I do disagree with the way you store your input but you can achieve your search with the following: String[,] arr = new string[2,4]; arr[0, 0] = “saif”; arr[0, 1] = “25”; arr[0, 2] = “ali”; arr[0, 3] = “17”; arr[1, 0] = “aakif”; arr[1, 1] = “11”; arr[1, 2] = “hassnain”; arr[1, 3] = … Read more
There are at least two, possibly three, things I can see wrong with it. It uses (low + high)/2. The addition may overflow to a negative number if the array is very large. If so, division by 2 will lead to a negative index. This can be fixed by using (low + high)>>>1. It is … Read more
Here binary search is being used to find the optimal solution.For binary search you need a range represented by high and low.You know the optimal answer lies within this range.In this particular problem low is represented by the minimum number of folders that a worker will have to look for, which is going to be … Read more
Basically I want to look at the position in where all my “X” are. You can use np.where to get the indices for each axis, then zip them to get index tuples for all the locations: >>> list(zip(*np.where(Arr == “X”))) [(3, 2), (5, 4), (6, 1)] If you want the (row, column) “locations”, you could … Read more
So from the sound of your question you are attempting to get a binary search working for your example. First, it is useful to know that a binary search only works reliably on sorted data. This is quite different than a standard linear search that will always find the value, regardless of order. As you … Read more
You need to update the left and right values not the middle..change it like this while (true) { middle = (left + right) / 2; int copmarison = Name.compareTo(StudentName[middle]); if (Name.equals(StudentName[middle])) { return middle; } else if (left > right) { return count; } else { if (copmarison > 0) { left = middle + … Read more
With binary search you typically search in a sorted random access data structure like an array, by discarding half of the array with each comparison. Hence, in k steps you effectively cover 2^k entries. This yields a complexity of at most log2(n) of n elements. With landau symbols, the base of the logarithm disappears because … Read more
Based on the phrasing of your question, I’m guessing that you are unfamiliar with the concept of pointers. In C, pointer is a type that stores the memory address of another variable. The & operator retrieves the memory address of a variable. int i = 4; // < An integer variable equal to 4 int* … Read more
The aim of the binary search is to find the index of a number inside a sorted array in a complexity log(n). FIRST STEP : initialization You start with your sorted array : {5,10,15,20,25,30,35,40,45,50} small : is the current smallest index. At the beginning it should be 0 for a zero-based array. big : is … Read more
Without tracing, it’s relatively straightforward to see what a given tree should print. The structure of the method means that a string will look like this: <content><left sub-tree><content><right sub-tree><content> So all you have to do is continually substitute that pattern for the left and right sub-trees (with empty strings for non-existent sub-trees) and get the … Read more