Design-1 solutions

[sundeep4sandy]

No description provided.

[super30admin]

Implement Hash Set (705DesignHashSet.js)

Your approach shows understanding of double hashing, but there are critical issues to address:

1. Initialization: Instead of initializing each bucket as an empty array, you should initialize them as null or leave them undefined. Then, in the add method, when you first access a bucket, you should create an array of the required size. This avoids wasting memory for unused buckets.

2. Pre-allocating Secondary Arrays: You need to determine the size of the secondary arrays. Since the key range is [0, 10^6], and you have 1009 primary buckets, the secondary arrays should be of size Math.ceil(1000001 / 1009) which is about 991. But note that the reference solution uses 1000 primary buckets and 1000 secondary buckets, except for the first bucket which has 1001. You can adopt a similar strategy.

3. Handling Edge Cases: For the first bucket (index 0), you might need an extra slot because the maximum key is 10^6. For example, if you have 1009 primary buckets, the secondary index for key 10^6 would be Math.floor(1000000 / 1009) = 990, which is within 0 to 990. But if you use 1000 primary buckets, the secondary index for key 10^6 would be 1000, which requires an array of size 1001 for the first bucket. So you need to adjust your secondary array sizes accordingly.

4. Boolean Storage: You are storing booleans, which is correct. But ensure that the secondary arrays are initialized with false for all elements. Currently, you are not initializing the secondary arrays, so when you set this.buckets[h1][h2] = true, other indices are undefined, which can be misinterpreted as false in conditions, but it's not consistent. For example, if (this.buckets[h1][h2]) will return false for undefined, but you want it to return false only if it was set to false or not set at all. However, when you remove, you set it to false, which is correct. But the array must be pre-allocated to avoid undefined.

5. Code Clarity: Your code is clear and commented. However, the hash2 function is named hash2, but it's actually the secondary index. The reference solution uses more descriptive names like getPrimaryHash and getSecondaryHash.

6. Efficiency Improvement: Pre-allocate the secondary arrays only when needed and with the correct size. For instance, in the add method, if the bucket is null, create a new array of the required size filled with false. For the first bucket (index 0), you might need a larger size.

Suggested correction for initialization and add method:

var MyHashSet = function () {
  this.size = 1000;
  this.buckets = new Array(this.size);
};

MyHashSet.prototype.hash1 = function (key) {
  return key % this.size;
};

MyHashSet.prototype.hash2 = function (key) {
  return Math.floor(key / this.size);
};

MyHashSet.prototype.add = function (key) {
  const h1 = this.hash1(key);
  const h2 = this.hash2(key);
  if (this.buckets[h1] === undefined) {
    // Determine the size of the secondary array
    const secondarySize = h1 === 0 ? this.size + 1 : this.size;
    this.buckets[h1] = new Array(secondarySize).fill(false);
  }
  this.buckets[h1][h2] = true;
};

Then, in remove and contains, you should check if the bucket exists before accessing it.

VERDICT: NEEDS_IMPROVEMENT

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Implement Min Stack (155MinStack.js)

Your solution is well-structured and meets all the problem requirements. Here are some strengths and minor suggestions:

Strengths:

• You have correctly implemented the MinStack with two stacks, ensuring constant time for all operations.
• The code is clean, with good variable names and clear logic.
• You handled the initial empty state of the minStack appropriately in the push method.

Areas for improvement:

• While your solution is correct, you might consider initializing the minStack with a sentinel value (like Infinity) to avoid the ternary check in push. However, your current approach is also valid and clear.
• In JavaScript, using arrays with push and pop is efficient, but note that the minStack is storing the minimum at each state, which is the standard approach. There is an alternative method that uses less space in some cases (by storing pairs of value and min), but your solution is standard and efficient.

Overall, your solution is excellent.

VERDICT: PASS