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Dec 14, 2023 · Learn how to print the duplicates in an array using nested loop, hash-map, or sorting algorithms. See examples, code, and time complexity analysis for C++, Java, Python, and Javascript.
- Array Duplicates | Practice | GeeksforGeeks
Given an array arr of size n which contains elements in...
- Find duplicates in O (n) time and O (1) extra space | Set 1
Given an array of n elements containing elements from 0 to...
- Array Duplicates | Practice | GeeksforGeeks
A medium-level problem that asks to find all the integers that appear twice in an array of length n and range [1, n]. The solution must run in O(n) time and use constant extra space.
Given an array arr of size n which contains elements in range from 0 to n-1, you need to find all the elements occurring more than once in the given array. Return the answer in ascending order. If no such element is found, return list containing
May 7, 2024 · Given an array of n elements containing elements from 0 to n-1, with any of these numbers appearing any number of times, find these repeating numbers in O(n) and using only constant memory space. Example: Input: n = 7 , array = {1, 2, 3, 1, 3, 6, 6} Output: 1, 3 and 6. Explanation: Duplicate element in the array are 1 , 3 and 6 Input: n = 6, array
- 7 min
- Brute Force
- Count Iterations
- Sorted Array
- Sum of The Elements
- Marker
- Runner Technique
- GeneratedCaptionsTabForHeroSec
The brute force solution is to implement two nested loops this way: O(n²) time complexity and O(1) space complexity.
Another solution is to have a data structure to count the number of iterations of each integer. This solution could be implemented either with an array or a hash table. An implementation in Java: The value of index i represent the number of iterations of i+1. This solution is O(n) time but O(n) space as we need an extra structure.
If we apply the simplification technique, we could try to find a solution with a sorted array. In this case, we would just have to compare each element with its right neighbor. In Java: O(1) in space but O(n log(n)) in time as we need to sort the collection up front.
A direction we may think about is to sum the elements of the array and to compare it with 1 + 2 + … + n. Let’s see an example: In this example, we are able to find the solution in O(n) time and O(1) space. Yet, it is working solely in the case where we have one duplicate. A counter-example: This direction is going nowhere. But sometimes we need to ...
There is something interesting to mention. So far, our solutions are not really making use of the fact that each integer is between 1 and n. Due to this interesting constraint, each value has its own corresponding index in the array. The solution is to consider this particular array as a sort of linked list. Any index is pointing to the value of th...
There is another solution which also considers the given array as a sort of linked list (again, this is possible because of the constraint that each integer is between 1 and n). Let’s analyze the example [1, 2, 3, 4, 2]: With this representation, we can simply say that a duplicate exists when a cycle does exist. Moreover, the duplicate is the entry...
Learn how to solve the problem of finding a duplicate in an array of n + 1 integers between 1 and n using different algorithms. Compare the time and space complexity, examples and code implementations in Java.
- Teiva Harsanyi
Learn 10 different ways to find duplicate elements in an array in JavaScript, with examples and performance comparison. Use nested loops, indexOf, lastIndexOf, includes, reduce, map, some, object, lodash and more.
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Jul 3, 2021 · There are multiple methods available to check if an array contains duplicate values in JavaScript. You can use the indexOf() method, the Set object, or iteration to identify repeated items in an array. Set Object. Set is a special data structure introduced in ES6 that stores a collection of unique values.
