time-complexity

问题 In the book Programming Interviews Exposed it says that the complexity of the program below is O(N), but I don't understand how this is possible. Can someone explain why this is? int var = 2; for (int i = 0; i < N; i++) { for (int j = i+1; j < N; j *= 2) { var += var; } } 回答1: You need a bit of math to see that. The inner loop iterates Θ(1 + log [N/(i+1)]) times (the 1 + is necessary since for i >= N/2 , [N/(i+1)] = 1 and the logarithm is 0, yet the loop iterates once). j takes the values (i

问题 In the book Programming Interviews Exposed it says that the complexity of the program below is O(N), but I don't understand how this is possible. Can someone explain why this is? int var = 2; for (int i = 0; i < N; i++) { for (int j = i+1; j < N; j *= 2) { var += var; } } 回答1: You need a bit of math to see that. The inner loop iterates Θ(1 + log [N/(i+1)]) times (the 1 + is necessary since for i >= N/2 , [N/(i+1)] = 1 and the logarithm is 0, yet the loop iterates once). j takes the values (i

问题 Consider 2 sequences X[1..m] and Y[1..n]. The memoization algorithm would compute the LCS in time O(m*n). Is there any better algorithm to find out LCS wrt time? I guess memoization done diagonally can give us O(min(m,n)) time complexity. 回答1: Gene Myers in 1986 came up with a very nice algorithm for this, described here: An O(ND) Difference Algorithm and Its Variations. This algorithm takes time proportional to the edit distance between sequences, so it is much faster when the difference is

问题 I have searched a bit on StackOverflow and have understood the complexity up to the point of the j-loop, which is O(n 2 ) . However with the nested addition of the k-loop, I am confused as why the complexity becomes O(n 3 ) . Can someone help me understand this? From my understanding, the i-loop have n iterations and the j-loop have 1+2+3+...+n iterations n*(n+1)/2 which is O(n 2 ) . for(i = 1; i < n; i++) { for(j = i+1; j <= n; j++) { for(k = i; k <= j; k++) { // Do something here... } } }

问题 Suppose that n records have keys in the range from 1 to k. Write an algorithm to sort the records in place in O(n+k) time. You may use O(k) storage outside the input array. Is your algorithm stable? if we use counting sort to we can do it in O(n+k) time and is stable but its not in place. if k=2 it can be done in place but its not stable (using two variables to maintain the indexes in the array for k=0 and k=1) but for k>2 i couldnt think of any good algo 回答1: First, let's rehash how counting

问题 Check if 2 tree nodes are related (i.e. ancestor-descendant) solve it in O(1) time, with O(N) space (N = # of nodes) pre-processing is allowed That's it. I'll be going to my solution (approach) below. Please stop if you want to think yourself first. For a pre-processing I decided to do a pre-order (recursively go through the root first, then children) and give a label to each node. Let me explain the labels in details. Each label will consist of comma-separated natural numbers like "1,2,1,4,5

问题 The problem I am given N arrays of C booleans. I want to organize these into a datastructure that allows me to do the following operation as fast as possible: Given a new array, return true if this array is a "superset" of any of the stored arrays. With superset I mean this: A is a superset of B if A[i] is true for every i where B[i] is true. If B[i] is false, then A[i] can be anything. Or, in terms of sets instead of arrays: Store N sets (each with C possible elements) into a datastructure

问题 I am not good at determining time and memory complexities and would appreciate it if someone could help me out. I have an algorithm, here and I am not sure what its time and memory complexities would be. Function sample(k) IF k < 2 Return 0 Return 1 + sample(k/2) What is its time and memory complexity and why? Thanks 回答1: Determining time and memory complexities amounts to counting how much of these two resources are used when running the algorithm, and seeing how these amounts change as the

问题 I know that worst case on mergesort is O(nlogn), the same as the average case. However, if the data are ascending or descending, this results to the minimum number of comparisons , and therefore mergesort becomes faster than random data. So my question is: What kind of input data produces the maximum number of comparisons that result to mergesort to be slower? The answer at this question says: For some sorting algorithms (e.g. quicksort), the initial order of the elements can affect the

问题 If the time complexity of my program is,say O(n^2) ,How do I express running time in terms of seconds for a large value of n,10^6 ? I need a rough estimate of that to know if optimization is required or I can proceed with my code....Time Limit is 0.6 seconds The question is not about calculation of Time complexity....It's about estimation of running time from Time complexity 回答1: There is no way to calculate or estimate the running time of a some piece of code based on its Big-O rating. Big-O