# Daily Challenge #273 - Remove Duplicates

In this challenge, you will remove the left-most duplicates from a list of integers and return the result.

// Remove the 3's at indices 0 and 3
// followed by removing a 4 at index 1
solve([3, 4, 4, 3, 6, 3]); // => [4, 6, 3]


Tests:
solve([3,4,4,3,6,3])
solve([1,2,1,2,1,2,3])
solve([1,1,4,5,1,2,1])
solve([1,2,1,2,1,1,3])

Good luck!

This challenge comes from KenKemau on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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### Discussion

This solution is in python

def Remove(duplicate):
final_list = []
for num in duplicate:
if num not in final_list:
final_list.append(num)
return final_list

print(Remove([3, 4, 4, 3, 6, 3]))


output

[3, 4, 6]


is
[3, 4, 6]
not
[4, 6, 3]

Fixed it

def Remove(duplicate):
final = []

for i in range(len(duplicate)):
if duplicate[i] not in duplicate[i+1:]:
final.append(duplicate[i])

return final

x = input("Enter the integers: ")
x = x.split()

print(Remove(x))



Hi, you could have use the set method, like this:

dev.to/rafaacioly/comment/12l2j

:)

Javascript in O(n) (more specifically 3n, looping three times on the length of the input, two identical reduce es and one filter):

const identity = (i) => i || i === 0;
return array;
};



Epic, but also this relies on the array being sparse in the first reduce (so, a map really), since new Array(Number.MAX_SAFE_INTEGER) is an error.

I'd express that as

const solve = arr => {
const vals = new Map(function * () {for (let i=0; i < arr.length; i++) yield [arr[i], i]} ())
const out = new Array(arr.length)
for (const [v, i] of vals) out[i] = v
return out.filter(x => typeof x !== "undefined")
}


which is just a funny version of

const solve = arr => {
const vals = new Map() // tfw no Map::with_capacity
for (let i=0; i < arr.length; i++) vals.set(arr[i], i)
const out = new Array(arr.length)
for (const [v, i] of vals) out[i] = v
return out.filter(x => typeof x !== "undefined")
}


var solve = (array) => new Set(array);


This isn't O(1)

(a => [...new Set(a)])([3, 4, 4, 3, 6, 3])


is
[3, 4, 6]
not
[4, 6, 3]

To get the correct result you'd have to do

[...new Set(a.reverse())].reverse()


which is... quite a lot.

module Main where

deduplicate :: [Int] -> [Int]
deduplicate [] = []
deduplicate (x:xs)
| elem x xs = deduplicate xs
| otherwise = x : deduplicate xs

