c++,search,stl,binary,binary-search

I'm quite certain the standard library doesn't include anything to do precisely what you're asking for. To get what you want, you'll probably want to start from std::lower_bound or std::upper_bound, and convert the iterator it returns into an index, then complement the index of the value wasn't found....

You've got different results because your variation works slightly faster in some cases. Binary search depends on how you split the data (i.e. if you've got 7 elements the first group may be 3 or 4 elements and here you may be more or less lucky). However, you start with...

The Mistake is right here: if((first-last)<2): #This always will be less than 2 Should be: if((last-first)<2): ...

This is probably the easiest way 1) read the file line by line and find the first name in your sorting method e.g. -read name_1. -read next name_2. If name_1 < name_2 then name_2 = name_1 and repeat. 2) read the file line by line again and find the second...

c#,math,binary-search,biginteger,newtons-method

Newton's method is N(x) = x - (x^n-a)/(n*x^(n-1))=( (n-1)*x + a/(x^(n-1)))/n This should work well also with integer operations. You may have to correct the result by plusminus 1. It is important to use a good initial value since far away from the root the convergence of Newtons method for...

TreeMap has a constructor that takes a Comparator as an argument. The comparator is used to order keys stored in the map. If you literally want to "count the number of comparisons carried out", you could write an instrumented comparator that counts the number of times it's called. public class...

I don't understand why we can use binary search. In order to use binary search the elements have to be sorted. In order words: for a given element e, we can't have any element less than e at its right side. But that is not the case in this...

The second definition of std::lower_bound is: template< class ForwardIt, class T, class Compare > ForwardIt lower_bound( ForwardIt first, ForwardIt last, const T& value, Compare comp ); So, you can just write your own function of comparing pairs and use it. Your template function will look like: template<class ForwardIt, class T,...

c#,list,readonly,binary-search

In this case, depends on personal preference. This is mine, as an example: private List<string> _MyList = new List<string>(); private void InitializeList() { //code here to fill list. //Keep in mind, that binary search works on sorted lists. _MyList.Sort(/* Place your object comparer here. */); } //Make a copy in...

I believe it always tries to find the middle by (0+n)/2 = (0+9)/2 = 4(Integer) In your case. So in case you want to search 25 itself, as per the algorithm you will find in the lower bound group, position 4 first as a match....

python,sorting,date,binary-search

You can use bisect.bisect_left() or bisect.bisect_right() (aliased to bisect.bisect()) to find that insertion point. It'll find that point in at most log N steps, using a binary search. The difference lies in what happens when you give it a date() that is in the list itself; bisect_left() will give the...

While I did not find a way to do a better type of search, I did manage to learn about embedded resources which considerably sped up the program. Scanning the entire file takes a fraction of a second now, instead of 5-10 seconds. Posting the following code: string searchfor =...

python,recursion,binary-search

These two lines don't do what you think: max = --midPointOfList min = ++midPointOfList Python doesn't have this type of increment operators, but they do parse and execute successfully. ++i parses as +(+i) and --i as -(-i). Both leave i unchanged and are effectively max = midPointOfList min = midPointOfList...

There is of course a generic version of binarySearch1. The reason why there are non-generic overloads is performance: value types would need to be boxed (and thus copied) in order to be able to use a generic binarySearch. Same goes for sort and other algorithms. 1 In fact there’s more...

arrays,algorithm,binary-search

If you don't know what the sorted first/last word on each page is then a binary search can only find what pair of pages may contain the word. If the "top" word in a page comes before the word you want then you can eliminate all pages before that, but...

c,function,sorting,binary-search

Change the condition while(l<=h) to while ( l < h ) And change this code snippet else if(key<A[m]) { h=m-1; } the following way else if(key<A[m]) { h = m; } The function could be defined like int binarysearch( const int A[], int n, int key ) { int l...

algorithm,binary-search,programming-pearls

If you have an array that is empty (n == 0), then a check of while(low + 1 != high) will only correctly terminate if low begins at -1 and high at 0. while((-1 + 1) != 0) //true If low began at 0 instead, or high began at -1...

