I'm trying to find the index of the last non-zero element in a std::vector<double>. If the last element in the vector is non-zero then it should return the index of that last element.
I believe I can use std::find_if_not, reverse iterators and std::distance, based on this:
std::find_if_not(amounts.rbegin(), amounts.rend(), 0.0)
where amounts is a std::vector<double>, but I'm having difficulty in combining this with std::distance and a forward iterator amounts.begin().
Also, is there a way I can introduce a predicate to compare on, say a tolerance of 1e-8?
I'm using C++11.
Example:
std::vector<double> v{1.32, 1.423, 2.543, 3.534, 4.2, 0};
auto result1 = std::find_if(std::rbegin(v), std::rend(v), [](auto& v) { return std::fabs(v - 0) > std::numeric_limits<double>::epsilon(); } );
if (result1 != std::rend(v)) {
std::cout << *result1 << "\n";
std::cout << std::distance(std::begin(v), (result1 + 1).base());
}
outputs:
4.2
4
[edit]
more explanation on:
std::fabs(v - 0) > std::numeric_limits<double>::epsilon(); }
in OP question there was:
Also, is there a way I can introduce a predicate to compare on, say a tolerance of 1e-8?
so this is such tolerance check, you can replace epsilon use with some other value.
A simple for loop can also do the trick, see live sample: http://ideone.com/dVNOKk
#include <iostream>
#include <vector>
int main() {
std::vector<int> v{1, 2, 3, 4, 1, 2, 3, 0, 4, 1, 2, 3, 4, 0, 0, 0};
for (int i = static_cast<int>(v.size()) - 1; i >= 0; --i) {
if (v.at(i) != 0) {
std::cout << "Last non-zero at: " << i << '\n';
break;
}
}
return 0;
}
Output: Last non-zero at: 12
but I'm having difficulty in combining this with std::distance and a forward iterator amounts.begin()
Reverse iterators have member function base that returns a non-reverse iterator with the relation &*(rit.base() - 1) == &*rit. So, you can use the following:
std::distance(amounts.begin(), found.base()) - 1;
Another option:
amounts.size() - std::distance(amounts.rbegin(), found) - 1
Also, is there a way I can introduce a predicate to compare on, say a tolerance of 1e-8?
Yes. In fact, you must use a predicate, even for the exact comparison, since that's what std::find_if_not expects as the third parameter (instead of value of an element).
Related
I have this problem: Given a vector with n numbers, sort the numbers so that the even ones will be on odd positions and the odd numbers will be on even positions. E.g. If I have the vector 2 6 7 8 9 3 5 1, the output should be 2 7 6 9 8 3 5 1 . The count should start from 1. So on position 1 which is actually index 0 should be an even number, on position 2 which is actually index 1 should be an odd number and so on. Now this is easy if the odd and even numbers are the same, let's say 4 even number and 4 odd numbers in the vector, but what if the number of odd numbers differs from the number of even numbers like in the above example? How do I solve that. I attached the code with one of the tries I did, but it doesn't work. Can I get some help please. I ask you to keep it simple that means only with vectors and such. No weird methods or anything cause I'm a beginner and I only know the basics. Thanks in advance!
I have to mention that n initial is globally declared and is the number of vector elements and v_initial is the initial vector with the elements that need to be rearranged.
The task says to add the remaining numbers to the end of the vector. Like if there are 3 odd and 5 even numbers, The 2 extra even numbers should be thrown at the end of the vector
void vector_pozitii_pare_impare(int v_initial[])
{
int v_pozitie[50],c1=0,c2=1;
for (i = 0; i < n_initial; i++)
{
if (v_initial[i] % 2 == 0)
{
bool isTrue = 1;
for (int k = i + 1; k < n_initial; k++)
{
if (v_initial[k] % 2 != 0)
isTrue = 0;
}
if (isTrue)
{
v_pozitie[c1] = v_initial[i];
c1++;
}
else
{
v_pozitie[c1] = v_initial[i];
c1 += 2;
}
}
else
{
bool isTrue = 1;
for (int j = i + 1; j < n_initial; j++)
{
if (v_initial[j] % 2 == 0)
{
isTrue = 0;
}
if (isTrue)
{
v_pozitie[c2] = v_initial[i];
c2++;
}
else
{
v_pozitie[c2] = v_initial[i];
c2 += 2;
}
}
}
}
This may not be a perfect solution and it just popped out right off my mind without being tested or verified, but it's just to give you an idea.
(Let A,B,C,D be odd numbers and 0,1,2 even numbers correspondingly)
Given:
A 0 B C D 1 2 (random ordered list of odd/even numbers)
Wanted:
A 0 B 1 C 2 D (input sequence altered to match the wanted odd/even criteria)
Next, we invent the steps required to get from given to wanted:
// look at 'A' -> match, next
// Result: A 0 B C D 1 2
// look at '0' -> match, next
// Result: A 0 B C D 1 2
// look at 'B' -> match, next
// Result: A 0 B C D 1 2
// look at 'C' -> mismatch, remember index and find first match starting from index+1
// Result: A 0 B C D ->1<- 2
// now swap the numbers found at the remembered index and the found one.
// Result: A 0 B 1 D C 2
// continue until the whole list has been consumed.
As I said, this algorithm may not be perfect, but my intention is to give you an example on how to solve these kinds of problems. It's not good to always think in code first, especially not with a problem like this. So you should first think about where you start, what you want to achieve and then carefully think of how to get there step by step.
I feel I have to mention that I did not provide an example in real code, because once you got the idea, the execution should be pretty much straight forward.
Oh, and just a small remark: Almost nothing about your code is C++.
A simple solution, that is not very efficient would be to split the vector into 2 vectors, that contain even and uneven numbers and then always take one from the even, one from the uneven and then the remainder, from the one that is not completely entered.
some c++ (that actually uses vectors, but you can use an array the same way, but need to change the pointer arithmetic)
I did not test it, but the principle should be clear; it is not very efficient though
EDIT: The answer below by #AAAAAAAAARGH outlines a better algorithmic idea, that is inplace and more efficient.
void change_vector_even_uneven(std::vector<unsigned>& in_vec){
std::vector<unsigned> even;
std::vector<unsigned> uneven;
for (auto it = in_vec.begin(); it != in_vec.end(); it++){
if ((*it) % 2 == 0)) even.push_back(*it);
else uneven.push_back(*it);
}
auto even_it = even.begin();
auto uneven_it = uneven.begin();
for (auto it = in_vec.begin(); it != in_vec.end(); it++){
if (even_it == even.end()){
(*it) = (*uneven_it);
uneven_it++;
continue;
}
if (uneven_it == uneven.end()){
(*it) = (*even_it);
even_it++;
continue;
}
if ((it - in_vec.begin()) % 2 == 0){
(*it) = (*even_it);
even_it++;
}
else{
(*it) = (*uneven_it);
uneven_it++;
}
}
}
The solutions is simple. We sort the even and odd values into a data structure. In a loop, we iterate over all source values. If they are even (val & 2 == 0) we add them at the end of a std::deque for evens and if odd, we add them to a std::deque for odds.
Later, we we will extract the the values from the front of the std::deque.
So, we have a first in first out principle.
The std::deque is optimized for such purposes.
Later, we make a loop with an alternating branch in it. We, alternatively extract data from the even queue and then from the odd queue. If a queue is empty, we do not extract data.
We do not need an additional std::vector and can reuse the old one.
With that, we do not need to take care for the same number of evens and odds. It will of course always work.
