find repeated element in array code example
Example 1: 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 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, 9, 4, 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;
}
};