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In the array implementation of lists, elements are stored into continuous locations. In order to add an element into the list at the end, we can insert it without any problem. But, assume if we desire to add the element at the starting or middle of the list, then we ought to rewrite all the elements after the position where the element ought to be added. We ought to shift (n)th element to (n+1)th position, where 'n' refer to number of elements in the list. The (n-1)thelement to (n)th position and it will continue till the ( r ) thelement to ( r + 1 )th position, where 'r' refer to the position of insertion. For doing this, thecount will be incremented.
From the above instance, if we desire to add element '35' after element '33'. We ought to shift 77 to the 8th position, 66 to the 7th position, so on, 44 to the 5th position.
Before Insertion
Count 1 2 3 4 5 6 7
11
22
33
44
55
66
77
Step 1
Count 1 2 3 4 5 6 7 8
Step 2
Step 3
Step 4
Step 5
35
The Space - Time Trade Off The best algorithm to solve a given problem is one that needs less space in memory and takes less time to complete its implementation. But in practic
1. Start 2. Get h 3. If h T=288.15+(h*-0.0065) 4. else if h T=216.65 5. else if h T=216.65+(h*0.001) 6. else if h T=228.65+(h*0.0028) 7. else if h T=270.65 8.
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