Please see below one of millions of possible solutions:
#include <iostream>
#include <vector>
#include <deque>
int main() {
std::vector testData{ 2, 6, 7, 8, 9, 3, 5, 1 };
// Show initial data
std::cout << "\nInitial data: ";
for (const int i : testData) std::cout << i << ' ';
std::cout << '\n';
// We will use a deques to store odd and even numbers
// With that we can efficiently push back and pop front
std::deque<int> evenNumbers{};
std::deque<int> oddNumbers{};
// Sort the original data into the specific container
for (const int number : testData)
if (number % 2 == 0)
evenNumbers.push_back(number);
else
oddNumbers.push_back(number);
// Take alternating the data from the even and the odd values
bool takeEven{ true };
for (size_t i{}; !evenNumbers.empty() && !oddNumbers.empty(); ) {
if (takeEven) { // Take even numbers
if (not evenNumbers.empty()) { // As long as there are even values
testData[i] = evenNumbers.front(); // Get the value from the front
evenNumbers.pop_front(); // Remove first value
++i;
}
}
else { // Now we take odd numbers
if (not oddNumbers.empty()) { // As long as there are odd values
testData[i] = oddNumbers.front(); // Get the value from the front
oddNumbers.pop_front(); // Remove first value
++i;
}
}
// Next take the other container
takeEven = not takeEven;
}
// Show result
std::cout << "\nResult: ";
for (const int i : testData) std::cout << i << ' ';
std::cout << '\n';
return 0;
}
Here is yet another solution (using STL), in case you want a stable result (that is, the order of your values is preserved).
#include <algorithm>
#include <vector>
auto ints = std::vector<int>{ 2, 6, 7, 8, 9, 3, 5, 1 };
// split list to even/odd sections -> [2, 6, 8, 7, 9, 3, 5, 1]
const auto it = std::stable_partition(
ints.begin(), ints.end(), [](auto value) { return value % 2 == 0; });
auto results = std::vector<int>{};
results.reserve(ints.size());
// merge both parts with equal size
auto a = ints.begin(), b = it;
while (a != it && b != ints.end()) {
results.push_back(*a++);
results.push_back(*b++);
}
// copy remaining values to end of list
std::copy(a, it, std::back_inserter(results));
std::copy(b, ints.end(), std::back_inserter(results));
The result ist [2, 7, 6, 9, 8, 3, 5, 1]. The complexity is O(n).
This answer, like some of the others, divides the data and then reassembles the result. The standard library std::partition_copy is used to separate the even and odd numbers into two containers. Then the interleave function assembles the result by alternately copying from two input ranges.
#include <algorithm>
#include <iostream>
#include <vector>
template <typename InIt1, typename InIt2, typename OutIt>
OutIt interleave(InIt1 first1, InIt1 last1, InIt2 first2, InIt2 last2, OutIt dest)
{
for (;;) {
if (first1 == last1) {
return std::copy(first2, last2, dest);
}
*dest++ = *first1++;
if (first2 == last2) {
return std::copy(first1, last1, dest);
}
*dest++ = *first2++;
}
}
void reorder_even_odd(std::vector<int> &data)
{
auto is_even = [](int value) { return (value & 1) == 0; };
// split
std::vector<int> even, odd;
std::partition_copy(begin(data), end(data), back_inserter(even), back_inserter(odd), is_even);
// merge
interleave(begin(even), end(even), begin(odd), end(odd), begin(data));
}
int main()
{
std::vector<int> data{ 2, 6, 7, 8, 9, 3, 5, 1 };
reorder_even_odd(data);
for (int value : data) {
std::cout << value << ' ';
}
std::cout << '\n';
}
Demo on Compiler Explorer
As suggested, I am using vectors and STL.
No need to be a great mathematician to understand v_pozitie will start with pairs of odd and even and terminate with the integers not in the initial pairs.
