Digit anagrams

Problem

Given an array of integers a, your task is to count the number of pairs i and j (where 0 ≤ i < j < a.length), such that a[i] and a[j] are digit anagrams.

Two integers are considered to be digit anagrams if they contain the same digits. In other words, one can be obtained from the other by rearranging the digits (or trivially, if the numbers are equal). For example, 12345 and 54231 are digit anagrams, but 321 and 782 are not (since they don't contain the same digits). 220 and 22 are also not considered as digit anagrams, since they don't even have the same number of digits.

Example

For a = [25, 35, 872, 228, 53, 278, 872], the output should be solution(a) = 4.

There are 4 pairs of digit anagrams:

a[1] = 35 and a[4] = 53 (i = 1 and j = 4),
a[2] = 872 and a[5] = 278 (i = 2 and j = 5),
a[2] = 872 and a[6] = 872 (i = 2 and j = 6),
a[5] = 278 and a[6] = 872 (i = 5 and j = 6).

Solution

Just read through comment in the code to help you understand

import (
    "strings"
    "sort"
)
func solution(a []int) int {
    anagramCount := 0
    // Create a map to store the frequency of digits in each number
    freqMap := make(map[string]int)
    for i := 0; i < len(a); i++ {
        // Convert the integer to a string and sort the digits
        numStr := strconv.Itoa(a[i])
        numArr := strings.Split(numStr, "")
        sort.Strings(numArr)
        sortedNumStr := strings.Join(numArr, "")
        // Check if we've seen this sorted string before
        if val, ok := freqMap[sortedNumStr]; ok {
            // If we have, increment the count and update the map
            anagramCount += val
            freqMap[sortedNumStr] = val + 1
        } else {
            // If we haven't seen this sorted string before, add it to the map with a count of 1
            freqMap[sortedNumStr] = 1
        }
    }
    return anagramCount
}