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# Minimum repeated addition of even divisors of N required to convert N to M

• Last Updated : 15 Jun, 2021

Given two numbers N and M, the task is to find the minimum operations required to convert N to M by repeatedly adding it with all even divisors of N except N. Print -1 if the conversion is not possible.

Examples:

Input: N = 6, M = 24
Output: 4
Explanation:
Step1: Add 2 (2 is an even divisor and 2!= 6) to 6. Now N becomes 8.
Step2: Add 4 (4 is an even divisor and 4!= 8) to 8. Now N becomes 12.
Step3: Add 6 (6 is an even divisor and 6!=12) to 12. Now N becomes 18.
Step4: Add 6 (6 is an even divisor and 6!=18) to 18. Now N becomes 24 = M.
Hence, 4 steps are needed.

Input: N = 9, M = 17
Output: -1
Explanation:
There are no even divisors for 9 to add, so we cannot convert N to M.

Naive Approach: The simplest solution is to consider all possible even divisors of a number and calculate the answer for them recursively and finally return the minimum of it.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// INF is the maximum value``// which indicates Impossible state``const` `int` `INF = 1e7;` `// Recursive Function that considers``// all possible even divisors of cur``int` `min_op(``int` `cur, ``int` `M)``{``    ``// Indicates impossible state``    ``if` `(cur > M)``        ``return` `INF;` `    ``// Return 0 if cur == M``    ``if` `(cur == M)``        ``return` `0;` `    ``// op stores the minimum operations``    ``// required to transform cur to M` `    ``// Initially it is set to INF that``    ``// means we can't transform cur to M``    ``int` `op = INF;` `    ``// Loop to find even divisors of cur``    ``for` `(``int` `i = 2; i * i <= cur; i++) {` `        ``// if i is divisor of cur``        ``if` `(cur % i == 0) {` `            ``// if i is even``            ``if` `(i % 2 == 0) {` `                ``// Finding divisors``                ``// recursively for cur+i``                ``op = min(op,``                         ``1 + min_op(cur + i, M));``            ``}` `            ``// Check another divisor``            ``if` `((cur / i) != i``                ``&& (cur / i) % 2 == 0) {` `                ``// Find recursively``                ``// for cur+(cur/i)``                ``op = min(``                    ``op,``                    ``1 + min_op(``                            ``cur + (cur / i),``                            ``M));``            ``}``        ``}``    ``}` `    ``// Return the answer``    ``return` `op;``}` `// Function that finds the minimum``// operation to reduce N to M``int` `min_operations(``int` `N, ``int` `M)``{``    ``int` `op = min_op(N, M);` `    ``// INF indicates impossible state``    ``if` `(op >= INF)``        ``cout << ``"-1"``;``    ``else``        ``cout << op << ``"\n"``;``}` `// Driver Code``int` `main()``{``    ``// Given N and M``    ``int` `N = 6, M = 24;` `    ``// Function Call``    ``min_operations(N, M);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// INF is the maximum value``// which indicates Impossible state``static` `int` `INF = (``int``) 1e7;` `// Recursive Function that considers``// all possible even divisors of cur``static` `int` `min_op(``int` `cur, ``int` `M)``{``    ``// Indicates impossible state``    ``if` `(cur > M)``        ``return` `INF;` `    ``// Return 0 if cur == M``    ``if` `(cur == M)``        ``return` `0``;` `    ``// op stores the minimum operations``    ``// required to transform cur to M` `    ``// Initially it is set to INF that``    ``// means we can't transform cur to M``    ``int` `op = INF;` `    ``// Loop to find even divisors of cur``    ``for` `(``int` `i = ``2``; i * i <= cur; i++)``    ``{` `        ``// if i is divisor of cur``        ``if` `(cur % i == ``0``)``        ``{` `            ``// if i is even``            ``if` `(i % ``2` `== ``0``)``            ``{` `                ``// Finding divisors``                ``// recursively for cur+i``                ``op = Math.min(op, ``1` `+ min_op(cur + i, M));``            ``}` `            ``// Check another divisor``            ``if` `((cur / i) != i && (cur / i) % ``2` `== ``0``)``            ``{` `                ``// Find recursively``                ``// for cur+(cur/i)``                ``op = Math.min(op, ``1` `+ min_op(``                                ``cur + (cur / i), M));``            ``}``        ``}``    ``}` `    ``// Return the answer``    ``return` `op;``}` `// Function that finds the minimum``// operation to reduce N to M``static` `void` `min_operations(``int` `N, ``int` `M)``{``    ``int` `op = min_op(N, M);` `    ``// INF indicates impossible state``    ``if` `(op >= INF)``        ``System.out.print(``"-1"``);``    ``else``        ``System.out.print(op + ``"\n"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// Given N and M``    ``int` `N = ``6``, M = ``24``;` `    ``// Function Call``    ``min_operations(N, M);``}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# Python3 program for the above approach` `# INF is the maximum value``# which indicates Impossible state``INF ``=` `int``(``1e7``);` `# Recursive Function that considers``# all possible even divisors of cur``def` `min_op(cur, M):``    ` `    ``# Indicates impossible state``    ``if` `(cur > M):``        ``return` `INF;` `    ``# Return 0 if cur == M``    ``if` `(cur ``=``=` `M):``        ``return` `0``;` `    ``# op stores the minimum operations``    ``# required to transform cur to M` `    ``# Initially it is set to INF that``    ``# means we can't transform cur to M``    ``op ``=` `int``(INF);` `    ``# Loop to find even divisors of cur``    ``for` `i ``in` `range``(``2``, ``int``(cur ``*``*` `1` `/` `2``) ``+` `1``):` `        ``# If i is divisor of cur``        ``if` `(cur ``%` `i ``=``=` `0``):` `            ``# If i is even``            ``if` `(i ``%` `2` `=``=` `0``):``                ` `                ``# Finding divisors``                ``# recursively for cur+i``                ``op ``=` `min``(op, ``1` `+` `min_op(cur ``+` `i, M));` `            ``# Check another divisor``            ``if` `((cur ``/` `i) !``=` `i ``and``                ``(cur ``/` `i) ``%` `2` `=``=` `0``):``                ` `                ``# Find recursively``                ``# for cur+(cur/i)``                ``op ``=` `min``(op, ``1` `+` `min_op(``                           ``cur ``+` `(cur ``/``/` `i), M));` `    ``# Return the answer``    ``return` `op;` `# Function that finds the minimum``# operation to reduce N to M``def` `min_operations(N, M):``    ` `    ``op ``=` `min_op(N, M);` `    ``# INF indicates impossible state``    ``if` `(op >``=` `INF):``        ``print``(``"-1"``);``    ``else``:``        ``print``(op);` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given N and M``    ``N ``=` `6``;``    ``M ``=` `24``;` `    ``# Function call``    ``min_operations(N, M);` `# This code is contributed by Amit Katiyar`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG {` `    ``// INF is the maximum value``    ``// which indicates Impossible state``    ``static` `int` `INF = (``int``)1e7;` `    ``// Recursive Function that considers``    ``// all possible even divisors of cur``    ``static` `int` `min_op(``int` `cur, ``int` `M)``    ``{``        ``// Indicates impossible state``        ``if` `(cur > M)``            ``return` `INF;` `        ``// Return 0 if cur == M``        ``if` `(cur == M)``            ``return` `0;` `        ``// op stores the minimum operations``        ``// required to transform cur to M` `        ``// Initially it is set to INF that``        ``// means we can't transform cur to M``        ``int` `op = INF;` `        ``// Loop to find even divisors of cur``        ``for` `(``int` `i = 2; i * i <= cur; i++)``        ``{` `            ``// if i is divisor of cur``            ``if` `(cur % i == 0)``            ``{` `                ``// if i is even``                ``if` `(i % 2 == 0)``                ``{` `                    ``// Finding divisors``                    ``// recursively for cur+i``                    ``op = Math.Min(op,1 +``                                  ``min_op(cur + i, M));``                ``}` `                ``// Check another divisor``                ``if` `((cur / i) != i && (cur / i) % 2 == 0)``                ``{` `                    ``// Find recursively``                    ``// for cur+(cur/i)``                    ``op = Math.Min(op, 1 +``                                  ``min_op(cur +``                                         ``(cur / i), M));``                ``}``            ``}``        ``}` `        ``// Return the answer``        ``return` `op;``    ``}` `    ``// Function that finds the minimum``    ``// operation to reduce N to M``    ``static` `void` `min_operations(``int` `N, ``int` `M)``    ``{``        ``int` `op = min_op(N, M);` `        ``// INF indicates impossible state``        ``if` `(op >= INF)``            ``Console.Write(``"-1"``);``        ``else``            ``Console.Write(op + ``"\n"``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``// Given N and M``        ``int` `N = 6, M = 24;` `        ``// Function Call``        ``min_operations(N, M);``    ``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`4`

Time Complexity: O(2N)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use Dynamic Programming and store the overlapping subproblems state in the naive approach to calculate the answer efficiently. Follow the below steps to solve the problem:

