1. Merge sort array A and create a copy (array B) 2. Take A[1] and find its position in sorted array B via a binary search. The number of inversions for this element will be one less than the index number of its position in B since every lower number that appears after the first...
1. Write a program to implement the Sort-and-Count algorithms in your favorite language, find the number of inversions in the given file. 2.In the lecture, we count the number of inversions in O(n log n) time, using the Merge-Sort idea. Is it possible to use the Quick-Sort idea ...
Pretty sure that task 1 can be solved using merge sort tree. You can also use lower_bound function normally if you have an array of pairs: pair<int,int> a[n], x; lower_bound(a, a+n, x); → Reply CF_contest_practice_2 4 weeks ago, # ^ | ← Rev. 2 +1 Usual lower_...
(nums,mid+1,high);returnmerge(nums,low,mid,high);}intmain(){intn,num;vector<int>nums;scanf("%d",&n);for(inti=0;i<n;i++){scanf("%d",&num);nums.push_back(num);}if(nums.size()<2){cout<<sum<<endl;return0;}tmp.resize(nums.size(),0);printf("%lld",mergeSort(nums,0,...
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I guess you are trying to find count of inversions in array which is if i < j and a[i] > a[j] This can be done using merge sort along with a simple modification. Link to understand sol → Reply tdpencil 11 months ago, # | 0 Statement's really unclear, can you clarify ...
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