Find All Duplicates in an Array

Problem Statement

You’re a duplicate detective with an array of integers from 1 to n, where some numbers appear twice. Find all duplicates without extra space (if possible). This medium-level challenge offers a hashing twist—or a clever in-place trick!

Example

Input: nums = [4, 3, 2, 7, 8, 2, 3, 1]

Output: [2, 3] (Both appear twice)

Input: nums = [1, 1, 2]

Output: [1]

Input: nums = [1]

Output: [] (No duplicates)

Code

public class Solution {
    public List findDuplicates(int[] nums) {
        List result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            int index = Math.abs(nums[i]) - 1;
            if (nums[index] < 0) result.add(index + 1);
            nums[index] = -nums[index];
        }
        return result;
    }
}
            

def find_duplicates(nums):
    result = []
    for num in nums:
        index = abs(num) - 1
        if nums[index] < 0:
            result.append(index + 1)
        nums[index] = -nums[index]
    return result
            

function findDuplicates(nums) {
    let result = [];
    for (let num of nums) {
        let index = Math.abs(num) - 1;
        if (nums[index] < 0) result.push(index + 1);
        nums[index] = -nums[index];
    }
    return result;
}
            

Explanation

• In-Place Insight: Use array as hash table; mark presence by negating values.
• Flow: For each num, mark its index; if already marked, it’s a duplicate.
• Example Walkthrough: [4,3,2,7,8,2,3,1] → mark 4,3,2,7,8, hit 2 again → [2,3].
• Hashing Alt: Set or map works but uses O(n) space.
• Constraint: Works because nums are 1 to n, fitting array indices.

Note