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C++ Data Structures Cheat Sheet by


Storing, data type of pointers and variables must be the same
Returns address of memory location of variable
data_type *pointer;
Initia­lize. Put * in front of name
Returns value at the memory location stored by pointer
Array variables are actually pointers to the first element in the array. The amount that your pointer moves from an arithmetic operation
Returns first element in array
Returns third element in array
◎ Variables that you declare are stored in a memory location in your
◎ The address of these memory locations can be stored in pointers
◎ Addresses are in hexade­cimal


//Append ::iterator to your data type declaration to create an iterator of
that data type
vector<int>::iterator it; // declares an iterator of vector<int>

// loops from the start to end of vi
for(vector<int>::iterator it = vi.begin(); it != vi.end(); it++) 
    cout << *it << " "; // outputs 1 2 3 4

deque<int> d;
deque<int>::iterator it;
it = d.begin(); //Points to first element 
it++; // Points to next element
it = d.end(); // Points to Last element
it--; // Points to previous element
cout << *it; // outputs the element it is pointing
Iterators are essent­ially pointers to an STL data structure


map<string, int> M;
M[“hello”] = 2;
M[“asd”] = 986;

M.count(“asd”); // returns 1
M.count(“doesnt_exist”); // returns 0

// Check for the existence of some key in the map - O(log N)
it = M.find(“asd”); //returns the iterator to “asd”
it = M.upper_bound("aaa");
it = M.lower_bound("aaa");
if (it == M.end())
    cout << "Does not exist\n";

for (auto it = mp.begin(); it != mp.end(); ++it) {
    cout << it.first << “, “ << it.second << “\n”;
A data structure that takes in any data[key]
Gives you the associated value stored through O(log N) magic

Best used when you need to lookup certain values in O(log N) time that are associated with other values


queue<int> q;

q.push(5); // Inserts/ Enqueue element at the back of the queue

q.front(); // Returns element atb the front of the queue
q.pop // Removes (Dequeues) element from the front of the queue

q.empty(); // Returns boolean value of whether queue is empty
First In First Out data structure where elements can only be added to the back and accessed at the front

Priority Queue

priority_queue<data_type> pq; // Largest at top
priority_queue<data_type, vector<data_type>, greater<data_type> >
pq; // Smallest at top

pq.push(5); // pushes element into it. Duplicates are allowed // Returns largest or smallest element
pq.pop() // Removes largest or smallest element

pq.size(); // Returns size
pq.empty(); Check if queue is empty
Like a queue except that only the element with the greatest priority (eg. larges­t/s­mal­lest) can be accessed

Fenwick tree

//Below here can mix & match
long long ft[100001]; // note: this fenwick tree is 1-indexed.
void fenwick_update(int pos, long long value) {
    while (pos <= N) {
        //cout<<"Fenwick Updating: "<<pos<<","<<value<<endl;
        ft[pos] += value;
        pos += pos&-pos;
long long fenwick_query(int pos) {
    long long sum = 0;
    while (pos) { // while p > 0
        sum += ft[pos];
        pos -= pos&-pos;
    return sum;
void fenwick_range_update(int pos_a, int pos_b, int val){
    //TLE way
    //for (int i=pos_a;i<=pos_b;i++){fenwick_update(i, val);}
    fenwick_update(pos_a, val);
    fenwick_update(pos_b+1, -val);
////PURQ////////////// ////////////////////////////////////////
long long fenwick_range_query(int pos_a, int pos_b) {
    return fenwick_query(pos_b) - fenwick_query(pos_a-1);

long long B1[100001];long long B2[100001];
void base_update(long long *ft, int pos, long long value){
  //Add largest power of 2 dividing x / Last set bit in number x
  for (; pos <= N; pos += pos&(-pos))
    ft[pos] += value;
void rurq_range_update(int a, int b,long long v){
  base_update(B1, a, v);
  base_update(B1, b + 1, -v);
  base_update(B2, a, v * (a-1));     
  base_update(B2, b + 1, -v * b);     
void rurq_point_update(int a, long long v){
long long base_query(long long *ft,int b){
  long long sum = 0; 
  for(; b > 0; b -= b&(-b))
    sum += ft[b];
  return sum;
// Return sum A[1...b]
long long rurq_query(int b){
  return base_query(B1, b) * b - base_query(B2, b);

//Return sum A[a...b]
long long rurq_range_query(int a,int b){
  return rurq_query(b) - rurq_query(a-1);


// Initialise
pair<data_type_1, data_type_2> variable;
// OR
pair<data_type_1, data_type_2> variable = make_pair(value1,value2);

