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cse 2050 exam 2 Questions with 100% Actual correct answers | verified | latest update | Graded A+ | Already Passed | Complete Solution

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cse 2050 exam 2 Questions with 100% Actual correct answers | verified | latest update | Graded A+ | Already Passed | Complete Solution

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  • June 24, 2024
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cse 2050 exam 2
recursion - ANS-technique of making a function fall itself, providing a way to break
complicated problems down into simple problems that are easier to solve

recursive function rules - ANS-1. there must be at least one base condition so that the
function eventually stops calling itself

2. the recursive step must make the function move towards the base case

function call stack - ANS-(often referred to just as the call stack or the stack) is
responsible for maintaining the local variables and parameters during function execution

calling a function pushes onto the stack, returning pops a frame off the stack

greedy algorithm - ANS-an algorithmic strategy that makes the best optimal choice at
each small stage with the goal of this eventually leading to a globally optimum solution

dynamic programming - ANS-storing previously calculated values to prevent repetitive
calculations and allow for quick retrievals as well as using smaller values to solve for
even larger ones

memoization - ANS-"top-down"
1. write the recursive function top-down
2. alter the function to check if we've already calculated the value
3. if so, use calculated value
4. if not, recursive call

pros of memoization - ANS-can be intuitive and memory saving, but has a larger
overhead because of recursion

tabulation - ANS-"bottom up"
start with solutions to the smallest problems and build solutions to the larger problems;
iterative solution

bubble sort - algorithm - ANS-def --sort(L):
has_swapped = True
while(has_swapped):
has_swapped = False

, for i in range(len(L)-1):
if L[i] > L[i+1]:
L[i], L[i+1] = L[i+1], L[i]
swaps += 1
has_swapped = True

if no elements swapped during the pass, can safely exit

insertion sort - algorithm - ANS-def --sort(L):
n = len(L)
i=0
while i < n:
j=n-i-1
while j < n - i - 1:
L[j], L[j+1] = L[j+1], L[j]
j += 1
i += 1

after iteration i, the last i elements are sorted; starts at position n-i-1 and goes to the end

selection sort - algorithm - ANS-def --sort(L):
n = len(L)
for i in range(n-1):
max_j = 0
for j in range(n-i):
if L[j] > L[max_j]:
max_j = j
if (n-i-1) != max_j:
L[n-i-1], L[max_j] = L[max_j], L[n-i-1]

binary search - ANS-a search algorithm that starts at the middle of a sorted set of
numbers and removes half of the data; this process repeats until the desired value is
found or all elements have been eliminated.

binary search - running time - ANS-asymptotic running time is o(log n)

tail recursion - ANS-a recursive function in which the recursive call is the last statement
that is executed by the function; nothing is left to execute after the recursion call

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