想象一个对象，以字符串作为键，以它们出现的频率计数作为值。

{"bravo charlie": 10, "alpha bravo charlie": 10, "delta echo foxtrot": 15, "delta echo": 7}

我正在尝试优化一个算法，以便A）任何作为另一个键的子字符串并且具有相同频率值的键都应该被删除。包含较长的键应该保留。B）允许只有一个单词的键保留，即使它包含在另一个键中
下面的成对比较方法可以工作，但在大型对象上会变得非常非常慢。例如，一个具有560k个键的对象需要大约30分钟才能完成成对比较：

// for every multi word key
// if a part of a longer key in candidates AND have same length delete
let keys = Object.keys(candidates)
    .sort((a, b) => b.length - a.length)
    .filter(key => {
        if (key.split(" ").length === 1) {
            return false
        }
        return true
    });
    // ^ order keys by length to speed up comparisons and filter out single words

    checkKeyI: for (const keyi of keys) {

        checkKeyJ: for (const keyj of keys) {

            // because we pre-sorted if length is less then we are done with possible matches
            if (keyj.length <= keyi.length) {
                continue checkKeyI
            }

            // keys must not match exactly
            if (keyj === keyi) {
                continue checkKeyJ
            }

            // keyi must be a substring of keyj
            if (!keyj.includes(keyi)) {
                continue checkKeyJ
            }

            // they must have the same freq occurr values
            if (candidates[keyj] === candidates[keyi]) {
                delete candidates[keyi]
                continue checkKeyI
            }

        }
    }

预期结果将是：

{"alpha bravo charlie": 10, "delta echo foxtrot": 15, "delta echo": 7}

因为bravo charlie被淘汰了。他们有没有什么明显或聪明的方法来加快速度？