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INSERTION
SORT
DR. HIMANI MITTAL
INSERTION SORT
• Complexity of Insertion sort is also N 2
like bubble sort. It
assumes that the first element is sorted and then tries to
put second element in place with respect to first. Then
first two are sorted and third element is placed either
before first or in middle of first and second or after second
element depending on the relation that a[3] is more of
less than a[2] and a[1]. And so on for all the elments.
• Insertion sort is so called because we insert the element in
its proper position in the first part of the array.
ALGORITHM
1 2 3 4 5 Index
5 3 2 7 1 Let this be the array to sort. I =2, J=1 , target = 3
3 5 2 7 1 Compare 5 > 3, true. Shift 5 and insert 3
3 5 7 1 Now I =3 , J=2, target = 2
3 5 7 1 Compare 5> 2 true so shift 5 to location 3, for J=2
3 5 7 1 Compare 3> 2 true so shift 3 to location 2, for J =1
2 3 5 7 1
J becomes 0 while loop is quit
Insert A[J+1] = target => A[0+1] = target
Now I = 4, J =3, target = 7
2 3 5 7 1 Compare 5>7 false so do not enter while loop
2 3 5 7 Now I = 5, J = 4, target=1
2 3 5 7 Compare 7>1, shift 7
2 3 5 7 Compare 5>1, shift 5
2 3 5 7 Compare 3>1, shift 3
2 3 5 7 Compare 2 >1, shift 2
1 2 3 5 7
J becomes zero so while loop is quit
Copy target at A[J+1]

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Insertion sort

  • 2. INSERTION SORT • Complexity of Insertion sort is also N 2 like bubble sort. It assumes that the first element is sorted and then tries to put second element in place with respect to first. Then first two are sorted and third element is placed either before first or in middle of first and second or after second element depending on the relation that a[3] is more of less than a[2] and a[1]. And so on for all the elments. • Insertion sort is so called because we insert the element in its proper position in the first part of the array.
  • 4. 1 2 3 4 5 Index 5 3 2 7 1 Let this be the array to sort. I =2, J=1 , target = 3 3 5 2 7 1 Compare 5 > 3, true. Shift 5 and insert 3 3 5 7 1 Now I =3 , J=2, target = 2 3 5 7 1 Compare 5> 2 true so shift 5 to location 3, for J=2 3 5 7 1 Compare 3> 2 true so shift 3 to location 2, for J =1 2 3 5 7 1 J becomes 0 while loop is quit Insert A[J+1] = target => A[0+1] = target Now I = 4, J =3, target = 7 2 3 5 7 1 Compare 5>7 false so do not enter while loop 2 3 5 7 Now I = 5, J = 4, target=1 2 3 5 7 Compare 7>1, shift 7 2 3 5 7 Compare 5>1, shift 5 2 3 5 7 Compare 3>1, shift 3 2 3 5 7 Compare 2 >1, shift 2 1 2 3 5 7 J becomes zero so while loop is quit Copy target at A[J+1]