find duplicate element in array code example

Example 1: js find duplicates in array

const names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl']

const count = names =>
  names.reduce((a, b) => ({ ...a,
    [b]: (a[b] || 0) + 1
  }), {}) // don't forget to initialize the accumulator

const duplicates = dict =>
  Object.keys(dict).filter((a) => dict[a] > 1)

console.log(count(names)) // { Mike: 1, Matt: 1, Nancy: 2, Adam: 1, Jenny: 1, Carl: 1 }
console.log(duplicates(count(names))) // [ 'Nancy' ]

Example 2: js find duplicates in array

[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) !== i) // [2, 4]

Example 3: Find the duplicate in an array of N integers.

// 287. Find the Duplicate Number
// Medium

// Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

// Example 1:

// Input: [1,3,4,2,2]
// Output: 2
// Example 2:

// Input: [3,1,3,4,2]
// Output: 3
// Note:

// You must not modify the array (assume the array is read only).
// You must use only constant, O(1) extra space.
// Your runtime complexity should be less than O(n2).
// There is only one duplicate number in the array, but it could be repeated more than once.

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int n=nums.size();
        int s=nums[0];
        int f=nums[nums[0]];
        while(s!=f) {
            s=nums[s];
            f=nums[nums[f]];
        }
        f=0;
        while(s!=f) {
            s=nums[s];
            f=nums[f];
        }
        return s;
        
    }
};