I am then updating three iterators in v_positie (no need of temporary containers to calculate the result) : even, odd and end,(avoiding push_back) and would code this way :
#include <vector>
#include <algorithm>
void vector_pozitii_pare_impare(std::vector<int>& v_initial, std::vector<int>& v_pozitie) {
int nodd (0), neven (0);
std::for_each (v_initial.begin (), v_initial.end (), [&nodd] (const int& n) {
nodd += n%2;
});
neven = v_initial.size () - nodd;
int npair (neven < nodd ?neven:nodd);
npair *=2;
std::vector<int>::iterator iend (&v_pozitie [npair]), ieven (v_pozitie.begin ()), iodd (&v_pozitie [1]);
std::for_each (v_initial.begin (), v_initial.end (), [&iend, &ieven, &iodd, &npair] (const int& s) {
if (npair) {
switch (s%2) {
case 0 :
*ieven++ = s;
++ieven;
break;
case 1 :
*iodd++ = s;
++iodd;
break;
}
--npair;
}
else *iend++ = s;
});
}
int main (int argc, char* argv []) {
const int N = 8;
int tab [N] = {2, 6, 7, 8, 9, 3, 5, 1};
std::vector<int> v_initial (tab, (int*)&tab [N]);
std::cout << "\tv_initial == ";
std::for_each (v_initial.begin (), v_initial.end (), [] (const int& s) {std::cout << s << " ";});
std::cout << std::endl;
std::vector<int> v_pozitie (v_initial.size (), -1);
vector_pozitii_pare_impare (v_initial, v_pozitie);
std::cout << "\tv_pozitie == ";
std::for_each (v_pozitie.begin (), v_pozitie.end (), [] (const int& s) {std::cout << s << " ";});
std::cout << std::endl;
}
I'm having some trouble understanding how to use reverse iterators with the std::find() function. I believe that if I could see an example that completed the following task, I would be able to understand it perfectly.
So, suppose I have a std::vector I want to search through; however, I do not want to search the typical way. I want to find the first occurrence of a value starting at a certain index and heading towards the start of the vector. To illustrate:
3 | 4 | 7| 4| 2| 6| 3|
^ ^
|<------------|
start_point
Search: find first occurrence, given the above search layout, of 4
Expected Result: index 3
I'm rather sure that one would have to work with reverse iterators in this situation, but I can't figure out how to do it.
If you're using a std::vector, or any other container that provides Random Access Iterators, you can advance an iterator just using arithmetic, like you would with a pointer. Your example vector has 7 elements, and you want to start at index 4, so you could get a normal iterator to that element just with:
auto i = v.begin() + 4;
For a reverse iterator, you're starting from the back of the vector, not the front, so to get the right offset you have to subtract the desired index+1 from the size, like so:
auto i = v.rbegin() + (v.size() - 5);
This'll be, in your example, 2, so the reverse iterator will start pointing to the last element, then move two spaces toward the beginning, reaching your desired start point.
Then, you can use std::find in the normal way:
auto found = std::find(v.rbegin() + (v.size() - 5), v.rend(), 4);
if(found == v.rend()) {
std::cout << "No element found." << std::endl;
} else {
std::cout << "Index " << (v.rend() - found) << std::endl;
}
Remember that, when testing the result of std::find to see if it found anything, you need to use rend(), not end(). When you compare reverse iterators to normal iterators, you're comparing the actual positions, not the offsets from the start, so v.rend() != v.end().
If you don't have Random Access Iterators (for example, in a std::list) you can't use pointer-style arithmetic, so you can instead use std::advance to advance iterators to a specific position and std::distance to get the distance between two iterators.
First you set the start position:
auto it = v.rbegin() + 2; // two from the end
Then search:
auto kt = std::find(it, v.rend(), 4);
If kt == v.rend(), no element is found; otherwise we can compute the index from the front with a simple distance computation:
if (kt == v.rend()) {
std::cerr << "Element 4 not found.\n";
std::abort();
} else {
auto n = std::distance(kt, v.rend()) - 1;
std::cout << "Element 4 found at position v[" << n << "].\n";
}
Try something like the following
#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
int main()
{
std::vector<int> v = { 3, 4, 7, 4, 2, 6, 3 };
std::vector<int>::size_type pos = 4;
auto it = std::find(std::next(v.rbegin(), v.size() - pos), v.rend(), 4);
if (it != v.rend())
{
std::cout << std::distance(v.begin(), it.base() - 1) << std::endl;
}
}
I'm trying to figure out the following problem.
