You can use a collections.OrderedDict in order to reduce the time complexity to O(n). Since it remembers the order of insertion the values resemble the various ids in order of their occurrence:
from collections
import OrderedDict
groups = OrderedDict()
for i, v in enumerate(idlist):
try:
groups[v].append(i)
except KeyError:
groups[v] = [i]Instead of a list, use a dict, which makes looking up for existence O(1):
def compiler(idlist):
groups = {}
for idx, val in enumerate(idlist):
if val in groups:
groups[val].append(idx)
else:
groups[val] = [idx]CPython:
func: op_implementation[10000 iterations] took: 1.3096 sec func: ordreddict_implementation[10000 iterations] took: 0.1866 sec func: defaultdict_implementation[10000 iterations] took: 0.3311 sec func: defaultdict_implementation_2[10000 iterations] took: 0.3817 sec func: dict_implementation[10000 iterations] took: 0.3231 sec
Pypy3:
func: op_implementation[10000 iterations] took: 0.2370 sec func: ordreddict_implementation[10000 iterations] took: 0.0243 sec func: defaultdict_implementation[10000 iterations] took: 0.1216 sec func: defaultdict_implementation_2[10000 iterations] took: 0.1299 sec func: dict_implementation[10000 iterations] took: 0.1175 sec
Pypy3 with 2000000 iterations:
func: op_implementation[200000 iterations] took: 4.6364 sec func: ordreddict_implementation[200000 iterations] took: 0.3201 sec func: defaultdict_implementation[200000 iterations] took: 2.2032 sec func: defaultdict_implementation_2[200000 iterations] took: 2.4052 sec func: dict_implementation[200000 iterations] took: 2.2429 sec
Last Updated : 27 Aug, 2019
The original list is: [1, 4, 5, 5, 5, 9, 1] The list of duplicate elements is: [3, 4, 6]
An empty list (l1) is taken to store the values of non-duplicate elements from the given list., An empty list (l1) is taken to store the values of non-duplicate elements from the given list. ,The list (l) with duplicate elements is taken to find the duplicates., The list (l) with duplicate elements is taken to find the duplicates.
l = [1, 2, 3, 4, 5, 2, 3, 4, 7, 9, 5]
l1 = []
for i in l:
if i not in l1:
l1.append(i)
else:
print(i, end = ' ')Thanks to the Enum module, in Elixir we can trivially remove duplicates from a list.,In the following example, we take a list of integers and pass it to the Enum.uniq/1 function which removes duplicates from the list without altering the original order of the remaining elements.,Alright, Elixir does the heavy lifting for us in this case, but how would we go about removing duplicates from a list in Elixir without using Enum.uniq/1? I mean from scratch, simply using recursion without relying on Enum, sets, :lists.usort/1, etc. ,head is our first element and then we are using a comprehension to generate a list without duplicates.
In the following example, we take a list of integers and pass it to the Enum.uniq/1 function which removes duplicates from the list without altering the original order of the remaining elements.
list = [1, 2, 2, 3, 3, 1, 2, 4] Enum.uniq(list) # Returns[1, 2, 3, 4]
If you are trying to only remove consecutive duplicate elements, then there is Enum.dedup/1:
list = [1, 2, 2, 3, 3, 1, 2, 4] Enum.dedup(list) # Returns[1, 2, 3, 1, 2, 4]
There are likely a few ways to implement this, but this what sprang to mind when I thought about it:
defmodule MyList do
def uniq([]), do: []
def uniq([head | tail]) do
[head |
for (x < -uniq(tail), x != head, do: x)
]
end
endHere is a simple example of how to use them:
for x < -[1, 2, 3, 4], do: x + x # Returns[2, 4, 6, 8]It also accepts filters, which allows us to specify a condition:
for x < -[1, 2, 3, 4], x < 3, do: x + x # Returns[2, 4]I am a bit confused here with the first solution.How does it guarantee to return duplicate values without first sorting the array? , It doesn't matter if two different values have the same hash if you are using a class such as HashSet or HashMap. The value of the hash indicates a set into which the entry will be placed, not the actual location of the entry in the set. At least, that is how I read the JavaDoc for the classes. , @sapan, you mean best way to learn Java8? or you mean using it on your real project?In case of first try solving programming problems like this using Java 8, for the second use Java 8 with collections and streams. , it will give us wrong answer if we have a element more than two times.for example - if java is three times in array then in answer it will print two times
for (int i = 0; i < names.length; i++) {
for (int j = i + 1; j < names.length; j++) {
if (names[i].equals(names[j])) {
// got the duplicate element
}
}
}Remote Work 2022 , Remote Work 2022
Let us look at the implementation of this using JavaScript
const arry = [1, 2, 1, 3, 4, 3, 5];
const toFindDuplicates = arry => arry.filter((item, index) => arr.indexOf(item) !== index)
const duplicateElementa = tofindDuplicates(arry);
console.log(duplicateElements);
// Output: [1, 3]In the above implementation, the output array can have duplicate elements if the elements have occurred more than twice in an array. To avoid this and for us to count the number of elements duplicated, we can make use of the use() method.
function toFindDuplicates(arry) {
const uniqueElements = new Set(arry);
const filteredElements = arry.filter(item => {
if (uniqueElements.has(item)) {
uniqueElements.delete(item);
} else {
return item;
}
});
return [...new Set(uniqueElements)]
}
const arry = [1, 2, 1, 3, 4, 3, 5];
const duplicateElements = toFindDuplicates(arry);
console.log(duplicateElements);
// Output: [1, 3]In JavaScript, an object consists of key-value pairs where keys are similar to indexes in an array and are unique. If one tries to add a duplicate key with a different value, then the previous value for that key is overwritten by the new value. We use this concept to compare the and find the duplicates.
toFindDuplicates();
function toFindDuplicates() {
let arry = [1, 2, 1, 3, 4, 3, 5];
let toMap = {};
let resultToReturn = false;
for (let i = 0; i < arry.length; i++) {
if (toMap[arry[i]]) {
resultToReturn = true;
// terminate the loop
break;
}
toMap[arr[i]] = true;
}
if (resultToReturn) {
console.log('Duplicate elements exist'
');
}
else {
console.log('Duplicates don'
t exist ');
}
}In this method, we compare each element of an array with all other elements of the array to check if two elements’ values are the same using nested for loop.
toFindDuplicates();
function toFindDuplicates(element, index) {
let arry = [1, 2, 1, 3, 4, 3, 5];
let resultToReturn = false;
for (let i = 0; i < arry.length; i++) { // nested for loop
for (let j = 0; j < arry.length; j++) {
// prevents the element from comparing with itself
if (i !== j) {
// check if elements' values are equal
if (arry[i] === arry[j]) {
// duplicate element present
resultToReturn = true;
// terminate inner loop
break;
}
}
}
// terminate outer loop
if (resultToReturn) {
break;
}
}
f(resultToReturn) {
console.log('Duplicate elements exist'
');
}
else {
console.log('Duplicates don'
t exist ');
}
}