# Digit anagrams

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Section titled Problem

## 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`).
Section titled Solution

## Solution

``````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
}
``````
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