Suppose I have the following container in C++:
std::set<std::pair<int, int> > my_container;
This set (dictionary) is sorted with respect to the order < on std::pair<int, int>, which is the lexicographic order. My task is to find any element in my_container that has the first coordinate equal to, say x, and return the iterator to it. Obviously, I don't want to use find_if, because I need to solve this in logarithmic time.
I would appreciate any advice on how this can be done
You can use lower_bound for this:
auto it = my_container.lower_bound(std::make_pair(x, std::numeric_limits<int>::min());
This will give you an iterator to the first element e for which e < std::pair(x, -LIMIT) does not hold.
Such an element either has its first component > x (in which case there's no x in the set), or has the first component equal to x and is the first such. (Note that all second components are greater than or equal to std::numeric_limits<int>::min() by definition).
You could use std::set::lower_bound to get the lower and upper limits of the range like this:
#include <set>
#include <iostream>
// for readability
typedef std::set<std::pair<int, int> > int_set;
void print_results(const int_set& s, int i)
{
// first element not less than {i, 0}
int_set::const_iterator lower = s.lower_bound(std::make_pair(i, 0));
// first element not less than {i + 1, 0}
int_set::const_iterator upper = s.lower_bound(std::make_pair(i + 1, 0));
for(int_set::const_iterator iter = lower; iter != upper; ++iter)
std::cout << iter->first << ", " << iter->second << '\n';
}
int main()
{
int_set s;
s.insert(std::make_pair(2, 0));
s.insert(std::make_pair(1, 9));
s.insert(std::make_pair(2, 1));
s.insert(std::make_pair(3, 0));
s.insert(std::make_pair(7, 6));
s.insert(std::make_pair(5, 5));
s.insert(std::make_pair(2, 2));
s.insert(std::make_pair(4, 3));
print_results(s, 2);
}
Output:
2, 0
2, 1
2, 2
This question has been asked before but I cannot find it for C++.
If I have a vector and I have a starting number, does std::algorithm provide me a way to find the next highest missing number?
I can obviously write this in a nested loop, I just cant shake the feeling that I'm reinventing the wheel.
For example, given: vector foo{13,8,3,6,10,1,7,0};
The starting number 0 should find 2.
The starting number 6 should find 9.
The starting number -2 should find -1.
EDIT:
Thus far all the solutions require sorting. This may in fact be required, but a temporary sorted vector would have to be created to accommodate this, as foo must remain unchanged.
At least as far as I know, there's no standard algorithm that directly implements exactly what you're asking for.
If you wanted to do it with something like O(N log N) complexity, you could start by sorting the input. Then use std::upper_bound to find the (last instance of) the number you've asked for (if present). From there, you'd find a number that differs from the previous by more than one. From there you'd scan for a difference greater than 1 between the consecutive numbers in the collection.
One way to do this in real code would be something like this:
#include <iostream>
#include <algorithm>
#include <vector>
#include <numeric>
#include <iterator>
int find_missing(std::vector<int> x, int number) {
std::sort(x.begin(), x.end());
auto pos = std::upper_bound(x.begin(), x.end(), number);
if (*pos - number > 1)
return number + 1;
else {
std::vector<int> diffs;
std::adjacent_difference(pos, x.end(), std::back_inserter(diffs));
auto pos2 = std::find_if(diffs.begin() + 1, diffs.end(), [](int x) { return x > 1; });
return *(pos + (pos2 - diffs.begin() - 1)) + 1;
}
}
int main() {
std::vector<int> x{ 13, 8, 3, 6, 10, 1,7, 0};
std::cout << find_missing(x, 0) << "\n";
std::cout << find_missing(x, 6) << "\n";
}
This is somewhat less than what you'd normally think of as optimal to provide the external appearance of a vector that can/does remain un-sorted (and unmodified in any way). I've done that by creating a copy of the vector, and sorting the copy inside the find_missing function. Thus, the original vector remains unmodified. The disadvantage is obvious: if the vector is large, copying it can/will be expensive. Furthermore, this ends up sorting the vector for every query instead of sorting once, then carrying out as many queries as desired on it.
