With pllk introducing 100 more problems to the CSES problemset, the race for shortest code is on. Here in this blog, I talk about the many different ways of shortening code to get that shortest code.

std::any_of

This is a very useful function that returns true iff any of the elements in the range [first, last) satisfies some condition. Let's say you want to figure out if an array contains at least one 9. You could just write the naive loop and waste time in contest.

bool ok = 0;
for(int i = 0; i < (int)(a).size(); ++i) {
	if(a[i] == 9) {
		ok = 1;
		break;
	}
}

but you could be smart and write this all in one line.

bool ok = any_of(a.begin(), a.end(), [](int x) { return x == 9; });

Time Complexity

This is linear aka $$$O(n)$$$.

std::all_of

This is another useful function that functions similar to std::any_of. The difference is that it returns true iff all the elements in the range [first, last) follow some condition.

Let's say you want to find if the array consists of all 9's.

bool ok = 1;
for(int i = 0; i < (int)(a).size(); ++i) {
	if(a[i] != 9) {
		ok = 0;
	}
}

Now this method saves a couple of lines which means that you'll get the fastest submit time. Guaranteed or I will give you your money back. :D

bool ok = all_of(a.begin(), a.end(), [](int x) { return x == 9; });

Time Complexity

Like the function that I mentioned, this is also $$$O(n)$$$.

std::none_of

This is yet another function that is close relatives of the two mentioned above. This function returns true iff all the elements does not follow some condition.

Let's say you want to find if the array doesn't contain 9.

Noobs will do:

bool ok = 1;
for(int i = 0; i < (int)(a).size(); ++i) {
	if(a[i] == 9) {
		ok = 0;
	}
}

Pros would do:

bool ok = none_of(a.begin(), a.end(), [](int x) { return x == 9; });

This next one is very useful and it appears quite frequently in problems on various judges.

Time Complexity

Linear, $$$O(n)$$$.

std::for_each

Thanks to Ta180m for pointing this out in the comments. This function applies a function hld to all of the elements in the range of [first, last).

Lets say that you want to increase all the elements in some vector by 5.

Naive way:

for(int i = 0; i < (int)(a).size(); ++i) {
	a[i] += 5;
}

A shorter way to write it would be:

auto change = [&](int& x) {
	x += 5;
};
for_each(a.begin(), a.end(), change);

As mentionted by purplesyringa, if you want to iterate from begin ... end it is better to use range-based for loops. std::for_each is only really useful when your iterating from something else other than begin ... end.

Time Complexity

$$$O(n \cdot X)$$$ where $$$X$$$ is the complexity of applying the function hld once.

std::count

This functions counts the number of elements in the range [first, last) that are equal to some variable val.

Noobinho:

int cnt = 0;
for(int i = 0; i < (int)(a).size(); ++i) {
	cnt += (a[i] == x);
}

Proinho:

int cnt = count(a.begin(), a.end(), x);

Time Complexity

$$$O(n)$$$.

std::find

This function returns the first iterator that compares equal to val.

Thanks to _rey for pointing this out to me. :)

NOTE: if using std::find on sets, use set::find instead. This guarantees it to be $$$O(\log n)$$$ where $$$n$$$ is the size of the set.

Nubs:

int hld = -1;
for(int i = 0; i < (int)(a).size(); ++i) {
	if(a[i] == x) {
		hld = i;
		break;
	}
}

Prubs:

int hld = find(a.begin(), a.end(), x) - a.begin();

Time Complexity

Linear, $$$O(n)$$$.

std::accumulate

Many coders use this function but I am still going to include it just in case you don't know the uses of it. This function returns the sum of init and all the elements from [first, last).

Let's say that you want to find the sum of all the elements in the vector. Beginners would do:

int sum = 0;
for(int i = 0; i < (int)(a).size(); ++i) {
	sum += a[i];
}

Now, we can use std::accumulate to do this in 1 line.

Thanks for N8LnR2Nqf8Q4FN for pointing this out to me.

