XOR between a Range of Integers

XOR between a Range of Integers - Easy

Problem : Given two integers a and b (a <= b ). Find the value of a ^ (a+1) ^ ... (b-1) ^ b. (^ is the xor operator).

This will be a short post. It's pretty trivial once you understand the pattern. I'd first refer you to this post first. In the alternate solutions, I've described how you can obtain the value of 1 ^ 2 ^ 3 .... ^ N.

Therefore, our solution simply applies that concept.

xor(b) ^ xor(a - 1) where xor function is implemented in Python as follows

For Python, this will work for negative integers as well, since the mod operator always returns a positive integer. For other languages, make sure you make appropriate changes to the xor() function. Until next time...

Comments

Find Increasing Triplet Subsequence - Medium

Problem - Given an integer array A[1..n], find an instance of i,j,k where 0 < i < j < k <= n and A[i] < A[j] < A[k]. Let's start with the obvious solution, bruteforce every three element combination until we find a solution. Let's try to reduce this by searching left and right for each element, we search left for an element smaller than the current one, towards the right an element greater than the current one. When we find an element that has both such elements, we found the solution. The complexity of this solution is O(n^2). To reduce this even further, One can simply apply the longest increasing subsequence algorithm and break when we get an LIS of length 3. But the best algorithm that can find an LIS is O(nlogn) with O( n ) space . An O(nlogn) algorithm seems like an overkill! Can this be done in linear time? The Algorithm: We iterate over the array and keep track of two things. The minimum value iterated over (min) The minimum increa...

Shortest Interval in k-sorted list - Hard

Given k-sorted array, find the minimum interval such that there is at least one element from each array within the interval. Eg. [1,10,14],[2,5,10],[3,40,50]. Output : 1-3 To solve this problem, we perform a k-way merge as described here . At each point of 'popping' an element from an array. We keep track of the minimum and maximum head element (the first element) from all the k-lists. The minimum and maximum will obviously contain the rest of the header elements of the k-arrays. As we keep doing this, we find the smallest interval (max - min). That will be our solution. Here's a pictorial working of the algorithm. And here's the Python code. Time Complexity : O(nlogk)

The 2 Missing Duplicate Numbers - Medium

This is a slightly trickier version of The Missing Duplicate Number . Instead of one single missing number, we have 2. Problem : Given an array of Integers. Each number is repeated even number of times. Except 2 numbers which occur odd number of times. Find the two numbers Ex. [1,2,3,1,2,3,4,5] The Output should be 4,5. The naive solution is similar to the previous solution. Use a hash table to keep track of the frequencies and find the elements that occur an odd number of times. The algorithm has a Time and Space Complexity of O(n). Say those two required numbers are a and b If you remember the previous post, we used the XOR over all the elements to find the required element. If we do the same here, we get a value xor which is actually a ^ b. So how can we use this? We know that a and b are different, so their xor is not zero. Infact their XOR value has its bit set for all its dissimilar bits in both a and b. (eg. 1101 ^ 0110 = 0011). It's important to notice ...

Lost in Compilation

Search This Blog