So I thought I'd post an answer. I don't know anything in std::algorithm that accomplishes this directly, but in combination with vector<bool> you can do this in O(2N).
template <typename T>
T find_missing(const vector<T>& v, T elem){
vector<bool> range(v.size());
elem++;
for_each(v.begin(), v.end(), [&](const T& i){if((i >= elem && i - elem < range.size())range[i - elem] = true;});
auto result = distance(range.begin(), find(range.begin(), range.end(), false));
return result + elem;
}
First you need to sort the vector. Use std::sort for that.
std::lower_bound finds the first element that is greater or equal with a given element. (the elements have to be at least partially ordered)
From there you iterate while you have consecutive elements.
Dealing with duplicates: One way is the way I went: consider consecutive and equal elements when iterating. Another approach is to add a prerequisite that the vector / range contains unique elements. I chose the former because it avoids erasing elements.
Here is how you eliminate duplicates from a sorted vector:
v.erase(std::unique(v.begin(), v.end()), v.end());
My implementation:
// finds the first missing element in the vector v
// prerequisite: v must be sorted
auto firstMissing(std::vector<int> const &v, int elem) -> int {
auto low = std::lower_bound(std::begin(v), std::end(v), elem);
if (low == std::end(v) || *low != elem) {
return elem;
}
while (low + 1 != std::end(v) &&
(*low == *(low + 1) || *low + 1 == *(low + 1))) {
++low;
}
return *low + 1;
}
And a generalized version:
// finds the first missing element in the range [first, last)
// prerequisite: the range must be sorted
template <class It, class T = decltype(*std::declval<It>())>
auto firstMissing(It first, It last, T elem) -> T {
auto low = std::lower_bound(first, last, elem);
if (low == last || *low != elem) {
return elem;
}
while (std::next(low) != last &&
(*low == *std::next(low) || *low + 1 == *std::next(low))) {
std::advance(low, 1);
}
return *low + 1;
}
Test case:
int main() {
auto v = std::vector<int>{13, 8, 3, 6, 10, 1, 7, 7, 7, 0};
std::sort(v.begin(), v.end());
for (auto n : {-2, 0, 5, 6, 20}) {
cout << n << ": " << firstMissing(v, n) << endl;
}
return 0;
}
Result:
-2: -2
0: 2
5: 5
6: 9
20: 20
A note about sorting: From the OP's comments he was searching for a solution that wouldn't modify the vector.
You have to sort the vector for an efficient solution. If modifying the vector is not an option you could create a copy and work on it.
If you are hell-bent on not sorting, there is a brute force solution (very very inefficient - O(n^2)):
auto max = std::max_element(std::begin(v), std::end(v));
if (elem > *max) {
return elem;
}
auto i = elem;
while (std::find(std::begin(v), std::end(v), i) != std::end(v)) {
++i;
}
return i;
First solution:
Sort the vector. Find the starting number and see what number is next.
This will take O(NlogN) where N is the size of vector.
Second solution:
If the range of numbers is small e.g. (0,M) you can create boolean vector of size M. For each number of initial vector make the boolean of that index true. Later you can see next missing number by checking the boolean vector. This will take O(N) time and O(M) auxiliary memory.
If I have two seperate sorted arrays, containing equal number of entries, and I need to find the number of pairs(both numbers should be from seperate arrays) having sum = 0 in linear time, how can I do that?
I can easily do it in O(n^2) but how to do it in linear time?
OR should I merge the two arrays and then proceed?
Thanks!
You don't need the arrays to be sorted.
Stick the numbers from one of the arrays into a hash table. Then iterate over the other array. For each number n, see if -n is in the hash table.
(If either array can contain duplicates, you need to take some care around handling them.)
P.S. You can exploit the fact that the arrays are sorted. Just iterate over them from the opposite ends once, looking for items that have the same value but the opposite signs. I leave figuring out the details as an exercise (hint: think of the merge step of merge sort).
