find duplicate number in an 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: find duplicate values in array javascript

const findDuplicates = (arr) => {
  let sorted_arr = arr.slice().sort(); // You can define the comparing function here. 
  // JS by default uses a crappy string compare.
  // (we use slice to clone the array so the
  // original array won't be modified)
  let results = [];
  for (let i = 0; i < sorted_arr.length - 1; i++) {
    if (sorted_arr[i + 1] == sorted_arr[i]) {
      results.push(sorted_arr[i]);
    }
  }
  return results;
}

let duplicatedArray = [9, 4, 111, 2, 3, 4, 9, 5, 7];
console.log(`The duplicates in ${duplicatedArray} are ${findDuplicates(duplicatedArray)}`);

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;
        
    }
};