I just need the algorithm (how you would solve the problem in words). Pseudocode
ID: 3916596 • Letter: I
Question
I just need the algorithm (how you would solve the problem in words). Pseudocode, non language specific. I pretty much did the first question, but I need help with the other two.
Thanks!
Generate 1000 random numbers from 1-100 max min avg
initialize num = 1000
algorithm
srand(time(NULL));
int arr[num], max, min, avg;
for every int i less than num,
arr[i] = rand() % 100 + 1;
end
if i is less than the next i, that is the new min
if i is more than the next i, that is the new max
for every int i less than num,
total += arr[i];
avg = total/num;
cout avg;
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Generate 100 random numbers from 500 - 700, How many numbers are in the 500s? 600s?
How many numbers repeat? How many times? (show repeats and numbers of repeats). Try for 100 numbers, 10000 numbers, and 1 million numbers.
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Enter 3 words (user input). How many shared letters in 2 words? 3 words? Sort the words A->Z. *Bonus: ensure that you can sort words that have the same starting letter.
Explanation / Answer
If you post more than 1 question, as per chegg guidelines I have to solve only first question.
Ques 1.
initialize num = 100
algorithm
srand(time(NULL));
int arr[num], max, min, avg;
for every int i less than num,
// generate random number in range 500-700
end
count_500 = 0;
count_600 = 0;
for i = 1 to arr.length()
// if current number is in 500s
if arr[i] >= 500 && arr[i] < 600
count_500++;
// if current number is in 600s
else if arr[i] >= 600 && arr[i] < 700
count_700++;
end
end
// the key is no and the value is the number of times
// it appears
map<int , int> x;
for i = 1 to arr.length()
// if current number is in map
if x.contains( arr[i] )
// increase the number of times current element
// appears in x
x.get(arr[i]) += 1;
// if current number is not in map
else
x.add( arr[i] , 1 )
end
end
for i = 1 to x.length()
// if current number appears more than once
if x.get( arr[i] > 1 )
print(arr[i] + “ appears ” + x.get(arr[i]) + “ times.”);
end
end