1. Initialize a dp table dp[i] = -1 for all N≤i≤M.
2. Consider all possible even divisors of a number and find the minimum from all of it.
3. Finally, store the result in dp[] and return the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// INF is the maximum value``// which indicates Impossible state``const` `int` `INF = 1e7;``const` `int` `max_size = 100007;` `// Stores the dp states``int` `dp[max_size];` `// Recursive Function that considers``// all possible even divisors of cur``int` `min_op(``int` `cur, ``int` `M)``{``    ``// Indicates impossible state``    ``if` `(cur > M)``        ``return` `INF;` `    ``if` `(cur == M)``        ``return` `0;` `    ``// Check dp[cur] is already``    ``// calculated or not``    ``if` `(dp[cur] != -1)``        ``return` `dp[cur];` `    ``// op stores the minimum operations``    ``// required to transform cur to M` `    ``// Initially it is set to INF that``    ``// meanswe cur can't be transform to M``    ``int` `op = INF;` `    ``// Loop to find even divisors of cur``    ``for` `(``int` `i = 2; i * i <= cur; i++) {` `        ``// if i is divisor of cur``        ``if` `(cur % i == 0) {` `            ``// if i is even``            ``if` `(i % 2 == 0) {` `                ``// Find recursively``                ``// for cur+i``                ``op = min(op, 1 + min_op(cur + i, M));``            ``}` `            ``// Check another divisor``            ``if` `((cur / i) != i && (cur / i) % 2 == 0) {` `                ``// Find recursively``                ``// for cur+(cur/i)``                ``op = min(op,``                         ``1 + min_op(cur + (cur / i), M));``            ``}``        ``}``    ``}` `    ``// Finally store the current state``    ``// result and return the answer``    ``return` `dp[cur] = op;``}` `// Function that counts the minimum``// operation to reduce N to M``int` `min_operations(``int` `N, ``int` `M)``{``    ``// Initialise dp state``    ``for` `(``int` `i = N; i <= M; i++) {``        ``dp[i] = -1;``    ``}` `    ``// Function Call``    ``return` `min_op(N, M);``}` `// Driver Code``int` `main()``{``    ``// Given N and M``    ``int` `N = 6, M = 24;` `    ``// Function Call``    ``int` `op = min_operations(N, M);` `    ``// INF indicates impossible state``    ``if` `(op >= INF)``        ``cout << ``"-1"``;``    ``else``        ``cout << op << ``"\n"``;` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG{` `// INF is the maximum value``// which indicates Impossible state``static` `int` `INF = (``int``) 1e7;``static` `int` `max_size = ``100007``;` `// Stores the dp states``static` `int` `[]dp = ``new` `int``[max_size];` `// Recursive Function that considers``// all possible even divisors of cur``static` `int` `min_op(``int` `cur, ``int` `M)``{``    ` `    ``// Indicates impossible state``    ``if` `(cur > M)``        ``return` `INF;` `    ``if` `(cur == M)``        ``return` `0``;``                ` `    ``// Check dp[cur] is already``    ``// calculated or not``    ``if` `(dp[cur] != -``1``)``        ``return` `dp[cur];` `    ``// op stores the minimum operations``    ``// required to transform cur to M` `    ``// Initially it is set to INF that``    ``// meanswe cur can't be transform to M``    ``int` `op = INF;` `    ``// Loop to find even divisors of cur``    ``for``(``int` `i = ``2``; i * i <= cur; i++)``    ``{``        ` `        ``// If i is divisor of cur``        ``if` `(cur % i == ``0``)``        ``{``            ` `            ``// If i is even``            ``if` `(i % ``2` `== ``0``)``            ``{``                ` `                ``// Find recursively``                ``// for cur+i``                ``op = Math.min(op,``                     ``1` `+ min_op(cur + i, M));``            ``}` `            ``// Check another divisor``            ``if` `((cur / i) != i &&``                ``(cur / i) % ``2` `== ``0``)``            ``{` `                ``// Find recursively``                ``// for cur+(cur/i)``                ``op = Math.min(op,``                     ``1` `+ min_op(cur + (cur / i), M));``            ``}``        ``}``    ``}` `    ``// Finally store the current state``    ``// result and return the answer``    ``return` `dp[cur] = op;``}` `// Function that counts the minimum``// operation to reduce N to M``static` `int` `min_operations(``int` `N, ``int` `M)``{``    ` `    ``// Initialise dp state``    ``for``(``int` `i = N; i <= M; i++)``    ``{``        ``dp[i] = -``1``;``    ``}` `    ``// Function call``    ``return` `min_op(N, M);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Given N and M``    ``int` `N = ``6``, M = ``24``;` `    ``// Function call``    ``int` `op = min_operations(N, M);` `    ``// INF indicates impossible state``    ``if` `(op >= INF)``        ``System.out.print(``"-1"``);``    ``else``        ``System.out.print(op + ``"\n"``);``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for the``# above approach`` ` `# INF is the maximum value``# which indicates Impossible state``INF ``=` `10000007``;``max_size ``=` `100007``;`` ` `# Stores the dp states``dp ``=` `[``0` `for` `i ``in` `range``(max_size)];`` ` `# Recursive Function``# that considers all``# possible even divisors``# of cur``def` `min_op(cur, M):` `    ``# Indicates impossible``    ``# state``    ``if` `(cur > M):``        ``return` `INF;`` ` `    ``if` `(cur ``=``=` `M):``        ``return` `0``;`` ` `    ``# Check dp[cur] is already``    ``# calculated or not``    ``if` `(dp[cur] !