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Stepping through insertion sort:
1) Assume the first item is sorted.
2) Find the next value to compare to the sorted value.
3) Shift over any necessary elements to make space for the value being added.
4) Insert value into sorted subset, and respect step1 to step3.
|INSERTION SORT (P)|
1: for (J <-- 2) To length[ P ]
2: do key <-- P[ J ] // Put P[ J ] into the sorted sequence P[ 1... J - 1 ]}
3: I <-- (J - 1)
4: while (J > 0), and
5: P[ I ] > 'key'
6: do P[ I + 1 ] <-- P[ I ]
7: J <-- I - 1
8: P[ I + 1 ] <-- 'key'
1) Each element Array [ i ] is taken one at a time from i = 0 to n - 1.
2) Before insertion of Array[ i ], the subarray from Array[ 0 ] to Array[ i - 1 ] is correctly ordered, while the subarray with elements Array[ i + 1 ]....Array[ n - 1 ] is unsorted.
1) Best situation: the data has already been sorted. In this complexity of the best situation is never executed the inner loop, and the outer loop is executed n-1 times and the total complexity is O(n).
2) Worst situation: In this situation data in reverse order. The inner loop of the complexity has executed the maximum number of times. Thus the complexity of the insertion sort in this worst possible case is quadratic or O( n power 2)
In this example array of the size is 6. And the position of the element is Array is certainly sorted. i.e 0 is sorted. when 0 compare to other given element. We found that all element is greater than 0 that means 0< all given element.
Question1: Take an array of size 5. Sort (8, -6, 3, 17, 5) using insertion sort.
|8||-6 ||3||17||5||-6 to be inserted|
|?||8||3||17||5||8 > -6, shift|
|-6||8||3||17||5|| insert -6|
|-6||8||3||17||5||3 to be inserted|
|-6||?||8||17||5||8 > 3, shift|
|-6||3||8||17||5||-6 < 3, insert 3|
|-6||3||8||17||5||17 to be inserted|
|-6||3||8||17||5||8 < 17, insert 17|
|-6||3||8||17||5||5 to be inserted|
|-6||3||8||?||17||17 > 5, shift|
|-6||3||?||8||17||8 > 5, shift|
|-6||3||5||8||17||3<5, insert 5|