 2:

Input: nums = [11,21,12]
Output: 2
Explanation: There are 2 beautiful pairs:
When i = 0 and j = 1: the first digit of nums[0] is 1, and the last digit of nums[1] is 1. Indeed, gcd(1,1) == 1.
When i = 0 and j = 2: the first digit of nums[0] is 1, and the last digit of nums[2] is 2. Indeed, gcd(1,2) == 1.
Thus, we return 2.

 

Constraints:

• 2 <= nums.length <= 100
• 1 <= nums[i] <= 9999
• nums[i] % 10 != 0

Solutions

Solution 1: Counting

We can use an array $\text{cnt}$ of length $10$ to record the count of the first digit of each number.

Iterate through the array $\text{nums}$. For each number $x$, we enumerate each digit $y$ from $0$ to $9$. If $\text{cnt}[y]$ is not $0$ and $\text{gcd}(x \mod 10, y) = 1$, then the answer is incremented by $\text{cnt}[y]$. Then, we increment the count of the first digit of $x$ by $1$.

After the iteration, return the answer.

The time complexity is $O(n \times (k + \log M))$, and the space complexity is $O(k + \log M)$. Here, $n$ is the length of the array $\text{nums}$, while $k$ and $M$ respectively represent the number of distinct numbers and the maximum value in the array $\text{nums}$.

Python3

class Solution:
    def countBeautifulPairs(self, nums: List[int]) -> int:
        cnt = [0] * 10
        ans = 0
        for x in nums:
            for y in range(10):
                if cnt[y] and gcd(x % 10, y) == 1:
                    ans += cnt[y]
            cnt[int(str(x)[0])] += 1
        return ans

Java

class Solution {
    public int countBeautifulPairs(int[] nums) {
        int[] cnt = new int[10];
        int ans = 0;
        for (int x : nums) {
            for (int y = 0; y < 10; ++y) {
                if (cnt[y] > 0 && gcd(x % 10, y) == 1) {
                    ans += cnt[y];
                }
            }
            while (x > 9) {
                x /= 10;
            }
            ++cnt[x];
        }
        return ans;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

C++

class Solution {
public:
    int countBeautifulPairs(vector<int>& nums) {
        int cnt[10]{};
        int ans = 0;
        for (int x : nums) {
            for (int y = 0; y < 10; ++y) {
                if (cnt[y] && gcd(x % 10, y) == 1) {
                    ans += cnt[y];
                }
            }
            while (x > 9) {
                x /= 10;
            }
            ++cnt[x];
        }
        return ans;
    }
};

Go

func countBeautifulPairs(nums []int) (ans int) {
	cnt := [10]int{}
	for _, x := range nums {
		for y := 0; y < 10; y++ {
			if cnt[y] > 0 && gcd(x%10, y) == 1 {
				ans += cnt[y]
			}
		}
		for x > 9 {
			x /= 10
		}
		cnt[x]++
	}
	return
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

TypeScript

function countBeautifulPairs(nums: number[]): number {
    const cnt: number[] = Array(10).fill(0);
    let ans = 0;
    for (let x of nums) {
        for (let y = 0; y < 10; ++y) {
            if (cnt[y] > 0 && gcd(x % 10, y) === 1) {
                ans += cnt[y];
            }
        }
        while (x > 9) {
            x = Math.floor(x / 10);
        }
        ++cnt[x];
    }
    return ans;
}

function gcd(a: number, b: number): number {
    if (b === 0) {
        return a;
    }
    return gcd(b, a % b);
}