Escrito por estudiantes que aprobaron Inmediatamente disponible después del pago Leer en línea o como PDF ¿Documento equivocado? Cámbialo gratis 4,6 TrustPilot
logo-home
Examen

WGU C949 EXAM QUESTIONS WELL ANSWERED LATEST

Puntuación
-
Vendido
-
Páginas
83
Grado
A+
Subido en
14-11-2025
Escrito en
2025/2026

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) {

Mostrar más Leer menos
Institución
C949
Grado
C949

Vista previa del contenido

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

Escuela, estudio y materia

Institución
C949
Grado
C949

Información del documento

Subido en
14 de noviembre de 2025
Número de páginas
83
Escrito en
2025/2026
Tipo
Examen
Contiene
Preguntas y respuestas

Temas

$12.99
Accede al documento completo:

¿Documento equivocado? Cámbialo gratis Dentro de los 14 días posteriores a la compra y antes de descargarlo, puedes elegir otro documento. Puedes gastar el importe de nuevo.
Escrito por estudiantes que aprobaron
Inmediatamente disponible después del pago
Leer en línea o como PDF

Conoce al vendedor

Seller avatar
Los indicadores de reputación están sujetos a la cantidad de artículos vendidos por una tarifa y las reseñas que ha recibido por esos documentos. Hay tres niveles: Bronce, Plata y Oro. Cuanto mayor reputación, más podrás confiar en la calidad del trabajo del vendedor.
joshuawesonga22 Liberty University
Seguir Necesitas iniciar sesión para seguir a otros usuarios o asignaturas
Vendido
110
Miembro desde
1 año
Número de seguidores
2
Documentos
14773
Última venta
1 día hace
Tutor Wes

Hi there! I'm Tutor Wes, a dedicated tutor with a passion for sharing knowledge and helping others succeed academically. All my notes are carefully organized, detailed, and easy to understand. Whether you're preparing for exams, catching up on lectures, or looking for clear summaries, you'll find useful study materials here. Let’s succeed together!

3.4

12 reseñas

5
4
4
1
3
4
2
2
1
1

Por qué los estudiantes eligen Stuvia

Creado por compañeros estudiantes, verificado por reseñas

Calidad en la que puedes confiar: escrito por estudiantes que aprobaron y evaluado por otros que han usado estos resúmenes.

¿No estás satisfecho? Elige otro documento

¡No te preocupes! Puedes elegir directamente otro documento que se ajuste mejor a lo que buscas.

Paga como quieras, empieza a estudiar al instante

Sin suscripción, sin compromisos. Paga como estés acostumbrado con tarjeta de crédito y descarga tu documento PDF inmediatamente.

Student with book image

“Comprado, descargado y aprobado. Así de fácil puede ser.”

Alisha Student

Preguntas frecuentes