main :: IO ()
main = do
print $deduplicate [3, 4, 4, 3, 6, 3] -- [4, 6, 3] print$ deduplicate [3, 4, 4, 3, 6, 3]      -- [4, 6, 3]
print $deduplicate [1, 2, 1, 2, 1, 2, 3] -- [1, 2, 3] print$ deduplicate [1, 1, 4, 5, 1, 2, 1]   -- [4, 5, 2, 1]
print $deduplicate [1, 2, 1, 2, 1, 1, 3] -- [2, 1, 3]  Test Wouldn't this have O(n^2) time complexity? This could be done in O(n), using a IntMap. EDIT: This is my attempt at writing a similar function but with O(n) time complexity (O(n*m) when m <= 64, where m is amount of elements in IntSet): import Data.Foldable (foldl') import Data.IntSet (empty, insert, notMember) deduplicate :: [Int] -> [Int] deduplicate = fst . foldl' takeUniq ([], empty) . reverse where takeUniq (xs, set) x | notMember x set = (x:xs, insert x set) | otherwise = (xs, set)  Hi and thanks for your reply. This looks like a very good solution. Wouldn't this have O(n^2) time complexity? Indeed this algorithm would have an O(n²) time complexity if the xs list would remain the same. I believe the time complexity is O(n log n) since we are decreasing the xs list each time in the solution I proposed. But I'm bad at time complexity so I wouldn't know. I didn't know about Data.IntSet I'll look into that. Thanks for sharing. You are forgetting OLog(n) steps used by IntMap internally, it is never O(n), Perhaps you're mistaking IntMap for Map? Here, in Map documentation most lookup and insertion operations are indeed O(log n): hackage.haskell.org/package/contai... But in IntMap documentation, you can see that the actual complexity is O(min(n, W)), where W is size of integer type (32 or 64): hackage.haskell.org/package/contai... This effectively means that after IntMap contains more than 64 elements, the time complexity is constant O(W) or O(1). Interesting .. I wasn't aware of that. In C with O(n2): #include <stdio.h> int solve(int* nums,int* newnums, int numsSize){ if(numsSize==0){ return 0; } int count = 0; for(int i=0;i<numsSize;i++){ int flag = 0; for(int j=0;j<count;j++){ if(nums[i]==newnums[j]){ flag = 1; break; } } if(flag == 0){ newnums[count++] = nums[i]; } } return count; } int main(void) { int ar[] = {1,2,1,2,1,1,3}; int n = sizeof(ar)/sizeof(ar[0]); int newar[n]; int len = solve(ar,newar,n); for(int i=0;i<len;i++){ printf("%d ",newar[i]); } printf("\n"); return 0; }  Thanks, I am new to Programming. In JavaScript (ES6) const originalArray = [3,4,4,3,6,3]; const distinctArray = [...new Set(originalArray)];  function solve(list) { return list.filter((a, b) => array.indexOf(a) == b) };  Simple JS O(n) solution (I think... 2n?) that will remove the left-most duplicates from a list of integers and return the result in the right order. const solve = arr => { let seen = []; for (let i = arr.length - 1; i > 0; i--) { seen.includes(arr[i]) ? _ : seen.unshift(arr[i]) } return seen; }  This is not O(n) because you're iterating over "seen" at every iteration of your for loop, so it's O(n^2) except in the degenerate case where all the elements are duplicates. Also, it isn't specified but unshift itself is also most likely an O(n) operation in most implementations. I had my doubts, but it seems the JS implementation uses a linear search algorithm. FWIW, I also had to check if Sets keep insertion ordering since in principle they shouldn't... but in JS they do! Good to know, I didn't know that was actual documented behavior of the Set. Here's my python solution O(n). The idea is to iterate over the input and keep track of the index of the last time you saw a given number. If you see a number, replace the old value with None, and do a second pass at the end to remove all the Nones. The two-phase approach is necessary to keep it O(n). If we remove items as we find them it could end up costing O(n^2). def solve(arr): last_seen = {} res = [] for val in arr: if val in last_seen: res[last_seen[val]] = None last_seen[val] = len(res) res.append(val) return list(v for v in res if v is not None)  ## javascript const solve = arr => { const vals = new Map() for (let i=0; i < arr.length; i++) vals.set(arr[i], i) const sparse = new Array(arr.length) for (const [v, i] of vals) sparse[i] = v const out = new Array(vals.length) let o = 0; for (const i in sparse) out[o++] = sparse[i] return out }  The question remains, does the for in kill the man what if it sorts lexically and not numerically? Should I have just for (const x of sparse) if (typeof x !== "undefined") out[o++] = x? Maybe function * isDef(it) { for (const x of it) if (typeof x !== "undefined") yield x } const solve = arr => { const vals = new Map() // tfw no Map::with_capacity for (let i=0; i < arr.length; i++) vals.set(arr[i], i) const sparse = new Array(arr.length) for (const [v, i] of vals) sparse[i] = v const out = new Array(vals.length) const search = isDef(sparse) for (let i = 0; i < vals.length; i++) out[i++] = search.next() return out }  ayy lmao Here is the simple solution with PHP: function solve($arr) {
$ansArr = [];$index = count($arr)-1; for(;$index >= 0; $index--) { if (in_array($arr[$index],$ansArr) == false) {
$ansArr[] =$arr[$index]; } } return array_reverse($ansArr);
}


Nim:

import sequtils, algorithm

proc solve*(data: seq[int]): seq[int] =
return data.reversed().deduplicate().reversed()



Here Goes, My Solution

pprint(solved)
set([3, 4, 6])
is not
[4,6,3]
which is the expected answer (you only keep the last occurrence of each duplicates, not the first)

Python 🐍

from typing import List

def solve(nums: List[int]): List[int]:
return list(set(nums))