Is there a way to increment the iterator if an 'if' condition is met

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std::advance
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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 0 to a 1 or 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 count.

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

Input: 0101010

Expected Output: 2

Output I get w/ this code: 3

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 010 sequences:

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 while loop:

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 010:

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."

Comments
  • Use a while loop. In for loop, the looping variable does not affect even if it's value is changed within.
  • for in python is different from for in 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 b would 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.