java,arrays,algorithm,big-o,binary-search

Instead of one binary search, use two: First find the first occurence of the value findValue. Let's call that index lo. I believe this is what your findIndex function does. Then find the first occurence of a value that is greater than findValue, or n if no such value exists....

java,arraylist,comparator,binary-search,suffix-array

Issue is over here: new SuffixArrayNaive().buildSuffixArray(baseString); String query="na"; new SuffixArrayNaive().search(query); In one Object of SuffixArrayNaive, you are creating suffix array (by populating list) while searching on another instance of SuffixArrayNaive, array (list) which is empty. Remember your list is define as non static: List<Elements> suffixArray = new ArrayList<>(); Meaning you...

big-o,time-complexity,complexity-theory,binary-search

As always, there can be a difference of the complexity of an algorithm and its implementation. If you implement binary search in a recursive function, that takes a copy of a part of the input array, this array has to be generated in O(n), so you are right, that this...

search,big-o,time-complexity,complexity-theory,binary-search

For binary search, the array should be arranged in ascending or descending order. In each step, the algorithm compares the search key value with the key value of the middle element of the array. If the keys match, then a matching element has been found and its index, or position,...

Live demo There is multiple things that you need to fix in the code, check my comments: var doSearch = function(array, targetValue) { var min = 0; var max = array.length - 1; var guess; var values; while (min <= max) { // "less or equal" otherwise some case won't...

javascript,arrays,binary-search

In your code, mid is always 1.5, because it's calculated before the loop. Instead, move the mid calculation within the loop, and calculate it as the rounded average of high and low: var arr = [1, 3, 5, 8]; var binary = function(arr, search) { var low = 0; var...

In the first case you calculate the value (low+high) which might be too huge to fit into an int, if low and high are both huge enough (say if both are equal to 2^30+1 /or even greater/). In the second case, you don't calculate (low+high), you do a small trick...

c,crash,binary-search,string-comparison

You appear to be trying to pass a pointer as an integer (see how ArrayPointer is declared as an int but you are casting it to a char **. You are then manipulating the integer with the binary search. This is why you are getting whacky results. Instead, you want...

python,recursion,binary-search

High and low appear to be indices for which part of the list to search. In the first example, 15 has an index of 3, so specifying a lower index of 4 means the 15 isn't included in the search space. In the second example, 84 has an index of...

import java.util.Scanner; public class Main { int[] array = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30}; public int binarySearch(int key, int l, int r) { if (r < l) { return -1; } int mid = (l + r) / 2; if...

java,collections,binary-search

You should send a Contact as argument that contains the name of the desired contact to search instead of sending just a String. Note that your ContactComparator compares Contacts not Contact and String. The code would look like this: public int ContactIndex(final String name) { Contact contactToSearch = new Contact();...

You have an unnecessary loop inside the function: while(first<=last) This is an infinite loop, because first and last are never reassigned inside the loop. You already use recursion by calling binarysearch inside the loop body. Either change the while to if, or remove the recursive calls and assign to first...

Just a forgotten return true. if (left > right) { return false; } middle = (left + right)/2; int compare = V.compareTo(A[middle]); System.out.println("Middle value: "+middle); if (compare == 0) { binaryOutput.setText("The book is: "+bookList[middle]); return true; // Missing! } if (compare < 0) { return binarySearch(A, left, (middle-1), V); }...

It doesn't seem to be doing anything for you; remove the call to midpoint, and point, and just have def binary_search(List,key,imin,imax,point): while (imax >= imin): imid = (imin + imax) / 2 (However, there are some things wrong with your code, and it won't work with just that change; You...

Basically, assume that You have a function f(x) which is monotonically increasing (decreasing) on the interval [a, b]. f(a) < C < f(b) You want to find x such that f(x) = C. Then you can use binary search to find x. Essentially, you half the possible interval for the...

python,arrays,search,binary-search

You need to change the & (bitwise "and") to and (logical "and") in your code: def find_pit(seq): first = 0 last = len(seq) - 1 origlast = last mid = 0 if len(seq) == 1 : return 0 else: #change next line to use logical and while first <= last...

