Show that the time complexity for the number of assignments of records for the Mergesort algorithm (Algorithms 2.2 and 2.4) is approximated by T (n) = 2nlgn.
package sort;
/*worst case of merge sort is NlogN
* Merge sort
* {20, 13, 4, 34, 5, 15, 90, 100, 75, 102}
* / \
* {20, 13, 4, 34, 5} {15, 90, 100, 75, 102}
* / \ / \
* {20, 13, 4}{34, 5} {15, 90, 100,} {75, 102}
* / \ / \ / \ / \
*{20, 13} {4} {34} {5} {15, 90} {100} {75} {102}
* / \ / \
*{20} {13} {15} {90}
*
*Height of above tree is Log(N)
*to merge the array two temp array it will take o(N) in worst
case
*so total time complexity it O(Nlog(N))
*
*/
public class MergeSort {
int ar[];
MergeSort(int ar[]){
this.ar=ar;
}
public void mergeSort(int ar[],int l,int r){
if(l<r){
int mid=(l+r)/2;
//take mid index
mergeSort(ar,l,mid); //call mergesort on first half
mergeSort(ar,mid+1,r); // and second half
marge(ar,l,r,mid); //no merge the two in sorted manner
}
}
public void marge(int ar[],int l,int r,int mid){
int t1[]=new int[mid-l+1]; //make
temp array of first hlaf size and
int t2[]=new int[r-mid]; //second
half size
int k=0; //index to 0
for(int i=l;i<=mid;i++){ //copy
fist half elements to first temp array
t1[k++]=ar[i];
}
k=0;
for(int i=mid+1;i<=r;i++){
//copy second half array to sec temp array
t2[k++]=ar[i];
}
k=0;
int k1=0; //from those two temp
array copy elements in sorted array in original array
while(k<t1.length &&
k1<t2.length){
if(t1[k]<=t2[k1]){ //if t1's element is less first copy it
ar[l++]=t1[k++];
}
else{
ar[l++]=t2[k1++]; //else t2's
}
}
while(k<t1.length){ //copy the
remaining elements
ar[l++]=t1[k++];
}
while(k1<t2.length){
ar[l++]=t2[k1++];
}
}
//driver program to test
public static void main(String[] args) {
int A[]={20, 13, 4, 34, 5, 15, 90,
100, 75, 102};
MergeSort m=new MergeSort(A);
m.mergeSort(A,0,A.length-1);
for(int
i=0;i<A.length;i++){
System.out.print(A[i]+" ");
}
}
}
output
4 5 13 15 20 34 75 90 100 102
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