reverse in O(log n) time

The basic framework for divide and conquer algorithm can be used for any operations like reverse, checking if array is sorted etc. The possible implementation of reversing a string/vector is given as follows which takes O(n) time where n is size of collection string/vector.

template<class BidirIt>
void reverse(BidirIt first, BidirIt last)
{
 while ((first != last) && (first != --last)) {
 std::iter_swap(first++, last);
 }
}

Better algorithm for this can be given as follows:

#include <iostream>
#include <algorithm>
#include <string>

template <typename T>
void reverse_rotate(T first, T last){
 if((first == last) || std::next(first) == last) return;
 T middle = first;
 std::advance(middle, std::distance(first, last)/2);
 reverse_rotate(first, middle);
 reverse_rotate(middle, last);
 std::rotate(first, middle, last);
}

int main(){
 std::string s = "This is super cool";
 reverse_rotate(s.begin(), s.end());
 std::cout << s << '\n';
}

is_sorted in O(log n) time

Checking if array is sorted is simple, compare consecutive elements to check if next element is greater than the previous element. This takes O(n) time. The possible implementation of std::is_sorted is given as follows:

template <class ForwardIterator>
 bool is_sorted (ForwardIterator first, ForwardIterator last)
{
 if (first==last) return true;
 ForwardIterator next = first;
 while (++next!=last) {
 if (*next<*first) // or, if (comp(*next,*first)) for version (2)
 return false;
 ++first;
 }
 return true;
}

Here I came with an algorithm which is like quick sort and uses O(log n) time. Here is the implementation.

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>

template <typename T>
bool check_sorted(T start, T finish, bool sorted = true){
 if(start == finish) return true;

if(std::next(start) == finish) return *start <= *finish;

T middle = start;
 std::advance(middle, std::distance(start, finish)/2);

sorted &= check_sorted(start, middle);
 sorted &= check_sorted(middle, finish-1);
 return sorted;
}

int main(){
 std::vector<int> v(20);
 std::iota(v.begin(), v.end(), 1);
 bool sorted = check_sorted(v.begin(), v.end());
 std::cout << "v is sorted? : " << sorted << '\n';

std::vector<int> v2 {20, 1, 3, 5, 2, 1, 4, 19, 18, 10};
 sorted = check_sorted(v2.begin(), v2.end());
 std::cout << "v2 is sorted?: " << sorted << '\n';

std::vector<int> v3 {1,1,1,1,1,1,1,1,1,1,1,1,1,1};
 sorted = check_sorted(v3.begin(), v3.end());
 std::cout << "v3 is sorted?: " << sorted << '\n';

std::vector<int> v4 {10,1,1,1,1,1,1,1,1,1,1,1,1,1};
 sorted = check_sorted(v4.begin(), v4.end());
 std::cout << "v4 is sorted?: " << sorted << '\n';
}

 

Remove dups in sorted vector

rawpixel-600789-unsplash

To remove duplicates of an array, use a hash map and array elements as keys, then extract keys. This requires extra space for storing the hash and time. Following algorithm does not need any extra space and time.

The algorithm is uses 3 iterators and moves the duplicate elements at the start of vector and erases the rest of vector. Pseudo code given as follows:

  1. take iterator ​`​start` that points to beginning of vector.
  2. take another iterator `i` that points to next element to start.
  3. move  i to next index until elements are duplicate.
  4. replace start with next element and increase start pointer.
  5. at the end erase elements from start+1 to end from vector.
#include <iostream>  // std::cout 
#include <vector>   
#include <cstdlib>  // std::srand
#include <ctime>    // std::time

int main(){
 std::vector<int> v(100);
 std::srand(std::time(nullptr)); // use current time as seed for random generator
 
// generate 1000 random numbers between 0 to 99
 for(int i = 0; i < 1000; i++){
   int random_variable = std::rand()%100;
   v.push_back(random_variable);
   std::cout << random_variable << ' ';
 }

// sort
std::sort(v.begin(), v.end());


std::vector<int>::iterator start = v.begin();
std::vector<int>::iterator end = v.end();
auto i = start+1; // std::vector<int>::iterator i = start+1;
 while(i != end-1){
   if(*start == *i){
     i++;
   }else{
     start++;
     *start = *i;
   }
 }

v.erase(start+1, end);
 for(const auto & i:v) std::cout << i << ' ';
}

Quick Sort using iterators

Lately I have been experimenting various algorithms using std iterators and tried to implement quick sort algorithm using just iterators instead of passing on the vector by reference. Following is the implementation of the algorithm.