NOTE: Be mindful for overflow when using std::accumulate. If you know that the sum will overflow, use 0LL otherwise use 0. If your not sure using 0LL won't hurt that much anyways.

int sum = accumulate(all(a), 0LL);

Time Complexity

$$$O(n)$$$

std::binary_search

This function is pretty useful in finding if the element val appears in the sorted range [first, last).

Let's say that you want to find if the element val appears in the sorted array. Standard binary search looks something like this.

bool ok = 0;
int l = 0, r = (int)(a).size(), target = val;
while(l < r) {
	int mid = (l + r) / 2;
	if(a[mid] < target) {
		l = mid + 1;
	} else if(a[mid] == target) {
		ok = 1;
		break;
	} else {
		r = mid;
	}
}

bool ok = binary_search(a.begin(), a.end(), val);

Time Complexity

$$$O(\log n)$$$

std::max_element

This function returns the max element in the range [first, last).

int cur = -INF;
for(int i = 0; i < (int)(a).size(); ++i) {
	cur = max(cur, a[i]);
}

Better way to do it is:

int cur = *max_element(a.begin(), a.end());

Time Complexity

$$$O(n)$$$

std::min_element

This is basically the same as std::max_element but returns the min element in the range [first, last).

int cur = INF;
for(int i = 0; i < (int)(a).size(); ++i) {
	cur = min(cur, a[i]);
}

int cur = *min_element(a.begin(), a.end());

Time Complexity

Linear in time, $$$O(n)$$$.

std::is_sorted

This function is for checking if the range [first, last) is sorted. Its pretty self-explanatory.

bool ok = 1;
for(int i = 1; i < (int)(a).begin(); ++i) {
	if(a[i] < a[i - 1]) {
		ok = 0;
	}
}

bool ok = is_sorted(a.begin(), a.end());

Alternatively, if you want to check if the range is sorted but in reverse order, you can first reverse the sequence and then check if it's sorted.

reverse(a.begin(), a.end());
bool ok = is_sorted(a.begin(), a.end());

Time Complexity

$$$O(n)$$$

std::lexicographical_compare

This function is used in many problems that compares two containers and checks which one is lexicographically first.

bool ok = lexigraphical_compare(a.begin(), a.end(), b.begin(), b.end());

Time Complexity

$$$O(n)$$$.

std::partial_sum

Thanks to Kavaliro for pointing this out to me. Imagine doing a prefix sum problem. You want to obviously compute the prefix sums. Naive way to do it would be

for(int i = 1; i < (int)(a).size(); ++i) {
	a[i] += a[i - 1];
}

Now, as Kavaliro pointed out, this could be done in 1 line.

partial_sum(a.begin(), a.end(), a.begin());

Suffix sums are much the same thing

partial_sum(a.rbegin(), a.rend(), a.rbegin());

Time Complexity

Both are linear, $$$O(n)$$$.

std::adjacent_difference

Thanks to elsantodel90 for pointing this out to me. Since I already included the std function for prefix sum, it "only natural to mention adjacent_difference too" — elsantode190. So here it is.

vector<int> hld = a;
for(int i = 1; i < si(hld); ++i) {
	hld[i] -= a[i - 1];
}

This takes 4 lines and much more characters than this:

adjacent_difference(a.begin(), a.end(), a.begin());

Saves a good amount of characters and shorter to type out.

Time Complexity

$$$O(n)$$$

std::iota

Thanks to purplesyringa for pointing this out to me! This function assigns every value from [first, last) successive values of val where val is the argument to the function. A useful scenario is when you have a DSU and you want to initialize it by setting each node's parent to itself. Now this can be done naively

for(int i = 0; i < n; ++i) {
    par[i] = i;
}

or it can be done in a much shorter way

iota(par.begin(), par.end(), 0);

Time Complexity

$$$O(n)$$$

Thats all for now, as I have school work to do. But I'll continue to update this blog with other useful functions. :D