Related Articles

# Find the minimum positive integer such that it is divisible by A and sum of its digits is equal to B

• Last Updated : 26 May, 2021

Given two integers A and B, the task is to find the minimum positive integer N such that N is divisible by A and the sum of the digits of N is equal to B. If number is not found then print -1.
Examples:

Input: A = 20, B = 30
Output: 49980
49980 is divisible by 20 and sum of its digit = 4 + 9 + 9 + 8 + 0 = 30
Input: A = 5, B = 2
Output: 20

Approach:

• Create empty queue q that stores the value of A and B and output number as a string and create integer type 2-D array visited[][] that stores the visited digit.
• Insert Node into queue and check if queue is non-empty.
• While the queue is non-empty, pop an element from the queue and for every digit from 1 to 9, concatenate the digit after the string num and check whether the number formed is the required number.
• If the required number is found, print the number.
• Else repeat the steps while the number is less than B and the queue is non-empty while pushing the non-visited number to the queue.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Array that stores visited digits``int` `visited[501][5001];` `// Structure for queue Node.``struct` `Node {``    ``int` `a, b;``    ``string str;``};` `// Function to return the minimum number such that it is``// divisible by 'a' and sum of its digits is equals to 'b'``int` `findNumber(``int` `a, ``int` `b)``{``    ``// Create queue``    ``queue q;` `    ``// Initially queue is empty``    ``Node temp = Node{ 0, 0, ``""` `};` `    ``// Initialize visited to 1``    ``visited[0][0] = 1;` `    ``// Push temp in queue``    ``q.push(temp);` `    ``// While queue is not empty``    ``while` `(!q.empty()) {` `        ``// Get the front of the queue and pop it``        ``Node u = q.front();``        ``q.pop();` `        ``// If popped element is the required number``        ``if` `(u.a == 0 && u.b == b)` `            ``// Parse int from string and return it``            ``return` `std::stoi(u.str);` `        ``// Loop for each digit and check the sum``        ``// If not visited then push it to the queue``        ``for` `(``int` `i = 0; i < 10; i++) {``            ``int` `dd = (u.a * 10 + i) % a;``            ``int` `ss = u.b + i;``            ``if` `(ss <= b && !visited[dd][ss]) {``                ``visited[dd][ss] = 1;``                ``q.push(Node{ dd, ss, u.str + ``char``(``'0'` `+ i) });``            ``}``        ``}``    ``}` `    ``// Required number not found return -1.``    ``return` `-1;``}` `// Driver code.``int` `main()``{``    ``int` `a = 25, b = 1;``    ``cout << findNumber(a, b);``    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;``class` `Solution``{`` ` `// Array that stores visited digits``static` `int` `visited[][]= ``new` `int``[``501``][``5001``];`` ` `// Structure for queue Node.``static` `class` `Node {``    ``int` `a, b;``    ``String str;``    ``Node(``int` `a1,``int` `b1,String s)``    ``{``        ``a=a1;``        ``b=b1;``        ``str=s;``    ``}``}`` ` `// Function to return the minimum number such that it is``// divisible by 'a' and sum of its digits is equals to 'b'``static` `int` `findNumber(``int` `a, ``int` `b)``{``    ``// Create queue``    ``Queue q= ``new` `LinkedList();`` ` `    ``// Initially queue is empty``    ``Node temp =``new`  `Node( ``0``, ``0``, ``""` `);`` ` `    ``// Initialize visited to 1``    ``visited[``0``][``0``] = ``1``;`` ` `    ``// Push temp in queue``    ``q.add(temp);`` ` `    ``// While queue is not empty``    ``while` `(q.size()!=``0``) {`` ` `        ``// Get the front of the queue and pop it``        ``Node u = q.peek();``        ``q.remove();`` ` `        ``// If popped element is the required number``        ``if` `(u.a == ``0` `&& u.b == b)`` ` `            ``// Parse int from string and return it``            ``return` `Integer.parseInt(u.str);`` ` `        ``// Loop for each digit and check the sum``        ``// If not visited then push it to the queue``        ``for` `(``int` `i = ``0``; i < ``10``; i++) {``            ``int` `dd = (u.a * ``10` `+ i) % a;``            ``int` `ss = u.b + i;``            ``if` `(ss <= b && visited[dd][ss]==``0``) {``                ``visited[dd][ss] = ``1``;``                ``q.add(``new` `Node( dd, ss, u.str + (``char``)(``'0'` `+ i) ));``            ``}``        ``}``    ``}`` ` `    ``// Required number not found return -1.``    ``return` `-``1``;``}`` ` `// Driver code.``public` `static` `void`  `main(String args[])``{``    ``//initialize visited``    ``for``(``int` `i=``0``;i<``500``;i++)``        ``for``(``int` `j=``0``;j<``500``;j++)``            ``visited[i][j]=``0``;``    ` `    ``int` `a = ``25``, b = ``1``;``    ``System.out.println(findNumber(a, b));``    ` `}``}``//contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach` `# Array that stores visited digits``visited ``=` `[[``0` `for` `x ``in` `range``(``501``)]``              ``for` `y ``in` `range``(``5001``)]` `# Structure for queue Node.``class` `Node:``    ` `    ``def` `__init__(``self``, a, b, string):``        ``self``.a ``=` `a``        ``self``.b ``=` `b``        ``self``.string ``=` `string` `# Function to return the minimum number``# such that it is divisible by 'a' and``# sum of its digits is equals to 'b'``def` `findNumber(a, b):` `    ``# Use list as queue``    ``q ``=` `[]` `    ``# Initially queue is empty``    ``temp ``=` `Node(``0``, ``0``, "")` `    ``# Initialize visited to 1``    ``visited[``0``][``0``] ``=` `1` `    ``# Push temp in queue``    ``q.append(temp)` `    ``# While queue is not empty``    ``while` `len``(q) > ``0``:` `        ``# Get the front of the queue``        ``# and pop it``        ``u ``=` `q.pop(``0``)` `        ``# If popped element is the``        ``# required number``        ``if` `u.a ``=``=` `0` `and` `u.b ``=``=` `b:` `            ``# Parse int from string``            ``# and return it``            ``return` `int``(u.string)` `        ``# Loop for each digit and check the sum``        ``# If not visited then push it to the queue``        ``for` `i ``in` `range``(``0``, ``10``):``            ``dd ``=` `(u.a ``*` `10` `+` `i) ``%` `a``            ``ss ``=` `u.b ``+` `i``            ` `            ``if` `ss <``=` `b ``and` `visited[dd][ss] ``=``=` `False``:``                ``visited[dd][ss] ``=` `1``                ``q.append(Node(dd, ss, u.string ``+` `str``(i)))` `    ``# Required number not found return -1.``    ``return` `-``1` `# Driver code.``if` `__name__ ``=``=` `"__main__"``:` `    ``a, b ``=` `25``, ``1``    ``print``(findNumber(a, b))``    ` `# This code is contributed by Rituraj Jain`
Output:
`100`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up