// Store values
variable.first = value;
variable second = value;
// Retrieve values
cout <<variable first << " " << variable.second << endl;

//Nesting pairs
pair<int, pair<int, int> > a;
a.first = 5;
a.second.first = 6;
a.second.second = 7;
Stores a pair of values


stack<int> s;

s.push(5); // push an element onto the stack - O(1)
s.pop(); // pop an element from the stack - O(1); // access the element at the top of the stack - O(1)
s.empty(); // whether stack is empty - O(1)
First-­In-­Las­t-Out data structure
Only Element at the top can be accessed / removed


// Initialize
vector<data_type> v;

v.push_back(value); // Add element to back
v.pop_back() // Remove last element
v.clear(); // Remove all elements

v[index] // Return element of index
v.back(); // Return last element

v.size(); // Return Size of vector
v.empty() // Return boolean value of whether vector is empty
Like arrays but re-siz­able. You can add and remove any number of elements from any position.

Sets and Multisets

set<int> s; set<int>::iterator it; 
multiset<int> s; multiset<int>::iterator it; 


it = s.find(8)
it = s.upper_bound(7);
it = s.lower_bound(7);

s.erase(10); //Remove element from set
s.erase(it) //Can also use iterators


// to loop through a set
for(it = s.begin(); it != s.end(); it++)
    cout << *it << " "; // outputs 2 7 10
In a set: All elements are sorted and no duplicates
Multisets can store duplicates though


deque<int> d;

// access an element / modify an element (0-indexed as well) - O(1)
d[0] = 2; // change deque from {5, 10} to {2, 10}
d[0]; // returns a value of 2

d.back(); // get back (last) element - O(1)
d.front(); // get front (first) element - O(1)
d.clear() // Remove all elements

d.push_back(5); // add an element to the back - O(1)
d.push_front(10); // add an element to the front - O(1)

d.pop_back(); // remove the back (last) element - O(1)
d.pop_front(); // remove the front (first) element - O(1)

d.size(); //Return size
d.empty // Whether queue is empty
A stack and queue combined.
...or a vector that and be pushed and popped from the front as well.

Deque = Double Ended Queue!

Segment Tree

struct node { 
    int start, end, mid, val, lazyadd;
    node left, right;  
    node(int _s, int _e) {  
        //Range of values stored
        start = _s; end = _e; mid = (start+end)/2;  
        //Min value stored
        val = 0;  lazyadd = 0;
        if (start!=end) {
            left = new node(start,mid);
            right = new node(mid+1,end);  
    int value(){
        if (start==end){
            val += lazyadd;lazyadd = 0;return val;
            val += lazyadd;
            // Propagate Lazyadd
            right->lazyadd += lazyadd; 
            left->lazyadd += lazyadd;
            lazyadd = 0;
            return val;
    void addRange(int lower_bound, int upper_bound, int val_to_add){
        if (start == lower_bound && end == upper_bound){
            lazyadd += val_to_add;
            if (lower_bound > mid){
                right->addRange(lower_bound, upper_bound, val_to_add);
            }else if (upper_bound <= mid){
                left->addRange(lower_bound, upper_bound, val_to_add);
                left->addRange(lower_bound, mid, val_to_add);
                right->addRange(mid+1, upper_bound, val_to_add);
            val = min(left->value(), right->value());
    // Update position to new_value // O(log N)
    void update(int pos, int new_val) { //position x to new value
        if (start==end) { val=new_val; return; }  
        if (pos>mid) right->update(pos, new_val);  
        if (pos<=mid) left->update(pos, new_val);
        val = min(left->val, right->val);  
    // Range Minimum Query // O(log N)
    int rangeMinimumQuery(int lower_bound, int upper_bound) {
        //cout<<"Node:"<<start<<" "<<end<<" "<<mid<<" "<<val<<endl;
        //If Query Range Corresponds////////////////
        if (start==lower_bound && end==upper_bound){
            return value();
        //Query Right Tree if range only lies there
        else if (lower_bound > mid){
            return right->rangeMinimumQuery(lower_bound, upper_bound);
        //Query Left Tree if range only lies there
        else if (upper_bound <= mid){
            return left->rangeMinimumQuery(lower_bound, upper_bound);
        //Query both ranges as range spans both trees
            return min(left->rangeMinimumQuery(lower_bound, mid),right->rangeMinimumQuery(mid+1, upper_bound));
} *root;

void init(int N){
    root = new node(0, N-1); // creates seg tree of size n
void update(int P, int V){
int query(int A, int B){
    int val = root->rangeMinimumQuery(A,B);
    return val;


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