I have only put 1 condition in second case for if map[x] > 0 then execute the code.

let result = [];
let freq = [];
let map = {};

for (let i = 0; i < queries.length; i++) {
    const [op, x] = queries[i];
    const f = map[x] || 0;

    if (op === 1) {
        map[x] = f + 1;
        freq[f] = (freq[f] || 0) - 1;
        freq[f + 1] = (freq[f + 1] || 0) + 1;
    }
    if (op === 2) {
        if (map[x] > 0) { 
            map[x] = f - 1;
            freq[f - 1] += 1;
            freq[f] -= 1;
        }
    }
    if (op === 3) {
        result.push(freq[x] > 0 ? 1 : 0);
    }
}
return result;