Change if(s!=1) middle=s/2; if(a[middle]==key) return 1; else if(key<a[middle])binarySearch(a,middle,key); else if(key>a[middle])binarySearch(&a[middle],middle,key); to if (s != 1){ middle = s / 2; if (a[middle] == key) return 1; else if (key<a[middle])binarySearch(a, middle, key); else if (key>a[middle])binarySearch(&a[middle], middle, key); } The variable middle is initialized only if s!=1. I have run this code...

Binary search(at least the way I implement it) relies on a simple property - a predicate holds true for one end of the interval and does not hold true for the other end. I always consider my interval to be closed at one end and opened at the other. So...

Your second algorithm suffers from a repeating cycle as soon as you encounter a partition where high == (low+1). When that happens, you essentially have mid = (low + low + 1)/2, which is equivalent to (2*low)/2 + 1/2. With integer division, this results in mid = low + 0....

You are misusing this method. Quoting its documentation (emphasis mine): Returns: index of the search key, if it is contained in the array; otherwise, (-(insertion point) - 1). The insertion point is defined as the point at which the key would be inserted into the array Which means that this...

java,recursion,spell-checking,binary-search

Ok, as someone commented, I think the problem is in the loops :P This is what we want to do when we read all the words: Create an ArrayList (better than Array, because you don't know exactly how many words you have in the text file). Then create a double...

algorithm,data-structures,computer-science,binary-search,linear-search

regarding 1&2, an absolute number as an answer would've been possible if an absolute number was provided as the size of the input. since the question asks about an arbitrarily sized array (of length n) then answer is also given in these terms. you can read more about big O...

List has a binary search method but since binary search is so easy to implement and since the IComparator you would need to define is so complex due to the range thing I suggest you implement the binary search method Something like this (NOT TESTED!) public static IPRange BinarySearch(List<IPRange> source,...

If I understand you correctly, use bisect.bisect_left to find the the matches and delete: from bisect import bisect_left for ele in y: ind = bisect_left(x, ele) if ind < len(x) -1 and x[ind][0] == ele[0]: del x[ind] If you look at the source you can see the code used for...

c,arrays,dynamic,binary-search

You can use realloc to change the size of the array. If the pointer is set to NULL, realloc will behave the same as malloc on the first allocation of memory. Consider using "%d". If there were to be any values with leading zeros such as 08, "%i" will try...

Arrays.copyRangeOf() uses System.arraycopy() which uses native code (could use memcopy for example - depending on JIT implementation) under the hood. The "magic" behind copying with System.arraycopy() is making one call to copy a block of memory instead of making n distinct calls. That means that using Arrays.copyOfRange() will definitely be...

c++,object,vector,binary-search

To understand why your code is not working you need to debug it or at least add some debug output. If you add something like cout << "Now compare: " << carList[middle].getMake() << endl; before if (carList[middle].getMake() == search) you'd see that sometimes (after first input) your program does not...

if-statement,binary-search,elixir

Theoretically you can use guard clauses, but they can make things much worse if you overdo it. Assuming you start with an implementation like this: defmodule MyEnum do def binsearch(collection, key) do binsearch(collection, key, 1, length(collection)) end defp binsearch(collection, key, lo, hi) do if hi < lo do -1 else...

It's possible, but you wouldn't need to actually binary-search through the data. You can use lower_bound to find the first element and then advance the resulting iterator until your key no longer matches your criteria, storing them in a <vector> or similar container to return them all....

java,algorithm,order,binary-search

You can use binary search on the value by counting how many entries in the multiplication table will be less than the value. This will require log(NM) iterations in the binary search so we need to be able to compute the count in O(N) for a total complexity of O(Nlog(NM))....