#include <vector>
#include <iostream>
#include <algorithm>

auto partition(std::vector<int>::iterator start, std::vector<int>::iterator end_it){
   auto partition_idx = start;
   for(auto i = start; i != end_it; i++){
     if(*i <= *end_it){
       std::iter_swap(i, partition_idx);
       partition_idx++;
     }
   }
   std::iter_swap(partition_idx, end_it);
   return partition_idx;
}

void quick_sort(std::vector<int>::iterator start, std::vector<int>::iterator end_it){
  int size = std::distance(start, end_it);
  if (size <= 1) return;
  auto partition_idx = partition(start, end_it);
  quick_sort(start, partition_idx-1);
  quick_sort(partition_idx, end_it);
}

int main(){
  std::vector<int> v{72, 23, 65, 53, 36, 90, 76, 50, 1, 20, 64, 18, 26, 39, 34, 91, 86, 21, 81, 7, 58, 27, 9, 61, 77, 2, 88, 70, 79, 3, 100, 44, 94, 51, 83, 48, 68, 13, 96, 30, 98, 24, 95, 67, 55, 8, 56, 19, 52, 85, 6, 69, 43, 16, 93, 74, 60, 41, 37, 84, 66, 99, 78, 57, 33, 14, 62, 35, 31, 46, 59, 38, 12, 92, 73, 25, 32, 22, 28, 63, 54, 97, 29, 47, 15, 40, 87, 80, 17, 71, 11, 4, 75, 89, 49, 10, 5, 45, 42, 82};
  quick_sort(v.begin(), v.end()-1); // one element before end
  for(const auto & i:v){
    std::cout << i << ' ';
  }
 return 0;
}

Count ones in binary representation

The task is to find ones in the binary representation of a number in given range. For an example: 1 is represented a 1 in binary so the number is 1’s in 1 is 1. Another example: 4 is represented as 100 in binary so number of 1’s in binary representation of 4 is 1.

In order to find 1’s in a very large range between 1 and 10,000,000 sounds complicated but can solved with following algorithm

#include <iostream>
#include <algorithm>
#include <cmath>

long long ccb(int col, int n) {
  double division = (n + 1) / double(2 << col);
  int sum = static_cast<int>(division);
  long long rest = std::max(0.0, (division - sum) * (2 << col) - (1 << col));
  return sum * (1 << col) + rest;
}

long long cab(int n) {
  long long bits = 0;
  for (int i = 0; i < 30; i++) {
  bits += ccb(i, n);
 }
 return bits;
}

long long countOnes(int left, int right) {
 return cab(right) - cab(left - 1);
}

Multiply large numbers as strings

This algorithm is very much like factorial. Multiplication of very large numbers represented as strings because they go out of the integer range.

#include <iostream>
#include <vector>
#include <string>
#include <sstream>

std::string multiply(std::string a, std::string b) {
  // take a vector of length 200 and initialize all of its elements to 0.
  std::vector<int> multiplication(200, 0); 

 int place = 0;
 for(int i = a.size() - 1 ; i >= 0; i--){
   int carry = 0;
   std::vector<int>::iterator it = multiplication.begin() + place;
   for(int j = b.size() - 1 ; j >= 0; j--){
     int a_element = a[i] - '0';
     int b_element = b[j] - '0';
     int prod = a_element*b_element + carry + *it;
     *it = prod % 10;
     carry = prod / 10;
     it++;
     while (carry){
       int current_carry = carry + *it;
       *it += current_carry % 10;
       it++;
       carry = current_carry / 10;
    }
  }
   place++;
 }

std::stringstream ss;
for(int i = multiplication.size(); i >= 0; i--){
 ss << multiplication[i];
}
 return ss.str();
}

Reverse a number

Algorithm for reversing the number is as follows:

#include <cmath> // for finding out the length of number
 
// find the length of the number to find out the highest power of 10
// then divide the number by 10 and take remainder to find digit 
// at the unit's place.
// multiply the digit with power of 10. repeat length times.
int reverse(int x) {
  int length = (int)log10(abs(x)); // magic formula to find length
  int reversed = 0;
  int remainder = 0;
  for(int i = length; i > 0 ; --i){
    remainder = x % 10;
    x = x / 10;
    reversed += remainder * pow(10, i);
  }
  reversed += x;
  return reversed;
 }