Number of Beautiful Pairs - Problem

Array Hash Table Math Counting Number Theory Easy

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [11, 21, 22, 12]

› Output: 3

💡 Note: Beautiful pairs: (0,1) first=1,last=1 gcd=1; (0,2) first=1,last=2 gcd=1; (0,3) first=1,last=2 gcd=1

Example 2 — No Beautiful Pairs

$ Input: nums = [42, 21]

› Output: 1

💡 Note: Only pair (0,1): first=4, last=1, gcd(4,1)=1, so we have 1 beautiful pair

Example 3 — All Beautiful

$ Input: nums = [31, 25, 72, 18]

› Output: 5

💡 Note: Most pairs have coprime first and last digits: (0,1),(0,2),(0,3),(1,2),(2,3) are beautiful

Constraints

2 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 9999

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is that two digits are coprime if their GCD equals 1. The brute force approach checks all pairs in O(n²) time, while the optimized approach uses a precomputed lookup table to avoid redundant GCD calculations. Best approach is frequency count optimization with Time: O(n²), Space: O(1).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each pair (i,j) where i < j, extract the first digit of nums[i] and last digit of nums[j], then check if their GCD equals 1. Count all such beautiful pairs.

Frequency Count Optimization

⏱️ Time: O(n) Space: O(1)

Instead of checking each pair individually, count how many numbers have each first digit and how many have each last digit. Then use a precomputed table of which digit pairs are coprime to calculate the result efficiently.

Brute Force — Algorithm Steps

Iterate through all pairs (i,j) where i < j
Extract first digit of nums[i] and last digit of nums[j]
Calculate GCD of these digits
Count pairs where GCD equals 1

Visualization

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Step-by-Step Walkthrough

Setup

Start with array and check all pairs

Extract Digits

Get first digit of nums[i] and last digit of nums[j]

Check GCD

Count pairs where GCD(first, last) = 1

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int getFirstDigit(int n) {
    if (n < 0) n = -n;
    char str[20];
    sprintf(str, "%d", n);
    return str[0] - '0';
}

int getLastDigit(int n) {
    if (n < 0) n = -n;
    return n % 10;
}

int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int solution(int* nums, int numsSize) {
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            int firstDigit = getFirstDigit(nums[i]);
            int lastDigit = getLastDigit(nums[j]);
            
            if (gcd(firstDigit, lastDigit) == 1) {
                count++;
            }
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    static int nums[1000];
    int numsSize = 0;
    parseArray(line, nums, &numsSize);
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all pairs, GCD calculation is O(1) for single digits

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting and calculations

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

845 Likes

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Number of Beautiful Pairs - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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