Yes. From n1570 (C99+amendments / C11): 7.22.5 Searching and sorting utilities 2 The implementation shall ensure that the second argument of the comparison function (when called from bsearch), or both arguments (when called from qsort), are pointers to elements of the array.302) The first argument when called from bsearch shall...

data-structures,queue,priority-queue,binary-search

Your loop, the binary search, only finds the index at which the new item should be inserted to maintain sorted order. Actually inserting it there is the hard part. Simply put, this takes linear time and there's nothing we can do about that. A sorted array is very fast for...

python,python-2.7,sorting,binary-search,lower-bound

I would use the bisect module (as it uses binary search, giving it a O(log n) complexity) to take a bisection of both sides, like so: my_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] import bisect def find_ge(a, low, high): i = bisect.bisect_left(a, low) g =...

c,data-structures,struct,binary-search

item_t create_node(char *input) should be item_t *create_node(char *input) What you return is a structure but you should be returning pointer of type struct item....

arrays,algorithm,binary-search

Suppose you're standing on a middle element lower than its left neighbor: element you element You look to your left. It looks like a hill. Suppose you climb that hill. What do you see? Well, there are three possibilities: 1. element you element element 2. you element element element 3....

it is what arunmoezhi said, you need to return your calls (recursive function) if not your result will banish and the next phase is returning false thats why you always get false. bool binarySearch(int key, int array[], int min, int max) { int midPoint = findMidPoint(min, max); if (max <...

Solution 1 I don't know if this is the same as Rem's solution, but you could solve this quite easily with a binary search tree. Initialize an empty binary search tree and then iterate in reverse order through the array. Insert each element in the binary search tree and then...

You're not excluding the element you just tested before the next loop. And your choice of variable names, Subtraction is dreadful. Its a midpoint. int first = 0; int last = size_-1; int mid = 0; while(last >= first) { mid = first + (last - first) / 2; if(value...

java,string,collections,binary-search

To perform case-insensitive binary search, use String::compareToIgnoreCase as a comparator: int i = Collections.binarySearch(list, key, String::compareToIgnoreCase); This will perform faster than comparing two strings reduced to the same case, because compareToIgnoreCase() compares chars one by one, reducing case of chars only if needed, which allows to return fast if strings...

java,data-structures,binary-search

First thing, if the element is not found, a negative value must be returned according to the documetation, or you cannot distinguish between found and not found. Ok, so why not just -insertion point? Imagine if the insertion point is 0 (the searched element is smaller than all existing ones),...

If you read the Javadoc of binarySearch you'll see that the array must be sorted : /** * Searches the specified array of longs for the specified value using the * binary search algorithm. The array must be sorted (as * by the {@link #sort(int[])} method) prior to making this...

java,arrays,autocomplete,binary-search

Your for-loop in allMatches(String prefix) has a wrong boolean predicate,instead of for (int i = 0; i > b - a; i++) it should be for (int i = 0; i < b - a; i++) ...

It depends on how you implement binary search. For example, one way to implement it like you describe is to have it search for the first element that is larger than or equal to your element. Then, when the binary search stops, the position it stops on is the answer...

c++,algorithm,list,binary-search

If you want to search multiple times I suggest putting the file content into a hash table. This gives you O(1) lookup which is optimal.

c++,algorithm,binary-search,floating-accuracy

The problem was I was setting minimal volume as high in the binary search, but I should use the maximal volume. The second problem was I was not passing maximal radius ^ 3 to the binary search function. Thanks for help

arrays,algorithm,search,binary-search,fibonacci

Because if we have a Fibonacci sequence : 0,1,1,2,3,5,8,13,... F(k - 1) + F(k - 3) is what falls in the middle between F(k - 1) and F(k), and that's what we want because it's divide and conquer.

c,arrays,runtime-error,binary-search

You are calling the bi_search function twice. Once in your if statement, and again in a printf statement. You should call it only once, and cache the value. main() { int n=0,ch=0; do { printf("\nEnter the value you want to search for--\n"); scanf("%d",&n); if(bi_search(n)==-1) // Calling it here printf("\nSORRY :(...

You can use bisect from the standard library. I'm not sure if I understood your problem correctly, but a solution may look something like this: frequency = bisect(AISmessages[messageIndex:], message+86400) Example: This gives you the number of items in the list a with values in a range of 30, starting from...

