Pre-requisite :
Segment tree or fenwick tree
Problem statement :
Given an array, you have to print for each position, how many elements appearing before it are smaller than it.Explanation:
The problem is a standard problem known as finding the number of inversions in an array. If we apply a bruteforce and count how many elements appearing before an element are smaller, this will time out (O(n^2)).We need a better approach. We can create a hash table where each positon will have 0 initially. For each element Ai, we can increment the value of hash[Ai] by 1. Since it is a hash table, if an element 'x' smaller than Ai appering before Ai is there, then hash[x] won't be 0 and hash[x] will appear before hash[Ai]. So finding the sum of values from hash[1] to hash[Ai - 1] will be equal to number of elements smaller than Ai appearing before Ai.
Consider this example.
A = {1, 3, 5};hash[] = {0, 0, 0, 0, 0, 0}
when we visit 1, hash[1] = 1
hash[] = {0, 1, 0, 0, 0, 0}
Sum of all values before 1 , i.e. hash[0] = 0, so elements smaller than 1 appearing before 1 is 0
when we visit 3, hash[3] = 1
Sum of all values before 3 , i.e. hash[0] + hash[1] + hash[2] = 1, so elements smaller than 3
appearing before 3 is 1
Now this approach is slow considering we have to find the sum of elements in linear time for each entry. We can make it logarithmic by using a segment tree or a fenwick tree. Both returns sum from first to Aith element in logarithmic time. Also we increment value in hash table in logarithmic time.
See the solution code which uses a fenwick tree.
Solution:
include<stdio.h>
include<string.h>
define MAX 1000009
int tree[MAX]; void init() { memset(tree,0,sizeof(tree)); } void add(int index) { while(index<max) {="" tree[index]++;="" index="index" +="" (index="" &="" -index);="" }="" }="" int="" sum(int="" index)="" {="" int="" s="0;" while(index=""> 0) { s+=tree[index]; index = index - (index & -index); } return s; } int main() { int test; scanf("%d",&test); while(test--) { init(); int n,input; scanf("%d",&n); while(n--) { scanf("%d",&input);
add(input); printf("%d ",sum(input-1)); } printf("\n"); } return 0;
}