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Find the minimum positive integer such that it is divisible by A and sum of its digits is equal to B in Python
Suppose we have two numbers A and B, we have to find the minimum positive number M so that M is divisible by A and the sum of the digits of M is same as B. So, if there is no such result, then return -1.
So, if the input is like A = 50, B = 2, then the output will be 200 as this is divisible by 50 and sum of its digit = 2 + 0 + 0 = 2.
To solve this, we will follow these steps −
Define one element type container, that contains two numbers a and b and one string
que := a new list
elem := a new element with (0, 0, blank string)
visited[0, 0] := 1
insert elem at the end of que
while size of que > 0, do
temp_elem := delete first element from que
if temp_elem.a is 0 and temp_elem.b is b, then
return integer of temp_elem.string
for i in range 0 to 9, do
x :=(temp_elem.a * 10 + i) mod a
y := temp_elem.b + i
if y <= b and visited[x, y] is False, then
visited[x, y] := 1
insert new element with x, y and temp_elem.string concatenate i into que
return -1
Example
Let us see the following implementation to get better understanding −
visited = [[0 for x in range(501)] for y in range(5001)] class Element: def __init__(self, a, b, string): self.a = a self.b = b self.string = string def get_number(a, b): que = [] elem = Element(0, 0, "") visited[0][0] = 1 que.append(elem) while len(que) > 0: temp_elem = que.pop(0) if temp_elem.a == 0 and temp_elem.b == b: return int(temp_elem.string) for i in range(0, 10): x = (temp_elem.a * 10 + i) % a y = temp_elem.b + i if y <= b and visited[x][y] == False: visited[x][y] = 1 que.append(Element(x, y, temp_elem.string + str(i))) return -1 a, b = 50, 2 print(get_number(a, b))
Input
50, 2
Output
200
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