Is there a way to increment the iterator if an 'if' condition is met
dereference iterator c++
random access iterator c++
map iterator c
forward iterator c++
list iterator c++
I'm solving this HackerRank challenge:
Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn't contain the substring '010'.
In one step, Alice can change a
1or vice versa. Count and print the minimum number of steps needed to make Alice see the string as beautiful.
So basically count the number of '010' occurrences in the string 'b' passed to the function.
I want to increment
i by 2 once the
if statement is true so that I don't include overlapping
'010' strings in my
And I do realize that I can just use the count method but I wanna know why my code isn't working the way I want to it to.
def beautifulBinaryString(b): count = 0 for i in range(len(b)-2): if b[i:i+3]=='010': count+=1 i+=2 return count
Output I get w/ this code:
If you want to do it using a for loop, you can add a delta variable to keep track of the number of positions that you have to jump over the current i value.
def beautifulBinaryString(b): count = 0 delta = 0 for i in range(len(b)-2): try: if b[i+delta:i+delta+3]=='010': count+=1 delta=delta+2 except IndexError: break return count
C++ Iterators, Iterator: a pointer-like object that can be incremented with ++, dereferenced with * , and compared Each STL container defines what class of iterators it can return. Note how much simpler this is than the equivalent for loop with cout and <<. But it does them in a special way. One example of a smart iterator is the transform iterator, that does not just give access to an element of a range with its operator*. Instead, it gives the result of applying a function f to the element of the range. Another example is the filter iterator.
You are counting overlapping sequences. For your input
0101010 you find
010 three times, but the middle
010 overlaps with the outer two
0101010 --- --- ---
You can't increment
i in a
for loop, because the
for loop construct sets
i at the top. Giving
i a different value inside the loop body doesn't change this.
Don't use a
for loop; you could use a
def beautifulBinaryString(b): count = 0 i = 0 while i < len(b) - 2: if b[i:i+3]=='010': count += 1 i += 2 i += 1 return count
A simpler solution is to just use
b.count("010"), as you stated.
Ways to increment Iterator from inside the For loop in Python , Definite iterations mean the number of repetitions is specified explicitly in advance. But have you ever wondered, what happens, if you try to increment the value of Increment an iterator standard map only bidirectional iterators, so there's no + n operation and what the most efficient way to compute the next iterator is
You don't need to count the occurrences; as soon as you find one occurrence, the string is "ugly". If you never find one, it's beautiful.
def is_beautiful(b): for i in range(len(b) - 2): if b[i:i+3] == '010': return False return True
You can also avoid the slicing by simply keeping track of whether you've started to see
seen_0 = False seen_01 = False for c in b: if seen_01 and c == '0': return False elif seen_1 and c == '1': seen_01 = True elif c == '0': seen_0 = True else: # c == 1, but it doesn't follow a 0 seen_0 = False seen_01 = False return True
Iterators in C++ STL, Iterators are used to point at the memory addresses of STL containers. 3. advance() :- This function is used to increment the iterator position till the specified It accepts 2 arguments, the container and iterator to position where the elements and its types in C/C++ · How to find index of a given element in a Vector in C++ An iterator is any object that, pointing to some element in a range of elements (such as an array or a container ), has the ability to iterate through the elements of that range using a set of operators (with at least the increment ( ++) and dereference ( *) operators). The most obvious form of iterator is a pointer: A pointer can point to elements in an array, and can iterate through them using the increment operator ( ++ ).
<iterator> - C++ Reference, Increments given iterator it by n elements. If n is negative, the iterator is decremented. In this case, InputIt must meet the requirements of You can also use an increment or decrement operator in an assignment statement: int a = 5; int b = a--; // both a and b are set to 4. Increment and decrement operators can be placed before (prefix) or after (postfix) the variable they apply to. If you place an increment or decrement operator before its variable, the operator is applied before the rest of the expression is evaluated.
std::advance, Return the nth successor of iterator it. LegacyBidirectionalIterator requirement that specifies that increment of an rvalue is guaranteed to work. increment() moves the iterator forward one Not only is it simpler to implement an iterator, there are far fewer places There is no way to know the return
std::next, But it does them in a special way. One example of a smart iterator is the transform iterator, that does not just give access to an element of a Add a nested iterator class to your linked list to sequentially access list elements. Your iterator supports begin () and end () functions along with overloaded not equal ("!="), dereferencing ("*"), and pre-incrementing ("++") operators. Use the list iterator to iterate, find, insert after, replace, and erase elements in the linked list."
- Use a
forloop, the looping variable does not affect even if it's value is changed within.
forin python is different from
forin C/C++. It is in iterator over an
iterable(a collection). So you can't skip an element.
- No, the actual challenge requires a count. Even if a True or False answer sufficed, then
return '010' in bwould have been much better.
- Not that the OP explained the problem very well, they only quoted a small part of the challenge explanation.
- Thank you guys so much, the while loop and using a delta variable answers really helped. Appreciate the quick response and thanks again.