Contains Duplicate

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

This is one of the cleanest hash table problems because the idea is simple: scan the array once, keep a set of values you have already seen, and stop immediately when a number shows up again. If you finish the scan without hitting a repeat, then all values were unique.

Budibadu uses this problem to build the habit of using a set for fast lookups instead of wasting time on nested loops. It is a small pattern, but it shows up everywhere.

Example 1:

Input: nums = [1,2,3,1]
Output: true
Explanation: The value 1 appears more than once.

Example 2:

Input: nums = [1,2,3,4]
Output: false
Explanation: Every number appears exactly once.

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true
Explanation: Several values repeat, so the answer is true.

Algorithm Flow

Recommendation Algorithm Flow for Contains Duplicate - Budibadu

Best Answers

java

import java.util.*;
class Solution {
    public boolean contains_duplicate(int[] nums) {
        Set<Integer> seen = new HashSet<>();
        for (int num : nums) {
            if (seen.contains(num)) return true;
            seen.add(num);
        }
        return false;
    }
}

javascript

function contains_duplicate(nums) {
    const seen = new Set();
    for (const num of nums) {
        if (seen.has(num)) return true;
        seen.add(num);
    }
    return false;
}

php

<?php
function contains_duplicate($nums) {
    $seen = [];
    foreach ($nums as $num) {
        if (isset($seen[$num])) return true;
        $seen[$num] = true;
    }
    return false;
}
?>

python

from typing import List

def contains_duplicate(nums: List[int]) -> bool:
    seen = set()
    for num in nums:
        if num in seen:
            return True
        seen.add(num)
    return False

rust

use std::collections::HashSet;
fn contains_duplicate(nums: Vec<i32>) -> bool {
    let mut seen = HashSet::new();
    for num in nums {
        if seen.contains(&num) {
            return true;
        }
        seen.insert(num);
    }
    false
}

typescript

function contains_duplicate(nums: number[]): boolean {
    var seen: { [key: string]: boolean } = {};
    for (var i = 0; i < nums.length; i++) {
        var key = String(nums[i]);
        if (seen[key]) return true;
        seen[key] = true;
    }
    return false;
}

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Contains Duplicate

Algorithm Flow

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