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 <bits/stdc++.h>using namespace std;// INF is the maximum value// which indicates Impossible stateconst int INF = 1e7;// Recursive Function that considers// all possible even divisors of curint 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 Mint 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 Codeint main(){ // Given N and M int N = 6, M = 24; // Function Call min_operations(N, M); return 0;} |
Java
// Java program for the above approachclass GFG{// INF is the maximum value// which indicates Impossible statestatic int INF = (int) 1e7;// Recursive Function that considers// all possible even divisors of curstatic 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 Mstatic 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 Codepublic 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 stateINF = int(1e7);# Recursive Function that considers# all possible even divisors of curdef 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 Mdef min_operations(N, M): op = min_op(N, M); # INF indicates impossible state if (op >= INF): print("-1"); else: print(op);# Driver Codeif __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 approachusing 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
<script>// Javascript program to implement// the above approach // INF is the maximum value// which indicates Impossible statelet INF = 1e7; // Recursive Function that considers// all possible even divisors of curfunction 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 let op = INF; // Loop to find even divisors of cur for (let 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 Mfunction min_operations(N, M){ let op = min_op(N, M); // INF indicates impossible state if (op >= INF) document.write("-1"); else document.write(op + "\n");}// Driver Code // Given N and M let N = 6, M = 24; // Function Call min_operations(N, M); </script> |
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:
- Initialize a dp table dp[i] = -1 for all N?i?M.
- Consider all possible even divisors of a number and find the minimum from all of it.
- 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 <bits/stdc++.h>using namespace std;// INF is the maximum value// which indicates Impossible stateconst int INF = 1e7;const int max_size = 100007;// Stores the dp statesint dp[max_size];// Recursive Function that considers// all possible even divisors of curint 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 Mint 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 Codeint 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 approachimport java.util.*;class GFG{// INF is the maximum value// which indicates Impossible statestatic int INF = (int) 1e7;static int max_size = 100007;// Stores the dp statesstatic int []dp = new int[max_size];// Recursive Function that considers// all possible even divisors of curstatic 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 Mstatic 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 Codepublic 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 stateINF = 10000007;max_size = 100007; # Stores the dp statesdp = [0 for i in range(max_size)]; # Recursive Function # that considers all # possible even divisors # of curdef 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 Mdef 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 approachusing System;class GFG{// INF is the maximum value// which indicates Impossible statestatic int INF = (int) 1e7;static int max_size = 100007;// Stores the dp statesstatic int []dp = new int[max_size];// Recursive Function that considers// all possible even divisors of curstatic 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 Mstatic 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 Codepublic 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
<script>// Javascript program for the above approach// INF is the maximum value// which indicates Impossible statevar INF = 10000000;var max_size = 100007;// Stores the dp statesvar dp = Array(max_size);// Recursive Function that considers// all possible even divisors of curfunction 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 var op = INF; // Loop to find even divisors of cur for (var 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 Mfunction min_operations(N, M){ // Initialise dp state for ( i = N; i <= M; i++) { dp[i] = -1; } // Function Call return min_op(N, M);}// Driver Code// Given N and Mvar N = 6, M = 24;// Function Callvar op = min_operations(N, M);// INF indicates impossible stateif (op >= INF) document.write( "-1");else document.write( op + "<br>");// This code is contributed by noob2000.</script> |
Output
4
Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(M)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a table to store the solution of the subproblems.
