# Minimize difference between any multiple of given three numbers with K

Given 4 integers A, B, C, and K the task is to find the minimum difference between any multiple of A, B, and C with K

Examples:

Input: A = 5, B =  4, C =  8, K = 9
output: 1
Explanation: Closest multiple of A, B and C greater than 9 is 10. Therefore, minimum difference is 10-9 = 1

Input: A = 6, B = 10, C = 9, K = 2
Output: 4
Explanation: Closest multiple of A, B and C greater than 2 is 6. Therefore, minimum difference is 6-2 = 4

Naive Approach: The task can be solved by using a for-loop to get the next multiples of A, B, and C, by keeping track of the closest multiple just greater than K Follow the below steps to solve the problem:

• Initialize 3 boolean variables say fa, fb, and fc to denote whether the multiples of A, B, and C becomes greater than K
• Start multiplying A, B, and C with cur initialized with 1
• As soon as, one of the three numbers becomes greater than K, store the difference between the nearest multiple and K accordingly

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the difference between` `// closest multiple of A, B, C with K` `void` `solve(``int` `A, ``int` `B, ``int` `C, ``int` `K)` `{`   `    ``// Stores the minimum difference` `    ``int` `ans = INT_MAX;`   `    ``// Check whether multiples of A, B and C becomes` `    ``// greater than K` `    ``bool` `fa = ``false``, fb = ``false``, fc = ``false``;`   `    ``// Start multiplication from 1` `    ``int` `cur = 1;` `    ``while` `(1) {`   `        ``// Finding multiples` `        ``A *= cur;` `        ``B *= cur;` `        ``C *= cur;`   `        ``// All the multiples becomes` `        ``// greater than K` `        ``if` `(fa && fb && fb)` `            ``break``;`   `        ``// Multiple of A` `        ``if` `(!fa) {`   `            ``// Valid multiple` `            ``if` `(A >= K) {`   `                ``// Minimize ans` `                ``ans = min(ans, A - K);` `                ``fa = ``true``;` `            ``}` `        ``}`   `        ``// Multiple of B` `        ``if` `(!fb) {`   `            ``// Valid multiple` `            ``if` `(B >= K) {`   `                ``// Minimize ans` `                ``ans = min(ans, B - K);` `                ``fb = ``true``;` `            ``}` `        ``}`   `        ``// Multiple of C` `        ``if` `(!fc) {`   `            ``// Valid multiple` `            ``if` `(C >= K) {`   `                ``// Minimize ans` `                ``ans = min(ans, C - K);` `                ``fc = ``true``;` `            ``}` `        ``}`   `        ``cur++;` `    ``}`   `    ``// Resultant answer` `    ``cout << ans << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `A = 6, B = 10, C = 9, K = 2;` `    ``solve(A, B, C, K);` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `class` `GFG{`   `// Function to find the difference between` `// closest multiple of A, B, C with K` `static` `void` `solve(``int` `A, ``int` `B, ``int` `C, ``int` `K)` `{`   `    ``// Stores the minimum difference` `    ``int` `ans = Integer.MAX_VALUE;`   `    ``// Check whether multiples of A, B and C becomes` `    ``// greater than K` `    ``boolean` `fa = ``false``, fb = ``false``, fc = ``false``;`   `    ``// Start multiplication from 1` `    ``int` `cur = ``1``;` `    ``while` `(``true``) {`   `        ``// Finding multiples` `        ``A *= cur;` `        ``B *= cur;` `        ``C *= cur;`   `        ``// All the multiples becomes` `        ``// greater than K` `        ``if` `(fa && fb && fb)` `            ``break``;`   `        ``// Multiple of A` `        ``if` `(!fa) {`   `            ``// Valid multiple` `            ``if` `(A >= K) {`   `                ``// Minimize ans` `                ``ans = Math.min(ans, A - K);` `                ``fa = ``true``;` `            ``}` `        ``}`   `        ``// Multiple of B` `        ``if` `(!fb) {`   `            ``// Valid multiple` `            ``if` `(B >= K) {`   `                ``// Minimize ans` `                ``ans = Math.min(ans, B - K);` `                ``fb = ``true``;` `            ``}` `        ``}`   `        ``// Multiple of C` `        ``if` `(!fc) {`   `            ``// Valid multiple` `            ``if` `(C >= K) {`   `                ``// Minimize ans` `                ``ans = Math.min(ans, C - K);` `                ``fc = ``true``;` `            ``}` `        ``}`   `        ``cur++;` `    ``}`   `    ``// Resultant answer` `    ``System.out.print(ans +``"\n"``);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `A = ``6``, B = ``10``, C = ``9``, K = ``2``;` `    ``solve(A, B, C, K);` `}` `}`   `// This code is contributed by shikhasingrajput`

