C++ Tips and Tricks 😀

General ⚙️

tie(w,x,y,z) = make_tuple(10,20,30,40) :- w=10,x=20,y=30,z=40
unordered_set,unordered_map are faster than set and maps.
set uses self balancing binery search trees therefore O(nlogn).
unordered_set uses hashing therefore O(1) same for the map and unordered_map.
auto keyword can be used for declaring variables, it automatically set the type of that variable.
lower_bound : find >= elem , upper_bound : find > element
when we want to find element equal to we use lower_bound else upper_bound
is_sorted : Returns true if the container is sorted else false.
In Global we can declare array of bool upto 10^8 or int array upto 10^7.
In Local we can declare array of bool upto 10^7 or int array upto 10^6.

Faster operations ⏩⏩

a>>1 == divide a by 2.
a<<1 == multiply a with 2.
if (a&1)==0 == number is even.
if (a&1)==1 == number is odd, make sure you use the brackets.
swap numbers == a=a^b;b=a^b;a=a^b;
We should use emplace_back to push pair in vector insteed of pb(mp()) or pb({}).

Bit magic / Bit mask 0️⃣1️⃣

__builtin_popcount(n) == returns no. of set bits in n.
find i’th bit –> (n & (1<<i))==0 then ith bit is 0 else 1.
set i’th bit –> ( n | (1<<i)) then ith bit set to 1.
clear i’th bit –> (n & ~(1<<i)) then ith bit set to 0
mask = &(1<<i) for find.
mask = |(1<<i) for set-1.
mask = & ~(1<<i) for set-0.
(n&(n-1)) used in many ways :
1. It reduces last set bit to 1. ex.1101->1100->1000->0000 then we can count no of set bits by count.
2. we can find if a number is power of 2 or not –>if (n&(n-1))==0 then it is power of two else not.

std::string 🧵

temp.find(B) != string::npos, find B is in string or not.
s.insert(pos, string), insert string at pos position.

std::vector 🔼

v.erase(iterator) to delete elem while traversing in vector. Remember to do it–.
auto it = lower_bound(v.begin(), v.end(),elem);
int index = it - v.begin(); OR index = it - v;

std::set 🍁

auto it = s.lower_bound(elem);
int elem = *it;
s.count() returns 1 if element is present in set else 0.
NOTE: for vector and other containers which contains duplicate elements count returns count of the number.

std::map 🗺️

mp.count() returns 1 if element is present in map else 0.
NOTE: for vector and other containers which contains duplicate elements it returns count of the number.

Other tips 🍹

We can use dec,hex,oct to print numbers in different bases.

 // bitset<bitcnt>(num)
 cout << bitset<12>(n) << "\n"; // Binary
 cout << oct << n << "\n"; // Octal
 cout << hex << n << "\n"; // Hexadecimal
 cout << dec << n << "\n"; // Decimal-normal

std::minmax_element == returns pair{minelement, maxelement}

 vector<int> x = {1, 2, 3, 4};
 auto it = minmax_element(all(x));
 cout << *it.first << ' ' << *it.second << endl;

 output:
 1 4

is_palindrome function using std::equal.

 bool is_palindrome(string s) {
     return equal(all(s) - s.size() / 2, s.rbegin());
 }