- Initialize the table with base cases
- Fill up the table iteratively
- Return the final solution
Implementation :
C++
#include <bits/stdc++.h>using namespace std;// Define a constant for impossible statesconst int INF = 1e7; int min_operations(int N, int M){ // Initialize dp table with INF int dp[M+1]; for (int i = 0; i <= M; i++) { dp[i] = INF; } // Base case dp[N] = 0; // Fill up dp table for (int i = N; i < M; i++) { // Skip impossible states if (dp[i] >= INF) { continue; } for (int j = 2; j * j <= i; j++) { if (i % j == 0) { if (j % 2 == 0) { int next = i + j; if (next <= M) { // Update the minimum number of operations to reach 'next' dp[next] = min(dp[next], dp[i] + 1); } } if ((i / j) != j && (i / j) % 2 == 0) { int next = i + (i / j); if (next <= M) { // Update the minimum number of operations to reach 'next' dp[next] = min(dp[next], dp[i] + 1); } } } } } // Return the answer if (dp[M] >= INF) { // It's impossible to reach M from N, so print -1 cout << "-1\n"; } else { // Print the minimum number of operations to reach M from N cout << dp[M] << "\n"; }}// Driver Codeint main(){ // Given N and M int N = 6, M = 24; // Function Call min_operations(N, M); return 0;}// this code is contributed by bhardwajji |
Java
import java.util.*;class Main { // Define a constant for impossible states static final int INF = 10000000; static int minOperations(int N, int M) { // Initialize dp table with INF int[] dp = new int[M + 1]; Arrays.fill(dp, INF); // Base case dp[N] = 0; // Fill up dp table for (int i = N; i < M; i++) { // Skip impossible states if (dp[i] >= INF) { continue; } for (int j = 2; j * j <= i; j++) { if (i % j == 0) { if (j % 2 == 0) { int next = i + j; if (next <= M) { // Update the minimum number of // operations to reach 'next' dp[next] = Math.min(dp[next], dp[i] + 1); } } if ((i / j) != j && (i / j) % 2 == 0) { int next = i + (i / j); if (next <= M) { // Update the minimum number of // operations to reach 'next' dp[next] = Math.min(dp[next], dp[i] + 1); } } } } } // Return the answer if (dp[M] >= INF) { // It's impossible to reach M from N, so print // -1 System.out.println("-1"); } else { // Print the minimum number of operations to // reach M from N System.out.println(dp[M]); } return 0; } // Driver Code public static void main(String[] args) { // Given N and M int N = 6, M = 24; // Function Call minOperations(N, M); }} |
Python3
# Define a constant for impossible statesINF = 10**7def min_operations(N: int, M: int) -> None: # Initialize dp table with INF dp = [INF] * (M+1) # Base case dp[N] = 0 # Fill up dp table for i in range(N, M): # Skip impossible states if dp[i] >= INF: continue for j in range(2, int(i**0.5)+1): if i % j == 0: if j % 2 == 0: next = i + j if next <= M: # Update the minimum number of operations to reach 'next' dp[next] = min(dp[next], dp[i] + 1) if (i // j) != j and (i // j) % 2 == 0: next = i + (i // j) if next <= M: # Update the minimum number of operations to reach 'next' dp[next] = min(dp[next], dp[i] + 1) # Return the answer if dp[M] >= INF: # It's impossible to reach M from N, so print -1 print("-1") else: # Print the minimum number of operations to reach M from N print(dp[M])# Driver Codeif __name__ == '__main__': # Given N and M N, M = 6, 24 # Function Call min_operations(N, M) |
C#
using System;public class MainClass{ // Define a constant for impossible states static readonly int INF = 10000000; static int minOperations(int N, int M) { // Initialize dp table with INF int[] dp = new int[M + 1]; Array.Fill(dp, INF); // Base case dp[N] = 0; // Fill up dp table for (int i = N; i < M; i++) { // Skip impossible states if (dp[i] >= INF) { continue; } for (int j = 2; j * j <= i; j++) { if (i % j == 0) { if (j % 2 == 0) { int next = i + j; if (next <= M) { // Update the minimum number of // operations to reach 'next' dp[next] = Math.Min(dp[next], dp[i] + 1); } } if ((i / j) != j && (i / j) % 2 == 0) { int next = i + (i / j); if (next <= M) { // Update the minimum number of // operations to reach 'next' dp[next] = Math.Min(dp[next], dp[i] + 1); } } } } } // Return the answer if (dp[M] >= INF) { // It's impossible to reach M from N, so print // -1 Console.WriteLine("-1"); } else { // Print the minimum number of operations to // reach M from N Console.WriteLine(dp[M]); } return 0; } // Driver Code public static void Main(string[] args) { // Given N and M int N = 6, M = 24; // Function Call minOperations(N, M); }} |
Javascript
// Define a constant for impossible statesconst INF = 1e7; function min_operations(N, M) { // Initialize dp table with INF let dp = Array(M+1).fill(INF); // Base case dp[N] = 0; // Fill up dp table for (let i = N; i < M; i++) { // Skip impossible states if (dp[i] >= INF) { continue; } for (let j = 2; j * j <= i; j++) { if (i % j == 0) { if (j % 2 == 0) { let next = i + j; if (next <= M) { // Update the minimum number of operations to reach 'next' dp[next] = Math.min(dp[next], dp[i] + 1); } } if ((i / j) != j && (i / j) % 2 == 0) { let next = i + (i / j); if (next <= M) { // Update the minimum number of operations to reach 'next' dp[next] = Math.min(dp[next], dp[i] + 1); } } } } } // Return the answer if (dp[M] >= INF) { // It's impossible to reach M from N, so print -1 console.log("-1"); } else { // Print the minimum number of operations to reach M from N console.log(dp[M]); }}// Driver Codelet N = 6, M = 24;// Function Callmin_operations(N, M); |
Output
4
Time Complexity : O(M * sqrt(M))
Auxiliary Space: O(M)
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