Binary Indexed Trees (Fenwick Tree)

from: http://algorithmsandme.in/2015/02/binary-indexed-trees/

Why Binary Indexed Tree?

Consider an array A : {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}, you have to find the sum of a range say (1,5). One way of doing this can be by keeping the sum of elements from index 0 to index i, where 0 <= i<= n. So the array with all the cumulative sum is “sum” :{1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136} and to calculate sum from 1 to 5 we can simply do sum[5] – sum[1].

Shuffle a set of numbers without duplicates.

Example:

Java Solution

LeetCode - Remove Element

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

Analysis

This problem is similar to another leetcode problem: Remove Duplicates from Sorted Array.

Leetcode: Remove Duplicates from Sorted Array

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn’t matter what you leave beyond the new length.

Analysis

We defined two indexes: pre and cur.

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.

Analysis

It’s easy to first sort the array, then pick up the Kth largest element. The time complexity is O(nlogn) if we use quick sort.

Min Heap Based Method

A better method is use a min heap or PriorityQueue with size K.

How to implement a calculator is a popular interview question. To answer this question well, you need to maser Stack data structures, convert an infix notation to RPN and evaluate reverse polish notation.

Since the input is usually in infix notation, e.g. “3 + 6 * 2”, it is difficult develop a program to evaluate it directly.

In practice, we can implement a calculator algorithm into two steps:

1. the first step is to convert your mathematical expressions, which is called infix notation, into Reverse Polish Notation (RPN),

[LeetCode] Evaluate Reverse Polish Notation (RPN)
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:

Analysis

Evaluate RPN is the second step to build a calculator. Once we have converted the infix expression into the RPN, we need to evaluate RPN to get the result.

The basic idea to evaluate reverse polish notation is to use a stack to process the strings.

We scan the array of the RPN expression.

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.
The input string does not contain leading or trailing spaces and the words are always separated by a single space.
For example,
Given s = "the sky is blue",
return "blue is sky the".
Could you do it in-place without allocating extra space?

Analysis

This problem is similar to the rotate array to right by k steps problem.

There are two steps:

1. use the space ‘ ‘ to determine the boundary of words.

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

How many different ways do you know to solve this problem?

Analysis

It is easy to use an intermediate array to solve this problem. Suppose the original array is A, the new array is B, we will copy

A[0:n-k -1] to B[k:n-1], and A[n - k:n-1] to B[0:k - 1], then we copy all the elements in B back to Array A using System.arrayCopy.