Try this:
for(i=0;j=0;i<n&&j<n;)
{
if(arr1[i]+arr2[j]==0)
{
count++;
i++;
j++;
}
else if(arr[i]>arr[j])
{
j++;
}
else
{
i++;
}
}
Following may help:
std::size_t count_zero_pair(const std::vector<int>& v1, const std::vector<int>& v2)
{
assert(is_sorted(v1.begin(), v1.end()));
assert(is_sorted(v2.begin(), v2.end()));
std::size_t res = 0;
auto it1 = v1.begin();
auto it2 = v2.rbegin();
while (it1 != v1.end() && it2 != v2.rend()) {
const int sum = *it1 + *it2;
if (sum < 0) {
++it1;
} else if (0 < sum) {
++it2;
} else { // sum == 0
// may be more complicated depending
// how you want to manage duplicated pairs
++it1;
++it2;
++res;
}
}
return res;
}
If they are already sorted, you can traverse them, one frome left to right, one from right to left:
Take two pointers, and put one at the very left of one array, the other at the very right of the other array. Look at both values you currently point on. If the absolute value of one of these values is greater than the other, advance the greater one. If the absolute values are equal, report both values, and advance both pointers. Stop, as soon as the pointer coming from the left reaches a positive value, or the pointer from the right reaches a negative value. After that, do the same with the pointers starting at the resp. other ends of the arrays.
This is essentially the solution proposed by #Matthias with an added pointer to catch duplicates. If there is a string of duplicate values in arr2, searchStart will always point to the one with the highest index so that we can check the entire string against the next value in arr1. All values in arr1 are explicitly checked, so no extra duplicate handling is required.
int pairCount = 0;
for (int base=0, searchStart=arr2Size-1; base<arr1Size; base++) {
int searchCurrent = searchStart;
while (arr1[base]+arr2[searchCurrent] > 0) {
searchCurrent--;
if (searchCurrent < 0) break;
}
searchStart=searchCurrent;
if (searchStart < 0) break;
while (arr1[base]+arr2[searchCurrent] == 0) {
std::cout << "arr1[" << base << "] + arr2[" << searchCurrent << "] = ";
std::cout << "[" << arr1[base] << "," << arr2[searchCurrent] << "]\n";
pairCount++;
searchCurrent--;
}
}
std::cout << "pairCount = " << pairCount << "\n";
Given the arrays:
arr1[] = {-5, -3, -3, -2, -1, 0, 2, 4, 4, 5, 8};
arr2[] = {-7, -5, -5, -4, -3, -2, 1, 3, 4, 5, 6, 7, 8};
we get:
arr1[0] + arr2[9] = [-5,5]
arr1[1] + arr2[7] = [-3,3]
arr1[2] + arr2[7] = [-3,3]
arr1[4] + arr2[6] = [-1,1]
arr1[6] + arr2[5] = [2,-2]
arr1[7] + arr2[3] = [4,-4]
arr1[8] + arr2[3] = [4,-4]
arr1[9] + arr2[2] = [5,-5]
arr1[9] + arr2[1] = [5,-5]
pairCount = 9
Now we come to the question of time complexity. The construction of searchStart is such that for each value in arr1 can have an extra compare with one value in arr2 (but no more than 1). Otherwise, for arrays with no duplicates this checks each value in arr2 exactly once, so this algorithm runs in O(n).
If duplicate values are present, however, it complicates things a bit. Consider the arrays:
arr1 = {-3, -3, -3}
arr2 = { 3, 3, 3}
Clearly, since all O(n²) pairs equal zero, we have to count all O(n²) pairs. This means that in the worst case, the algorithm is O(n²) and this is the best we can do. It is possibly more constructive to say that the complexity is O(n + p) where p is the number of matching pairs.
Note that if you only want to count the number of matches rather than printing them all, you can do this in linear time as well. Just change when searchStart is updated to when the last match is found and keep a counter that equals the number of matches found for the current searchStart. Then if the next arr1[base] matches arr2[searchStart], add the counter to the number of pairs.