This part will be your abort condition: if(low>high) return false; at the end - if no element has been found - the recursion will be called with low beeing greater than high - and that aborts immediately. look at the two calls: if(A[mid]<key) return Bsearch( A, mid+1, A.length, key); if(A[mid]>key)...

sorting,quicksort,binary-search,mergesort

After some Googling and testing, I finally got the answer: In case of Binary Search, when I tried it for the input 20,21 and 30, searching for the element 20, I finally landed upon a typical case wherein p=r=0. If the base condition included here is p>=r, then the result...

time,big-o,time-complexity,binary-search-tree,binary-search

Update Yes, You are absolutely right. In the worst case function() will be invoked 31 times, and each invocation requires time O(n), hence the running time of your algorithm is simply given by 31 * O(n) = O(n). Solution for the original question where x = function(mid) Your question is...

Some suggestion: 1) First, try to find the middle element (623 in the example) of the sorted short list, the result is index 2) Divide the short list into lower half (which are all elements less than middle element) and upper half (which are all elements greater than middle element)....

java,recursion,stack-overflow,binary-search

} else if (sArray.get(mid).compareTo(key) > 0) { return bSearch(sArray, key, lowIndex, mid + 1); } else { return bSearch(sArray, key, mid - 1, highIndex); } This would stuck when key cannot be found. It should be } else if (sArray.get(mid).compareTo(key) > 0) { return bSearch(sArray, key, lowIndex, mid - 1);...

algorithm,search,dynamic-programming,binary-search,divide-and-conquer

Drop first egg from floors 1, 3, 6, ... (the partials sums of 1, 2, 3, ...). Then do linear search between last two floors.

Change your function prototypte to template<typename T> T searchBinary(const std::vector<std::vector<T> >& vectorList, const std::vector<T>& compareVector) { i.e. pass by constant reference not by value. This will avoid two vector copies. You can refactor using a single conditional test < per iteration. (You'll also need to change the while conditional)....

If the array is not that large which means additional memory space is acceptable. I suggest you separate the array before binary search just because this make problem more easy. odd_list = x[::2] even_list = x[1::2] What's more, you can even use bisect for sorted array directly. Otherwise, I don't...

java,loops,user-input,binary-search

From what I understood, you want to search an inputted value in an array and return its index. So you first need to get the value from user input then search it in your array like this: public class Main { public int Search(int[] a, int x) { int low...

c,arrays,recursion,binary-search

The following line has a serious problem: else if(x = array[middle]) return(1); Instead of comparing x to array[middle], you are assigning the value of array[middle] to x. Provided this value is nonzero, it will always evaluate to true, and so your function will always return at that point. You should...

You can't write key < a[mid] like that in Java, instead, you call the comparator's compare method: if (comparator.compare(key, a[mid]) < 0) ... It works like this: the compare method return positive/zero/negative value depending on how the comparison ended. I remember it like this: a OP b --> comparator.compare(a, b)...

algorithm,time-complexity,binary-search,divide-and-conquer

You are correct. Unless the array is sorted, you still have to examine every element in each half (and each quarter and each eighth as you recur). The only way it can be O(log N) is if you can discard half the search space each recursion level (such as searching...

binary-search,c++-standard-library

binary_search is (IMO) poorly named. It tells you via a binary search whether an item exists in your collection. It does not tell you where. lower_bound tells you where the item should go in the collection, but does not actually verify that the item exists at that place in the...

c,optimization,binary-search,linear-search

You should look at the generated instructions to see (gcc -S source.c), but generally it comes down to these three: 1) N is too small. If you only have a 8 different branches, you execute an average of 4 checks (assuming equally probable cases, otherwise it could be even faster)....

c++,binary-search,lower-bound,upperbound

The lower and upper bound of a binary search are the lowest and highest position where the value could be inserted without breaking the ordering. (In the C++ standard library, these bounds will be represented by iterators referencing the element before which the value could be inserted, but the concept...