``=` `-``1``):``        ``return` `dp[cur];`` ` `    ``# op stores the minimum``    ``# operations required to``    ``# transform cur to M`` ` `    ``# Initially it is set``    ``# to INF that meanswe cur``    ``# can't be transform to M``    ``op ``=` `INF;`` ` `    ``i ``=` `2``    ` `    ``# Loop to find even``    ``# divisors of cur``    ``while``(i ``*` `i <``=` `cur):`` ` `        ``# if i is divisor of cur``        ``if` `(cur ``%` `i ``=``=` `0``):`` ` `            ``# if i is even``            ``if``(i ``%` `2` `=``=` `0``):`` ` `                ``# Find recursively``                ``# for cur+i``                ``op ``=` `min``(op, ``1` `+` `min_op(cur ``+``                                        ``i, M));           `` ` `            ``# Check another divisor``            ``if` `((cur ``/``/` `i) !``=` `i ``and``                ``(cur ``/``/` `i) ``%` `2` `=``=` `0``):`` ` `                ``# Find recursively``                ``# for cur+(cur/i)``                ``op ``=` `min``(op, ``1` `+` `min_op(cur ``+``                        ``(cur ``/``/` `i), M))``        ` `        ``i ``+``=` `1`    ` ` `    ``# Finally store the current state``    ``# result and return the answer``    ``dp[cur] ``=` `op;``    ``return` `op`` ` `# Function that counts the minimum``# operation to reduce N to M``def` `min_operations(N, M):``    ` `    ``# Initialise dp state``    ``for` `i ``in` `range``(N, M ``+` `1``):``        ``dp[i] ``=` `-``1``;`` ` `    ``# Function Call``    ``return` `min_op(N, M);` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``# Given N and M``    ``N ``=` `6``    ``M ``=` `24`` ` `    ``# Function Call``    ``op ``=` `min_operations(N, M);`` ` `    ``# INF indicates impossible state``    ``if` `(op >``=` `INF):``        ``print``(``-``1``)``    ``else``:``        ``print``(op)` `# This code is contributed by rutvik_56`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG{` `// INF is the maximum value``// which indicates Impossible state``static` `int` `INF = (``int``) 1e7;``static` `int` `max_size = 100007;` `// Stores the dp states``static` `int` `[]dp = ``new` `int``[max_size];` `// Recursive Function that considers``// all possible even divisors of cur``static` `int` `min_op(``int` `cur, ``int` `M)``{``    ` `    ``// Indicates impossible state``    ``if` `(cur > M)``        ``return` `INF;` `    ``if` `(cur == M)``        ``return` `0;``                ` `    ``// Check dp[cur] is already``    ``// calculated or not``    ``if` `(dp[cur] != -1)``        ``return` `dp[cur];` `    ``// op stores the minimum operations``    ``// required to transform cur to M` `    ``// Initially it is set to INF that``    ``// meanswe cur can't be transform to M``    ``int` `op = INF;` `    ``// Loop to find even divisors of cur``    ``for``(``int` `i = 2; i * i <= cur; i++)``    ``{       ``        ``// If i is divisor of cur``        ``if` `(cur % i == 0)``        ``{           ``            ``// If i is even``            ``if` `(i % 2 == 0)``            ``{               ``                ``// Find recursively``                ``// for cur+i``                ``op = Math.Min(op, 1 +``                              ``min_op(cur + i, M));``            ``}` `            ``// Check another divisor``            ``if` `((cur / i) != i &&``                ``(cur / i) % 2 == 0)``            ``{``                ``// Find recursively``                ``// for cur+(cur/i)``                ``op = Math.Min(op, 1 +``                              ``min_op(cur +``                                     ``(cur / i), M));``            ``}``        ``}``    ``}` `    ``// Finally store the current state``    ``// result and return the answer``    ``return` `dp[cur] = op;``}` `// Function that counts the minimum``// operation to reduce N to M``static` `int` `min_operations(``int` `N,``                          ``int` `M)``{   ``    ``// Initialise dp state``    ``for``(``int` `i = N; i <= M; i++)``    ``{``        ``dp[i] = -1;``    ``}` `    ``// Function call``    ``return` `min_op(N, M);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{   ``    ``// Given N and M``    ``int` `N = 6, M = 24;` `    ``// Function call``    ``int` `op = min_operations(N, M);` `    ``// INF indicates impossible state``    ``if` `(op >= INF)``        ``Console.Write(``"-1"``);``    ``else``        ``Console.Write(op + ``"\n"``);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`4`

Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(M)

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