O(n2) ,how to optimized two loop? - c++

Question

I want to compare the elements in array whether there is a similar element is present or not?
I am using two loops and it gives TLE can anyone help me in this?
code:-
int main() {
int arr=[1, 2, 3, 3, 4, 5, 6, 5, 6]
int max=0;
int k;
int c=0;
for (int i=0;i<n;i++) {
for (int j= i+1;j<n;j++) {
if (a[i]==a[j]) {
c++;
if (k>max) {
max= k;
}
}
}
}
}
how to optimize this?

Answer 1

The std::sort function time complexity is nlog(n) (According to Wikipedia, it's Intro Sort)
So one approach for your question is to:
Sort the array (It's better to use std::vector instead of C-Style array)
Compare every element with the next one (Which is going to be O(n))
If the two elements are equal, so show it or add it to a std::set to use it later
Here is my implementation for the algorithm that I mentioned before:
#include <vector>
#include <iostream>
#include <set>
#include <algorithm>
int main() {
std::vector<int> data{1, 2, 3, 4, 6, 2, 7, 1, 5, 4};
std::set<int> duplicatedValues;
std::sort(data.begin(), data.end());
for (int i{0}; i < data.size()-1; ++i)
if (data[i] == data[i+1])
duplicatedValues.insert(data[i]);
for (int value : duplicatedValues)
std::cout << value << " ";
std::cout << std::endl;
}
So time complexity for code above is going to be nlog(n)
Note: I should add that if you have n data and all of them are between range of 0 to n-1, We have O(n) for that, take a look at here.
Second Approach: (By using std::unordered_map)
(It takes a little shorter time than the previous approach, but roughly 2 times more memory!)
We can add every element in a dictionary and if we had that same data again in our array, increase its number in dictionary; and finally, show elements that have number more than 1.
#include <iostream>
#include <unordered_map>
#include <vector>
#include <stdlib.h>
int main() {
std::vector<int> data{1, 2, 8, 2, 5, 4, 1, 9, 2, 3, 8};
std::unordered_map<int, int> duplicatedData;
for (int i{0}; i < data.size(); ++i) {
if (duplicatedData.find(data[i]) == duplicatedData.end()) {
duplicatedData[data[i]] = 1;
}
else {
duplicatedData[data[i]] += 1;
}
}
for (auto x : duplicatedData) {
if (x.second > 1) {
std::cout << x.first << " ";
}
}
std::cout << std::endl;
}
EDIT: I recorded a video for this, with performance measurement in here

Answer 2

i will do it like
#include <algorithm>
#include <vector>
#include <execution>
#include <atomic>
#include <iostream>
int main()
{
int arr[] = {1,2,3,5,6};
std::atomic<bool> com_tr = false;
std::for_each(std::execution::par, &arr[0], (&arr[0] + sizeof(arr)/sizeof(int)),
[&](auto a)
{ com_tr = std::count(&arr[0], (&arr[0] + sizeof(arr)/sizeof(int)), a) == 1 ? (bool)com_tr : !false; });
if (com_tr) std::cout << "have one " << std::endl;
}
compiled with c++ -std=c++17 -O3 '/home/alex/etr/b.cpp' -o t -ltbb
but if it's really all you need to accomplish better would be std::any_of()