Problem: Inversion Count for an array indicates – how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum. Formally speaking, two elements a[i] and a[j] form an inversion if a[i] > a[j] and
Continue Reading “Find the Inversion Count of an Array [ Microsoft, Flipkart]”
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#include <stdio.h> #include <iostream> using namespace std; int arr[100000]; void merge(int lw,int mid,int hi) { int L[mid-lw+1]; int R[hi-mid+1]; int tmp; for(int i=0;i<mid-lw+1;i++) L[i]=arr[lw+i]; for(int i=0;i<hi-(mid+1)-1;i++) R[i]=arr[mid+i+1]; int i = 0; int j = 0; int k = lw; while(i<(mid-lw+1) && j<(hi-mid)) { if(L[i]<=R[j]) { arr[k]=L[i]; i++; k++; } else if(L[i]>R[j]) { arr[k]=R[j]; k++; j++; } } while(i<(mid-lw+1)) { arr[k++]=L[i++]; } while(j<(hi-mid)) { arr[k++]=R[j++]; } } void mergesort(int lw,int hi) { if(hi>lw) { int mid=lw+(hi-lw)/2; mergesort(lw,mid); mergesort(mid+1,hi); merge(lw,mid,hi); } } int main() { int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&arr[i]); mergesort(0,n-1); for(int i=0;i<n;i++) cout<<arr[i]<<" "; } return 0; } |
Problem: Find the minimum difference between any pair in an unsorted array.
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#include <stdio.h> #include <algorithm> #include <iostream> using namespace std; int arr[105]; int res[100005]; int main() { int t,n,k,a; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=0;i<105;i++) arr[i]=0; for(int i=0;i<n;i++) { scanf("%d",&a); arr[a]++; } //sort(arr,arr+n); k=0; for(int i=0;i<105;i++) { if(arr[i]==0) continue; for(int j=1;j<=arr[i];j++) res[k++]=i; } int MN=res[1]-res[0]; for(int i=2;i<n;i++) { if(res[i]-res[i-1]<MN) MN=res[i]-res[i-1]; } cout<<MN<<endl; } return 0; } |