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Tekslate

Published Date

21st September, 2018

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1079

Suppose an array A with ‘n’ elements A[0],A[1],A[2]….. is in memory. The selection sort algorithm in data structures for sorting ‘A’ works as follows **Pass 1:** First find the smallest element in the list and put it into the first position. Then find the second smallest element in the list of ‘N’ elements A[0],A[1],A[2]------ and then interchange A[Loc] and A[0] is sorted. **Pass 2:** Find the location LOC of the smallest in the sub list of n-1 elements A[1],A[2],A[3] then A[0],A[1] sorted **Pass N-1:** Find the location LOC of the smaller of the elements A[N-1], A[N] and then interchange A[LOC] and A[N-1] then : A[0],A[1],A[2]----- is sorted.

Implementation of selection sort # include<stdio.h> void selection sort(int a[],int n); void main() { int a[20],n,i; clrscr(); printf(“Enter n value \n”); scanf(“%d”,&n); printf(“Enter those numbers \n”); for (i=0;i<n;i++) { scanf(“%d”,&a[i]); } selection sort(a,n); printf(“Following are the elements in sorting orger: \n”); for(i=0;i<n;i++) { printf(“%5d”,a[i]); } } void selection sort(int a[], int n) { int i, min Loc, t,j; for(i=0;i<n-1;i++) { MinLOC=i; for(j=i+1; j<n;j++) { if(a[j]<a[minLoc]) minLoc=j; } t=a[i]; a[i]=a[MinLoc]; a[MinLoc]=t; } }

Complexity of the selection sort algorithm f(n)=(n-1)+(n-2)+-------- +2+1=n(n-1)/2=0(n^{2})

Algorithm | Worst cape | Average cape |

selection sort | n(n-1)/2=o(n^{2}) |
n(n-1)/2=o(n^{2)} |

Enter n value : 7 Enter those numbers : 60 30 20 10 50 70 40 Following are the elements in the sorting order: 10 20 30 40 50 60 70

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