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WGU C949 EXAM QUESTIONS WELL ANSWERED LATEST

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WGU C949 EXAM QUESTIONS WELL ANSWERED LATEST def find(lst, item, low, high, indent): """ Finds index of string in list of strings, else -1. Searches only the index range low to high Note: Upper/Lower case characters matter """ print(indent, 'find() range', low, high) range_size = (high - low) + 1 mid = (high + low) // 2 if item == lst[mid]: # Base case 1: Found at mid print(indent, 'Found person.') pos = mid elif range_size == 1: # Base case 2: Not found print(indent, 'Person not found.') pos = 0 else: # Recursive search: Search lower or upper half if item lst[mid]: # Search lower half print(indent, 'Searching lower half.') pos = find(lst, item, low, mid, indent + ' ') else: # Search upper half print(indent, 'Searching upper half.') pos = find(lst, item, mid+1, high, indent + ' ') print(indent, 'Returning pos = %d.' % pos) return pos attendees = [] d('Adams, Mary') d('Carver, Michael') d('Domer, Hugo') - Answers True def find(lst, item, low, high, indent): """ Finds index of string in list of strings, else -1. Searches only the index range low to high Note: Upper/Lower case characters matter """ print(indent, 'find() range', low, high) range_size = (high - low) + 1 mid = (high + low) // 2 if item == lst[mid]: # Base case 1: Found at mid print(indent, 'Found person.') pos = mid elif range_size == 1: # Base case 2: Not found print(indent, 'Person not found.') pos = 0 else: # Recursive search: Search lower or upper half if item lst[mid]: # Search lower half print(indent, 'Searching lower half.') pos = find(lst, item, low, mid, indent + ' ') else: # Search upper half print(indent, 'Searching upper half.') pos = find(lst, item, mid+1, high, indent + ' ') print(indent, 'Returning pos = %d.' % pos) return pos attendees = [] d('Adams, Mary') d('Carver, Michael') d('Domer, Hugo') - Answers True def find(lst, item, low, high, indent): """ Finds index of string in list of strings, else -1. Searches only the index range low to high Note: Upper/Lower case characters matter """ print(indent, 'find() range', low, high) range_size = (high - low) + 1 mid = (high + low) // 2 if item == lst[mid]: # Base case 1: Found at mid print(indent, 'Found person.') pos = mid elif range_size == 1: # Base case 2: Not found print(indent, 'Person not found.') pos = 0 else: # Recursive search: Search lower or upper half if item lst[mid]: # Search lower half print(indent, 'Searching lower half.') pos = find(lst, item, low, mid, indent + ' ') else: # Search upper half print(indent, 'Searching upper half.') pos = find(lst, item, mid+1, high, indent + ' ') print(indent, 'Returning pos = %d.' % pos) return pos attendees = [] d('Adams, Mary') d('Carver, Michael') d('Domer, Hugo') - Answers False A recursive function with parameter n counts up from any negative number to 0. An appropriate base case would be n == 0. - Answers True A recursive function can have two base cases, such as n == 0 returning 0, and n == 1, returning 1. - Answers True n factorial (n!) is commonly implemented as a recursive function due to being easier to understand and executing faster than a loop implementation. - Answers False In the code below, suppose str1 is a pointer or reference to a string. The code only executes in constant time if the assignment copies the pointer/reference, and not all the characters in the string. srt2 = str1 - Answers True Certain hardware may execute division more slowly than multiplication, but both may still be constant time operations. - Answers True The hardware running the code is the only thing that affects what is and what is not a constant time operation. - Answers False Computational complexity analysis allows the efficiency of algorithms to be compared. - Answers True Two different algorithms that produce the same result have the same computational complexity. - Answers False Runtime and memory usage are the only two resources making up computational complexity. - Answers False Nearly every algorithm has a best-case time complexity when N = 0. - Answers False An algorithm's best and worst case scenarios are always different. - Answers False GetEvent(list, listSize) { i = 0 evensList = Create new, empty list while (i listSize) {

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Voorbeeld van de inhoud

WGU C949 EXAM QUESTIONS WELL ANSWERED LATEST 2025-2026

def find(lst, item, low, high, indent):

"""

Finds index of string in list of strings, else -1.

