LeetCode #2748 — EASY

Number of Beautiful Pairs

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[2,5,1,4]

Example 2

[11,21,12]

Step 02

Core Insight

What unlocks the optimal approach

Since nums.length is small, you can find all pairs of indices and check if each pair is beautiful.
Use integer to string conversion to get the first and last digit of each number.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2748: Number of Beautiful Pairs
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);
    }
}

// Accepted solution for LeetCode #2748: Number of Beautiful Pairs
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)
}

# Accepted solution for LeetCode #2748: Number of Beautiful Pairs
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

// Accepted solution for LeetCode #2748: Number of Beautiful Pairs
fn count_beautiful_pairs(nums: Vec<i32>) -> i32 {
    use std::collections::HashSet;

    fn gcd(a: i32, b: i32) -> i32 {
        let mut a = a;
        let mut b = b;

        loop {
            let tmp = a % b;
            if tmp == 0 {
                return b;
            }

            a = b;
            b = tmp;
        }
    }

    let pairs: Vec<(i32, i32)> = nums
        .into_iter()
        .map(|n| n.to_string().bytes().collect::<Vec<u8>>())
        .map(|bs| ((bs[0] - b'0') as i32, (*bs.last().unwrap() - b'0') as i32))
        .collect();

    let mut cache = HashSet::new();
    let len = pairs.len();
    let mut ret = 0;
    for i in 0..len {
        for j in (i + 1)..len {
            if cache.contains(&(pairs[i].0, pairs[j].1)) || gcd(pairs[i].0, pairs[j].1) == 1 {
                cache.insert((pairs[i].0, pairs[j].1));
                ret += 1;
            }
        }
    }

    ret
}

fn main() {
    let nums = vec![2, 5, 1, 4];
    let ret = count_beautiful_pairs(nums);
    println!("ret={ret}");
}

#[test]
fn test_count_beautiful_pairs() {
    {
        let nums = vec![2, 5, 1, 4];
        let ret = count_beautiful_pairs(nums);
        assert_eq!(ret, 5);
    }
    {
        let nums = vec![11, 21, 12];
        let ret = count_beautiful_pairs(nums);
        assert_eq!(ret, 2);
    }
    {
        let nums = vec![31, 25, 72, 79, 74];
        let ret = count_beautiful_pairs(nums);
        assert_eq!(ret, 7);
    }
}

// Accepted solution for LeetCode #2748: Number of Beautiful Pairs
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × (k + log M)

Space

O(k + log M)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.