## Python3

 `# Python Program to implement` `# the above approach`   `# Function to find the difference between` `# closest multiple of A, B, C with K` `def` `solve(A, B, C, K):`   `    ``# Stores the minimum difference` `    ``ans ``=` `10` `*``*` `9`   `    ``# Check whether multiples of A, B and C becomes` `    ``# greater than K` `    ``fa ``=` `False` `    ``fb ``=` `False` `    ``fc ``=` `False`   `    ``# Start multiplication from 1` `    ``cur ``=` `1` `    ``while` `(``1``):`   `        ``# Finding multiples` `        ``A ``*``=` `cur` `        ``B ``*``=` `cur` `        ``C ``*``=` `cur`   `        ``# All the multiples becomes` `        ``# greater than K` `        ``if` `(fa ``and` `fb ``and` `fb):` `            ``break`   `        ``# Multiple of A` `        ``if` `(``not` `fa):`   `            ``# Valid multiple` `            ``if` `(A >``=` `K):`   `                ``# Minimize ans` `                ``ans ``=` `min``(ans, A ``-` `K)` `                ``fa ``=` `True`   `        ``# Multiple of B` `        ``if` `(``not` `fb):`   `            ``# Valid multiple` `            ``if` `(B >``=` `K):`   `                ``# Minimize ans` `                ``ans ``=` `min``(ans, B ``-` `K)` `                ``fb ``=` `True`   `        ``# Multiple of C` `        ``if` `(``not` `fc):`   `            ``# Valid multiple` `            ``if` `(C >``=` `K):`   `                ``# Minimize ans` `                ``ans ``=` `min``(ans, C ``-` `K)` `                ``fc ``=` `True`   `        ``cur ``+``=` `1`   `    ``# Resultant answer` `    ``print``(ans)`   `# Driver Code` `A ``=` `6` `B ``=` `10` `C ``=` `9` `K ``=` `2` `solve(A, B, C, K)`   ` ``# This code is contributed by saurabh_jaiswal.`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {`   `// Function to find the difference between` `// closest multiple of A, B, C with K` `static` `void` `solve(``int` `A, ``int` `B, ``int` `C, ``int` `K)` `{`   `    ``// Stores the minimum difference` `    ``int` `ans = Int32.MaxValue;`   `    ``// Check whether multiples of A, B and C becomes` `    ``// greater than K` `    ``bool` `fa = ``false``, fb = ``false``, fc = ``false``;`   `    ``// Start multiplication from 1` `    ``int` `cur = 1;` `    ``while` `(``true``) {`   `        ``// Finding multiples` `        ``A *= cur;` `        ``B *= cur;` `        ``C *= cur;`   `        ``// All the multiples becomes` `        ``// greater than K` `        ``if` `(fa && fb && fb)` `            ``break``;`   `        ``// Multiple of A` `        ``if` `(!fa) {`   `            ``// Valid multiple` `            ``if` `(A >= K) {`   `                ``// Minimize ans` `                ``ans = Math.Min(ans, A - K);` `                ``fa = ``true``;` `            ``}` `        ``}`   `        ``// Multiple of B` `        ``if` `(!fb) {`   `            ``// Valid multiple` `            ``if` `(B >= K) {`   `                ``// Minimize ans` `                ``ans = Math.Min(ans, B - K);` `                ``fb = ``true``;` `            ``}` `        ``}`   `        ``// Multiple of C` `        ``if` `(!fc) {`   `            ``// Valid multiple` `            ``if` `(C >= K) {`   `                ``// Minimize ans` `                ``ans = Math.Min(ans, C - K);` `                ``fc = ``true``;` `            ``}` `        ``}`   `        ``cur++;` `    ``}`   `    ``// Resultant answer` `    ``Console.Write(ans +``"\n"``);` `}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `A = 6, B = 10, C = 9, K = 2;` `        ``solve(A, B, C, K);` `    ``}` `}`   `// This code is contributed by sanjoy_62.`

## Javascript

 ``

Output

`4`

Time Complexity: O(min(K/A, K/B, K/C))
Auxiliary Space: O(1)

Efficient Approach: Above approach can be generalized into a formula: min(⌈K/A⌉*A, ⌈K/B⌉*B, ⌈K/C⌉*C) − K.

• ⌈K/A⌉*A gives the nearest multiple of A just greater than K
• ⌈K/B⌉*B gives the nearest multiple of B just greater than K
• ⌈K/C⌉*C gives the nearest multiple of C just greater than K
• Difference of minimum of all these with K gives the desired result

