### MXRCI's blog

By MXRCI, history, 2 months ago,

My thought process : I thought like what we do in inversion count we add smaller numbers at index of fenwick and then count the sum from (sorted index+1 to n-1) , we can do same in this case like making array of (nums1[i] — nums2[i] — diff) and then finding the correct index (using binary search) at a particular i and then finding sum till that index to get count. but i guess i cant implement it correctly can u please help

Fenwicktree struct

struct fenwicktree{
int n;
vector<int> fn;

void init(int _n){
n = _n+1;
fn.resize(n,0);
}

void get(int x,int sum){
for(x++;x<n;x+=(x&(-x))){
fn[x] += sum;
}
}

// sum from 1 to idx
int _sum(int x){
int ans = 0;
for(x++;x>0;x-=(x&(-x))){
ans += fn[x];
}
return ans;
}

int sum(int left , int right){
return _sum(right) - _sum(left-1);
}
};


Logic:

class Solution {
public:
long long numberOfPairs(vector<int>& nums1, vector<int>& nums2, int diff) {
int n = nums1.size();
fenwicktree tree;
tree.init(n);
vector<pair<int,int>> v;
vector<int> a;
for(int i = 0 ; i < n ; i++){
v.push_back({nums1[i]-nums2[i]-diff,i});
a.push_back({nums1[i]-nums2[i]-diff});
}

sort(v.begin(),v.end());
sort(a.begin(),a.end());

int ans = 0;
for(int i = 0 ; i  < n ; i++){
int sum = v[i].first;
int idx = v[i].second;

int j = lower_bound(a.begin(),a.end(),sum+diff)-a.begin();
ans += tree._sum(j);
tree.get(idx,1);
}
return ans;
}
};



• 0

 » 2 months ago, # |   0 You don't need fenwick tree for this. Rearranging the equation given, you get:(nums1[i] — nums2[i]) — (nums1[j] — nums2[j]) <= diffLooking at the equation, you should realise that you only need one array "nums3" where nums3[i] = nums1[i] — nums2[i]. Now all you need to do is find number of pair (i, j) where nums3[i] — nums3[j] <= diff. So first, sort nums3 in ascending order. Then, for each element a in nums3, binary search the first element b such that a-b <= diff, or instead b >= a-diff. Notice, that every element after b will also work, so add (length of num3 — position of b) to some variable "ans". After you iterate through the entire list, you get your answer.
•  » » 2 months ago, # ^ |   0 You forgot about i < j.