The Brute force solution is O(n^2), compute the distance between each pair and return the smallest. We can calculate the smallest distance in O(nLogn) time using Divide and Conquer strategy. In this post, a O(n x (Logn)^2) approach is discussed. We will be discussing a O(nLogn) ...
3. we can also use divide and conquer: reference: http://www.geeksforgeeks.org/divide-conquer-set-6-search-row-wise-column-wise-sorted-2d-array/
Divide-and-conquer works when you either don’t want collaboration (such as war, or two children yelling at each other) or you do want handoffs, such as in a factory, where work moves from one machine to another. Divide-and-conquer for knowledge work creates too much WIP (work in prog...
// Count of Smaller Numbers After Self// The smaller numbers on the right of a number are exactly #jumps from its right to its left during a stable sort.// Merge sort with added tracking of those right-to-left jumps// http://www.geeksforgeeks.org/counting-inversions/// two elements ...
public static void main(String[] args){ int N = args.length; array = new int[N]; for(int i = 0;i < N;i++){ array[i] = Integer.parseInt(args[i]); } mark a = merge(0,N-1); System.out.println(a.lom + " " + a.him +" " + a.value); }...