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the difference between` `// closest multiple of A, B, C with K` `void` `solve(``int` `A, ``int` `B, ``int` `C, ``int` `K)` `{` `    ``// Stores the multiples of A, B, C` `    ``// just greater than K` `    ``int` `fa, fb, fc;`   `    ``// Multiple of A` `    ``if` `((K % A) != 0) {`   `        ``fa = ((K / A) + 1) * A;` `    ``}` `    ``else` `{` `        ``fa = (K / A) * A;` `    ``}`   `    ``// Multiple of B` `    ``if` `((K % B) != 0) {`   `        ``fb = ((K / B) + 1) * B;` `    ``}` `    ``else` `{` `        ``fb = (K / B) * B;` `    ``}`   `    ``// Multiple of C` `    ``if` `((K % C) != 0) {`   `        ``fc = ((K / C) + 1) * C;` `    ``}` `    ``else` `{` `        ``fc = (K / C) * C;` `    ``}`   `    ``// Store the resultant answer` `    ``int` `ans;` `    ``ans = min(fa - K, min(fc - K, fb - K));` `    ``cout << ans << endl;` `}`   `// Drive code` `int` `main()` `{` `    ``int` `A = 6, B = 10, C = 9, K = 2;` `    ``solve(A, B, C, K);` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `public` `class` `GFG {` `    `  `    ``// Function to find the difference between` `    ``// closest multiple of A, B, C with K` `    ``static` `void` `solve(``int` `A, ``int` `B, ``int` `C, ``int` `K)` `    ``{` `      `  `        ``// Stores the multiples of A, B, C` `        ``// just greater than K` `        ``int` `fa, fb, fc;` `    `  `        ``// Multiple of A` `        ``if` `((K % A) != ``0``) {` `    `  `            ``fa = ((K / A) + ``1``) * A;` `        ``}` `        ``else` `{` `            ``fa = (K / A) * A;` `        ``}` `    `  `        ``// Multiple of B` `        ``if` `((K % B) != ``0``) {` `    `  `            ``fb = ((K / B) + ``1``) * B;` `        ``}` `        ``else` `{` `            ``fb = (K / B) * B;` `        ``}` `    `  `        ``// Multiple of C` `        ``if` `((K % C) != ``0``) {` `    `  `            ``fc = ((K / C) + ``1``) * C;` `        ``}` `        ``else` `{` `            ``fc = (K / C) * C;` `        ``}` `    `  `        ``// Store the resultant answer` `        ``int` `ans;` `        ``ans = Math.min(fa - K, Math.min(fc - K, fb - K));` `        ``System.out.println(ans);` `    ``}`   `    ``// Drive code` `    ``public` `static` `void` `main (String[] args)` `    ``{` `        ``int` `A = ``6``, B = ``10``, C = ``9``, K = ``2``;` `        ``solve(A, B, C, K);` `    ``}`   `}`   `// This code is contributed by AnkThon`

## Python3

 `# Python3 program for the above approach`   `# Function to find the difference between` `# closest multiple of A, B, C with K` `def` `solve( A, B, C, K) :`   `    ``# Stores the multiples of A, B, C` `    ``# just greater than K`   `    ``# Multiple of A` `    ``if` `((K ``%` `A) !``=` `0``) :` `        ``fa ``=` `((K ``/``/` `A) ``+` `1``) ``*` `A;` `    ``else` `:` `        ``fa ``=` `(K ``/``/` `A) ``*` `A;`   `    ``# Multiple of B` `    ``if` `((K ``%` `B) !``=` `0``) :`   `        ``fb ``=` `((K ``/``/` `B) ``+` `1``) ``*` `B;` `    `  `    ``else` `:` `        ``fb ``=` `(K ``/``/` `B) ``*` `B;`   `    ``# Multiple of C` `    ``if` `((K ``%` `C) !``=` `0``) :`   `        ``fc ``=` `((K ``/``/` `C) ``+` `1``) ``*` `C;` `    ``else` `:` `        ``fc ``=` `(K ``/``/` `C) ``*` `C;`   `    ``# Store the resultant answer` `    ``ans ``=` `min``(fa ``-` `K, ``min``(fc ``-` `K, fb ``-` `K));` `    `  `    ``print``(ans);`   `# Drive code` `if` `__name__ ``=``=` `"__main__"` `:`   `    ``A ``=` `6``; B ``=` `10``; C ``=` `9``; K ``=` `2``;` `    ``solve(A, B, C, K);`   `    ``# This code is contributed by AnkThon`

## C#

 `// C# program for the above approach` `using` `System;` `public` `class` `GFG {`   `    ``// Function to find the difference between` `    ``// closest multiple of A, B, C with K` `    ``static` `void` `solve(``int` `A, ``int` `B, ``int` `C, ``int` `K)` `    ``{`   `        ``// Stores the multiples of A, B, C` `        ``// just greater than K` `        ``int` `fa, fb, fc;`   `        ``// Multiple of A` `        ``if` `((K % A) != 0) {`   `            ``fa = ((K / A) + 1) * A;` `        ``}` `        ``else` `{` `            ``fa = (K / A) * A;` `        ``}`   `        ``// Multiple of B` `        ``if` `((K % B) != 0) {`   `            ``fb = ((K / B) + 1) * B;` `        ``}` `        ``else` `{` `            ``fb = (K / B) * B;` `        ``}`   `        ``// Multiple of C` `        ``if` `((K % C) != 0) {`   `            ``fc = ((K / C) + 1) * C;` `        ``}` `        ``else` `{` `            ``fc = (K / C) * C;` `        ``}`   `        ``// Store the resultant answer` `        ``int` `ans;` `        ``ans = Math.Min(fa - K, Math.Min(fc - K, fb - K));` `        ``Console.WriteLine(ans);` `    ``}`   `    ``// Drive code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``int` `A = 6, B = 10, C = 9, K = 2;` `        ``solve(A, B, C, K);` `    ``}` `}`   `// This code is contributed by ukasp.`

## Javascript

 ``

Output

`4`

Time Complexity: O(1)
Auxiliary Space: O(1)

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