Contains Duplicate

Problem

Given an integer array nums, return true if any value appears at least twice, and false if every element is distinct.

Intuition

The brute-force instinct is to compare every pair. That works, but it asks the same question over and over: "have I seen this value already?"

That question has a direct data-structure answer: a HashSet.

As you scan left to right, either the current value is new, or it is already present in the set. The moment it is already present, you have found a duplicate and can return immediately.

Approach

use std::collections::HashSet;

impl Solution {
    pub fn contains_duplicate(nums: Vec<i32>) -> bool {
        let mut seen = HashSet::with_capacity(nums.len());
        for x in nums {
            if !seen.insert(x) {
                return true;
            }
        }
        false
    }
}

function containsDuplicate(nums: number[]): boolean {
  const seen = new Set<number>();
  for (const x of nums) {
    if (seen.has(x)) return true;
    seen.add(x);
  }
  return false;
}

Edge cases

• Empty array — no duplicate exists, so return false.
• Negative values or large values — hashing does not care about numeric range.
• Duplicate at the front — early return means best-case work can be much smaller than n.

Takeaway

If the real question is "have I seen this before?", the answer is usually a set.

This is the smallest possible HashSet problem, but the reflex matters. The same lookup pattern drives #1 Two Sum, #242 Valid Anagram, and #128 Longest Consecutive Sequence.