Java.Collections.Will HashMap work if all added keys have the same hashCode()?

✅ Yes, HashMap will still work

If all keys have the same hashCode(), the HashMap will:

1. Map every key to the same bucket (because index = hash & (table.length - 1) will always be the same).
2. Store all entries in that one bucket.
3. Use equals() to differentiate between keys inside that bucket.

So functionality-wise:

map.put(key1, "value1");
map.put(key2, "value2");
map.get(key2); // ✅ will still return "value2"

As long as equals() works correctly, the HashMap will behave correctly.

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🚨 But performance will degrade badly

If all keys fall into one bucket:

• That bucket becomes a linked list (or a red-black tree if ≥8 entries and table ≥64 slots).
• So every get(), put(), and remove() in that bucket becomes:
  • O(n) time in Java 7 and earlier (linked list)
  • O(log n) in Java 8+ (treeified bucket)

That’s a huge step down from the expected O(1) performance of a normal HashMap.

💣 Example of worst-case hashCode():

class BadKey {
    int id;
    public int hashCode() { return 42; } // constant hash!
    public boolean equals(Object o) {
        return o instanceof BadKey && ((BadKey)o).id == this.id;
    }
}

All keys hash to the same bucket → massive collision.

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✅ How does Java mitigate this?

Since Java 8, buckets with ≥8 entries and a table size ≥64 will automatically convert from a linked list to a red-black tree to prevent performance degradation:

• put() and get() then become O(log n) for that bucket.

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🧠 TL;DR

Condition	Behavior
All keys have same hashCode()	HashMap still works ✅
But performance degrades	From O(1) to O(n) or O(log n)
Java 8+ optimization	Treeifies bucket at threshold

So yes — HashMap is functionally safe even with bad hash codes, but if performance matters, always use well-distributed hashCode() implementations.