We define mid with the statement on the third line: int mid = (first + last) / 2; As for the others, we define first and last by calling the function. For example: a = bsearch(my_list, 1, 100, 55); according to int bsearch(const int list[], int first, int last, int...

python,python-3.x,binary-search

print("Enter "+int(n)+" Elements") TypeError: Can't convert 'int' object to str implicitly Do you see the error? Try this: print("Enter "+str(n)+" Elements") ...

Using Collections.binarySearch(...) When you run a Collections.binarySearch(...); on a list the objects in that list must either implement Comparable, or you will have to pass a Comparator into the binarySearch(...) method; Using a Comparator as an example you could do the following; class KlantComparator implements Comparator<Klant> { @Override public int...

I seriously don't understand why you put that while(found==0) loop in the middle of the function, but it is definitely leading to the endless loop. Simply try removing it. In order to know whether a solution was found, we can make the method return -1 in this particular case. This...

visual-studio-2013,cuda,binary-search,thrust

It turns out that the compiler was actually smart enough to realize that I wasn't doing anything with the host-side "findTarget" routine so it completely cut it out of the compiled code - i.e. it wasn't even being executed (hence explaining why dramatically increasing the simulation count did nothing and...

Make the search end case based on first index p and last index r, int BinarySearch(int A[], int p, int r, int value) { int q = (p + r) / 2; if( p > r ) { return 0; } if (A[q] == value) { return q; //value found...

You might have been advised that, to reduce the number of operations to compute the mid. (indexMax - indexMin)/2 + indexMin equals (indexMax + indexMin)/2...

algorithm,search,data-structures,big-o,binary-search

Big-oh notation does not care about constants. In fact, it doesn't care about anything but the dominating term in an expression. So even if your algorithm does 4 * log n operations of a certain type, it is still O(log n). As long as it is a constant times f(n),...

python,recursion,binary-search

if word in wordlist[len(wordlist)/2:] will make Python search through half of your wordlist, which is kinda defeating the purpose of writing a binary search in the first place. Also, you are not splitting the list in half correctly. The strategy for binary search is to cut the search space in...

If your List is sorted by start time. You could run a Binary search with a custom comparer to find out where the range could possibly be located. protected List<TimeRange> GetBoundedTimeRanges(List<TimeRange> timeRanges, DateTime startTime, DateTime endTime) { var startSearch = timeRanges.BinarySearch(new TimeRange(startTime, startTime), new TimeRangeComparer()); if (startSearch < 0) {...

I present a solution faster than the raw functions taken from the bisect library Solution With Optimised Binary Search def search(a, x): right = 0 h = len(a) while right < h: m = (right+h)//2 if x < a[m]: h = m else: right = m+1 # start binary search...

c,algorithm,binary-search,upperbound,last-occurrence

n.m.'s 2-search approach will work, and it keeps the optimal time complexity, but it's likely to increase the constant factor by around 2, or by around 1.5 if you begin the second search from where the first search ended. If instead you take an "ordinary" binary search that finds the...

In the second else if part you need to set last instead of first - else if(obj.compareTo(array[middle]) < 0) last = middle - 1; ...

c,arrays,binary-tree,binary-search-tree,binary-search

The problem is here: void balanceTree(struct tree *root) Although this function works, the new root that is created is NOT returned to the caller because root has been passed by value. You need to change the function to: struct tree *balanceTree(struct tree *root) and add this at the end: return...

How about this? int ans = -1; bsearch() if f(x) > t then high = x-1 else if f(x)<= t then low = x+1, ans = x Another method: Just use your current bsearch to find the x where f(x) is smallest value > t, then what you want is...

When you call recursiveBinarySearch( ) recursively, you call it with a return statement, because if the recursive function returns something, the function that calls the recursive function should return the same value. The code should be: int recursiveBinarySearch(int* a, int p, int r, int x) { if(p>r) return -1; else...

arr_in = [-1, 1,2,4,5] arr_in.bsearch{ |x| 2 - x } #=> 2 arr_in.bsearch{ |x| -1 - x } #=> -1 arr_in.bsearch{ |x| 3 - x } #=> nil Binary search uses block's result as a hint which part of array (left or right side) should be chosen for searching on...