Find the minimum number of moves for Sum(A) <= B

Question

You are given two non-negative integers A and B. Let the function Sum(A) be the sum of digits of the number A.
You can perform any number of moves (possibly zeros).

In one move you can increase the value of A by 1.

Find the minimum number of moves you have to perform to make sum(A) <= B.

Constraints:

1 <= A <= 10^12

1 <= B <= 108

Sample Test Cases:

1. A = 1, B = 1       Output = 0
2. A = 9, B = 2       Output = 1
3. A = 555, B = 10    Output = 45

Inputs are given in string.

How to solve this problem in python?

Here is the code I have tried:

def countmoves(A, B):

    int_A = int(A)
    int_B = int(B)
    count = 0
    
    while True:
        digits = [int(i) for i in str(int_A)]
        s_m = sum(digits)
        if s_m <= int_B:
            print(f"count: {count}")
            return count
        else:
            int_A += 1
            count += 1

But my code is getting Timed Out. How can I optimize it?

Asked By: JamesHunt

||

Source

Answer 1

You could do it by finding the rightmost non-zero number and adding a value such that the element to the left is increased by 1 and all to the right (inclusive) become 0:

def countmoves(A, B): #nin17()
    count = 0
    B = int(B)
    while True:
        total = sum(int(i) for i in A)
        if total <= B:
            return count
        else:
            index = max(A.rfind(i) for i in ['1', '2', '3', '4', '5', '6', '7', '8', '9'])
            right = A[index:]
            moves = 10**len(right)-int(right)
            A = f'{int(A)+moves}'
            count += moves

Or with a for loop instead, looping from right to left through A:

def countmoves(A, B): #nin172()
    count = 0
    B = int(B)
    for i, j in enumerate(range(len(A)-1, -2, -1), 1):
        total = sum(int(i) for i in A)
        if total <= B:
            return count
        else:
            moves = 10**i-int(A[j:])
            A = f'{int(A)+moves}'
            count += moves

Timings (with given test cases):

for A, B in (('1', '1'), ('9', '2'), ('555', '10')):
    %timeit op(A, B)
    %timeit nin17(A, B) #while
    %timeit nin172(A, B) #for

Output:

600 ns ± 9.37 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each) #OP
473 ns ± 2.1 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each) #Nin17 while
732 ns ± 1.94 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each) #Nin17 for
1.12 µs ± 8.3 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each) #OP
2.6 µs ± 44.3 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each) #Nin17 while
1.58 µs ± 2.49 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each) #Nin17 for
26.6 µs ± 63.3 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each) #OP
5.18 µs ± 7.37 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each) #Nin17 while
2.89 µs ± 17.3 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each) #Nin17 for

Timings with larger values:

A = '363513974325'
B = '28'
assert op(A, B) == nin17(A, B)
assert nin17(A, B) == nin172(A, B)
%timeit op(A, B)
%timeit nin17(A, B)
%timeit nin172(A, B)

Output:

45.2 ms ± 66.4 µs per loop (mean ± std. dev. of 7 runs, 10 loops each) #OP
19.3 µs ± 197 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each) #Nin17 while
12.2 µs ± 18.4 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each) #Nin17 for

Answered By: Nin17

Answer 2

def digit_sum(A):
    return sum(list(map(int, str(A).strip())))

def countmoves(A, B):
    final_moves = 0
    place_holder = 0

    while digit_sum(A) > B:
        place_holder += 1
        remainder = A % (10 ** place_holder)
        if (remainder):
            moves = (10 ** place_holder) - remainder
            A = A + moves
            final_moves += moves

    return(final_moves)

Answered By: chawlajc

Find the minimum number of moves for Sum(A) <= B

Question:

Answers: