Are two words anagrams?

def is_anagram(s, t):
    if len(s) != len(t):
        return False
    count = [0] * 26
    for a, b in zip(s, t):
        count[ord(a) - 97] += 1      # up for s
        count[ord(b) - 97] -= 1      # down for t
    return all(c == 0 for c in count)

#include <string>
using namespace std;

bool solve(const string& s, const string& t) {
    if (s.size() != t.size()) return false;
    int count[26] = {0};
    for (size_t i = 0; i < s.size(); ++i) {
        count[s[i] - 'a']++;     // up for s
        count[t[i] - 'a']--;     // down for t
    }
    for (int c : count) if (c != 0) return false;
    return true;
}

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) return false;
        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++;   // up for s
            count[t.charAt(i) - 'a']--;   // down for t
        }
        for (int c : count) if (c != 0) return false;
        return true;
    }
}

function isAnagram(s, t) {
  if (s.length !== t.length) return false;
  const count = new Array(26).fill(0);
  for (let i = 0; i < s.length; i++) {
    count[s.charCodeAt(i) - 97]++;   // up for s
    count[t.charCodeAt(i) - 97]--;   // down for t
  }
  return count.every(c => c === 0);
}

fn solve(s: &str, t: &str) -> bool {
    if s.len() != t.len() { return false; }
    let mut count = [0i32; 26];
    for (a, b) in s.bytes().zip(t.bytes()) {
        count[(a - b'a') as usize] += 1;   // up for s
        count[(b - b'a') as usize] -= 1;   // down for t
    }
    count.iter().all(|&c| c == 0)
}

import qualified Data.Map.Strict as M

solve :: String -> String -> Bool
solve s t = length s == length t && all (== 0) (M.elems counts)
  where
    -- +1 for each letter of s, -1 for each letter of t
    counts = M.fromListWith (+) ([(c, 1 :: Int) | c <- s] ++ [(c, -1) | c <- t])