How to make this code loop again and again?

Question

arr = [1,2,3]
Count = 0
step-1 sort the array
Step-2 check if min(arr)==max(arr): break
else:
Step-3- iterate over the arr in range (len(arr)-1):
arr[i]+=1
Count+=1
step-4 now again check if min(arr) == max(arr): break
else: go to step 1 and repeat until all the elements become equal...

For eg:
[1,2,3] #here 1(min) and 3(max)..so by leaving the max element..we have to make the min element equal to to 3..so this is how we do it..
1 2 3.
2 3 3. 1step
3 4 3. 2step
4 4 4. < Every element is equal now...and it was a 3 step process..so the Count = 3

I am having problems in implementation..so i need a lil help..

I made a working solution, but it take too long to give a solution for a very large array..like in range(1000)
Link>

https://code.sololearn.com/c1W10i4qNI3w/?ref=app

Answer 1

Oh...but new version is better

Answer 2

Well check out the previous question i asked, i tried a recursive solution but recursion is a new thing for me right now..Lily Mea

Answer 3

are all numbers different?

Answer 4

do u mean u want to increase other values in the list till it equal the max value in the list?

Answer 5

Yes..Lily Mea ...but the max value keeps changing..but if the code is looping over the previous max value..the loop doesn't care about the new max value until it is finished...i don't how to change the max value while in a loop..

Answer 6

i mean let say u have a list [1,2,4,5]
u want to keep increasing 1,2,4 till it becomes 5 and stop the loop?

Answer 7

why dont u try recursion

Answer 8

See how it counts:
1 2 4 5
2 3 5 5. (added 1 to every elem..except the max)
3 4 6 5. (+1)
4 5 7 5 (+1)
5 6 8 5 < the min element became equal to max..but the max changed to '8', so i need to make them equal again..
after sorting..
5 5 6 8
6 6 7 8
7 7 8 8
8 8 9 8...again max changed to 9, so again
Sort
8 8 8 9
9 9 9 9

it took 10 steps to equalize..so the count = 10

Lily Mea

Answer 9

in recursion u put the algorithm to a function and keep calling it till it break

for example
arr = [1,2,3,8,2,3]

def func(arr) :
if(min(arr) == max(arr)):
return;

#when this if statement true,the recursion break, else

ind = arr.index(min(arr))
#get the index of min value

arr[ind] += 1
#adding 1

func(arr)
#call this function back

func(arr)
#call the func
print(arr)
#print the result

but still,the larger the list, the larger time it takes to run...
im new to this programming thing too,but i think that probably work

Answer 10

Well, it did make every element equal to 5..but that's not a solution for the question..Great effort tho!👍 Lily Mea

Answer 11

The solution that you gave, got accepted..and it was really fast..thanks again..
Oma Falk

Answer 12

So can you implement the new version?
And paste your solution here? Oma Falk