You are given an array of transactions transactions where transactions[i] = [fromi, toi, amounti] indicates that the person with ID = fromi gave amounti $ to the person with ID = toi.
Return the minimum number of transactions required to settle the debt.
The question below is similar to an N way partition problem (instead of 3 way partition problem
Then the problem can be converted to the following:
Given N accounts/persons who have non zero amounts,find the M = max number of sets where each set s has sum(s)==0.Each of these sets must be min_set. In other words, min_set CANNOT be divided to setA and setB where sum(setA) == sum(setB)==0.
I am adding a few problems pertaining to the use of this data structure
Next greater element
Points to remember:
- We need to process the array from right to left
- Using stack we end up processing each element just 2 times , hence the complexity is just O(n)
void nextgreaterelement()…
Problems
1.https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/ — why does the heuristic in bfs / the condition to enter into the bfs queue matter ?
Merge Sort (Count Inversions)
Complexity : Worst Case O(n log n)
Quick Sort
Complexity : Worst Case O(n²)
int partition(int l,int r,vector<int>&v){int pivot=(l+r)/2;while(l<=r){ while(v[l]<v[pivot]) l++; while(v[r]>v[pivot]) r--; if(l<=r) { swap(v[l],v[r]); l++; r--; }}return l;}void quicksort(int l,int r,vector<int>&v){ int idx=partition(l,r,v); if(l<idx-1) quicksort(l,idx-1,v); if(idx<r) quicksort(idx,r,v);}void quick(){ vector<int>v; v.push_back(10); v.push_back(20); v.push_back(9); v.push_back(3); v.push_back(4); quicksort(0,v.size()-1,v); for(int i=0;i<v.size();i++) cout<<v[i]<<" "; cout<<endl;}
Points to remember:
- How do you make the partition algorithm more efficient
- Applications ? (kth largest element, getting k top elements etc. )