Searches only the index range low to high

Note: Upper/Lower case characters matter

"""

print(indent, 'find() range', low, high)

range_size = (high - low) + 1

mid = (high + low) // 2

if item == lst[mid]: # Base case 1: Found at mid

print(indent, 'Found person.')

pos = mid

elif range_size == 1: # Base case 2: Not found

print(indent, 'Person not found.')

pos = 0

else: # Recursive search: Search lower or upper half

if item < lst[mid]: # Search lower half

print(indent, 'Searching lower half.')

pos = find(lst, item, low, mid, indent + ' ')

else: # Search upper half

print(indent, 'Searching upper half.')

pos = find(lst, item, mid+1, high, indent + ' ')

print(indent, 'Returning pos = %d.' % pos)

return pos

,attendees = []

attendees.append('Adams, Mary')

attendees.append('Carver, Michael')

attendees.append('Domer, Hugo')

attendees.ap - Answers True

def find(lst, item, low, high, indent):

"""

Finds index of string in list of strings, else -1.

Searches only the index range low to high

Note: Upper/Lower case characters matter

"""

print(indent, 'find() range', low, high)

range_size = (high - low) + 1

mid = (high + low) // 2

if item == lst[mid]: # Base case 1: Found at mid

print(indent, 'Found person.')

pos = mid

elif range_size == 1: # Base case 2: Not found

print(indent, 'Person not found.')

pos = 0

else: # Recursive search: Search lower or upper half

if item < lst[mid]: # Search lower half

print(indent, 'Searching lower half.')

pos = find(lst, item, low, mid, indent + ' ')

else: # Search upper half

,print(indent, 'Searching upper half.')

pos = find(lst, item, mid+1, high, indent + ' ')

print(indent, 'Returning pos = %d.' % pos)

return pos

attendees = []

attendees.append('Adams, Mary')

attendees.append('Carver, Michael')

attendees.append('Domer, Hugo')

attendees.ap - Answers True

def find(lst, item, low, high, indent):

"""

Finds index of string in list of strings, else -1.

Searches only the index range low to high

Note: Upper/Lower case characters matter

"""

print(indent, 'find() range', low, high)

range_size = (high - low) + 1

mid = (high + low) // 2

if item == lst[mid]: # Base case 1: Found at mid

print(indent, 'Found person.')

pos = mid

elif range_size == 1: # Base case 2: Not found

print(indent, 'Person not found.')

pos = 0

else: # Recursive search: Search lower or upper half

, if item < lst[mid]: # Search lower half

print(indent, 'Searching lower half.')

pos = find(lst, item, low, mid, indent + ' ')

else: # Search upper half

print(indent, 'Searching upper half.')

pos = find(lst, item, mid+1, high, indent + ' ')

print(indent, 'Returning pos = %d.' % pos)

return pos

attendees = []

attendees.append('Adams, Mary')

attendees.append('Carver, Michael')

attendees.append('Domer, Hugo')

attendees.ap - Answers False

A recursive function with parameter n counts up from any negative number to 0. An appropriate
base case would be n == 0. - Answers True

A recursive function can have two base cases, such as n == 0 returning 0, and n == 1, returning 1.
- Answers True

n factorial (n!) is commonly implemented as a recursive function due to being easier to
understand and executing faster than a loop implementation. - Answers False

In the code below, suppose str1 is a pointer or reference to a string. The code only executes in
constant time if the assignment copies the pointer/reference, and not all the characters in the
string.

srt2 = str1 - Answers True

Certain hardware may execute division more slowly than multiplication, but both may still be
constant time operations. - Answers True

The hardware running the code is the only thing that affects what is and what is not a constant
time operation. - Answers False

Computational complexity analysis allows the efficiency of algorithms to be compared. -
Answers True

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