notes

if(s[i] == 'someChar') is correct 
if(s[i] == "someChar") is wrong 

anotherStr = strName.substr(start_index, upto_how_many_chars)

isdigit(char) returns whether char is digit or not 
eg: isdigit(s[i])
- used for checking digits in a string 
- similar functions 
- islower(s[i])
- isupper(s[i])

while multiplying a constant number will some var, if it is large, add LL 
eg: long long int ans = 2LL * (maxEl - minEl);

finding minimum and maximum el 
int minEl = a[0];
int maxEl = a[0];
for (int i = 0; i < n; i++)
{
    minEl = min(minEl, a[i]);
    maxEl = max(maxEl, a[i]);
}

ceil function only works for float 
eg ceil(n/2)
say n = 7 and n is int 
cout << ceil(n/2) gives value 3
cout << ceil((float)n/2) gives value 4

subarray and substring are continous
subsequence is not continous

odd = odd + even 
eg : odd = 3 + even 

Frequency array for lowercase str 
vector<char> freqArr(26);
for(auto ch : s)
    freqArr[ch - 'a']++;


if(x & 1) can also be used for checking odd number condition
or if(x%2!=0)


if in a problem, it's given take the result without rounding, 
dont consider floor or ceil, take the decimal point answer

If x>y then it will remain x>y no matter how many times we perform the operation.

Multiple different cases could be just one simple case 

in any problem, where you replace a[i] with some a[i]+1/2 or something which involves a[i], u sort the array

for inserting more than two elements in max func:
max({max1, max2, max3..})

Property of XOR: 
if we XOR a number with itself by even no of times 
    then we get result as 0
else if we XOR a number with itself by odd no of times 
    then we get result = number 
eg : 10 XOR 10 = 1010 XOR 1010 = 0000 = 0
10 XOR 10 = 0
(10 XOR 10 XOR 10) = 10

Real life example of race condition::
#include <iostream>
#include <string>

using namespace std;

int main()
{
    double num;
    string mystr;

    cout << "Please enter a number: " << "\n";
    cin >> num;
    cout << "Your number is: " << num << "\n";
    cout << "Please enter your name: \n";
    cout << "So your name is " << mystr << "?\n";
    cout << "Have a nice day. \n";

}

In the above case, when running the program it never stopped to ask for the string. It just skipped over it.
remember the clock-wala problem ?you faced similar problem
So you run into a race condition because you assume the stream is clear each time you ask for a input.

The solution to above problem is always use cin.ignore(numeric_limits<streamsize>::max(), '\n'); before using getline()
or use cin.ignore() after taking a number input and before taking an int input 


If you are precomputing something in a function, alwyas call that function in main, else there would no point in precomputing if you call it in solve()

Any no of 2^n+1 will have binary format as 1.0.1 i.e a no starting from 1 and ending with 1 and rest will be zeros

if X = gcd(A, B) then A % X = 0

inversion as a concept:
j > i & a[i] > a[j] i.e if you plot indices on X-axis and elements on Y-axis, you will get a negative slope
eg: arr[] = {5, 4, 3, 2, 1}

a number is divisible by 3 if the sum of the digits is divisible by 3 and a number is divisible by 99 if the sum of the digits is divisible by 9.

Any no of (2^n)-1 will have binary format as 111.. i.e n no of ones
eg 7 = 111, n=3

^ - XOR
a^b = c
then a^c = b

perfect squares have odd factors
Any no of 2^n+1 will have binary format as 1.0.1 i.e a no starting from 1 and ending with 1 and rest will be zeros

if X = gcd(A, B) then A % X = 0

inversion as a concept:
j > i & a[i] > a[j] i.e if you plot indices on X-axis and elements on Y-axis, you will get a negative slope
eg: arr[] = {5, 4, 3, 2, 1}

a number is divisible by 3 if the sum of the digits is divisible by 3 and a number is divisible by 99 if the sum of the digits is divisible by 9.

Any no of (2^n)-1 will have binary format as 111.. i.e n no of ones
eg 7 = 111, n=3

^ - XOR
a^b = c
then a^c = b

perfect squares have odd factors

the parity of an integer is the (non-negative) remainder obtained when dividing it by 2. For example, the parity of 246 is 0 and the parity of 11 is 1. In other words, an even number has parity 0 and an odd number has parity 1

n & (n - 1) drops the lowest set bit: https://leetcode.com/problems/number-of-1-bits/discuss/55255/C%2B%2B-Solution%3A-n-and-(n-1).

You can reverse a rows of any matrix by doing: 
for(int i = 0; i < n; i++)
	reverse(begin(matrix[i]), end(matrix[i]));

for reversing cols of a matrix:
reverse(begin(matrix), end(matrix));

Subarray question? 
Do arr contain only positive els: Go for sliding window 
else prefix sum

For unsetting the bits of a specific no, do an XOR with it
For setting bits of a specific no, do an OR with it
eg: https://leetcode.com/problems/longest-nice-subarray/submissions/

For reversing a vector of pair
sort(engineers.rbegin(), engineers.rend());

Getting binary representation of dec in string format: one-liner
string binaryRep = bitset<8>(decimalNo).to_string();

When a graph problem involves finding a shortest path, BFS should be used over DFS. This is because with BFS, all nodes at distance x from start will be visited before any node at distance x + 1 will be visited. Once we find the target (end), we know that we found it in the shortest number of steps possible

For finding if a key is present in map mp 
Please use mp.count(key) instead of 
mp.find(key) == mp.end()