Given two integers N and M, the task is to calculate the minimum number of moves to change N to M, where In one move it is allowed to add any divisor of the current value of N to N itself except 1. Print “-1” if it is not possible.
Example:
Input: N = 4, M = 24
Output: 5
Explanation: The given value of N can be converted into M using the following steps: (4)+2 -> (6)+2 -> (8)+4 -> (12)+6 -> (18)+6 -> 24. Hence, the count of moves is 5 which is the minimum possible.Input: N = 4, M = 576
Output: 14
BFS Approach: The given problem has already been discussed in Set 1 of this article which using the Breadth First Traversal to solve the given problem.
Approach: This article focused on a different approach based on Dynamic Programming. Below are the steps to follow:
- Create a 1-D array dp[], where dp[i] stores the minimum number of operations to reach i from N. Initially, dp[N+1… M] = {INT_MAX} and dp[N] = 0.
- Iterate through the range [N, M] using a variable i and for each i, iterate through all the factors of the given number i. For a factor X, the DP relation can be define as follows:
dp[i + X] = min( dp[i], dp[i + X])
- The value stored at dp[M] is the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the minimum count of // operations to convert N to M int minOperationCnt(int N, int M) { // Stores the DP state of the array int dp[M + 1]; // Initialize each index with INT_MAX for (int i = N + 1; i <= M; i++) { dp[i] = INT_MAX; } // Initial condition dp[N] = 0; // Loop to iterate over range [N, M] for (int i = N; i <= M; i++) { // Check if this position // can be reached or not if (dp[i] == INT_MAX) { continue; } // Loop to iterate through all divisors // of the current value i for (int j = 2; j * j <= i; j++) { // If j is a divisor of i if (i % j == 0) { if (i + j <= M) { // Update the value of dp[i + j] dp[i + j] = min(dp[i + j], dp[i] + 1); } // Check for value i / j; if (i + i / j <= M) { // Update the value of dp[i + i/j] dp[i + i / j] = min(dp[i + i / j], dp[i] + 1); } } } } // Return Answer return (dp[M] == INT_MAX) ? -1 : dp[M]; } // Driver Code int main() { int N = 4; int M = 576; cout << minOperationCnt(N, M); return 0; } |
Java
// Java implementation for the above approach class GFG { // Function to find the minimum count of // operations to convert N to M public static int minOperationCnt(int N, int M) { // Stores the DP state of the array int[] dp = new int[M + 1]; // Initialize each index with INT_MAX for (int i = N + 1; i <= M; i++) { dp[i] = Integer.MAX_VALUE; } // Initial condition dp[N] = 0; // Loop to iterate over range [N, M] for (int i = N; i <= M; i++) { // Check if this position // can be reached or not if (dp[i] == Integer.MAX_VALUE) { continue; } // Loop to iterate through all divisors // of the current value i for (int j = 2; j * j <= i; j++) { // If j is a divisor of i if (i % j == 0) { if (i + j <= M) { // Update the value of dp[i + j] dp[i + j] = Math.min(dp[i + j], dp[i] + 1); } // Check for value i / j; if (i + i / j <= M) { // Update the value of dp[i + i/j] dp[i + i / j] = Math.min(dp[i + i / j], dp[i] + 1); } } } } // Return Answer return (dp[M] == Integer.MAX_VALUE) ? -1 : dp[M]; } // Driver Code public static void main(String args[]) { int N = 4; int M = 576; System.out.println(minOperationCnt(N, M)); } } // This code is contributed by saurabh_jaiswal. |
Python3
# python implementation for the above approach import math INT_MAX = 2147483647 # Function to find the minimum count of # operations to convert N to M def minOperationCnt(N, M): # Stores the DP state of the array dp = [0 for _ in range(M + 1)] # Initialize each index with INT_MAX for i in range(N+1, M+1): dp[i] = INT_MAX # Initial condition dp[N] = 0 # Loop to iterate over range [N, M] for i in range(N, M+1): # Check if this position # can be reached or not if (dp[i] == INT_MAX): continue # Loop to iterate through all divisors # of the current value i for j in range(2, int(math.sqrt(i))+1): # If j is a divisor of i if (i % j == 0): if (i + j <= M): # Update the value of dp[i + j] dp[i + j] = min(dp[i + j], dp[i] + 1) # Check for value i / j; if (i + i // j <= M): # Update the value of dp[i + i/j] dp[i + i // j] = min(dp[i + i // j], dp[i] + 1) # Return Answer if dp[M] == INT_MAX: return -1 else: return dp[M] # Driver Code if __name__ == "__main__": N = 4 M = 576 print(minOperationCnt(N, M)) # This code is contributed by rakeshsahni |
C#
// C# implementation for the above approach using System; class GFG { // Function to find the minimum count of // operations to convert N to M public static int minOperationCnt(int N, int M) { // Stores the DP state of the array int[] dp = new int[M + 1]; // Initialize each index with INT_MAX for (int i = N + 1; i <= M; i++) { dp[i] = int.MaxValue; } // Initial condition dp[N] = 0; // Loop to iterate over range [N, M] for (int i = N; i <= M; i++) { // Check if this position // can be reached or not if (dp[i] == int.MaxValue) { continue; } // Loop to iterate through all divisors // of the current value i for (int j = 2; j * j <= i; j++) { // If j is a divisor of i if (i % j == 0) { if (i + j <= M) { // Update the value of dp[i + j] dp[i + j] = Math.Min(dp[i + j], dp[i] + 1); } // Check for value i / j; if (i + i / j <= M) { // Update the value of dp[i + i/j] dp[i + i / j] = Math.Min(dp[i + i / j], dp[i] + 1); } } } } // Return Answer return (dp[M] == int.MaxValue) ? -1 : dp[M]; } // Driver Code public static void Main() { int N = 4; int M = 576; Console.Write(minOperationCnt(N, M)); } } // This code is contributed by saurabh_jaiswal. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find the minimum count of // operations to convert N to M function minOperationCnt(N, M) { // Stores the DP state of the array let dp = new Array(M + 1) // Initialize each index with INT_MAX for (let i = N + 1; i <= M; i++) { dp[i] = Number.MAX_VALUE; } // Initial condition dp[N] = 0; // Loop to iterate over range [N, M] for (let i = N; i <= M; i++) { // Check if this position // can be reached or not if (dp[i] == Number.MAX_VALUE) { continue; } // Loop to iterate through all divisors // of the current value i for (let j = 2; j * j <= i; j++) { // If j is a divisor of i if (i % j == 0) { if (i + j <= M) { // Update the value of dp[i + j] dp[i + j] = Math.min(dp[i + j], dp[i] + 1); } // Check for value i / j; if (i + i / j <= M) { // Update the value of dp[i + i/j] dp[i + i / j] = Math.min(dp[i + i / j], dp[i] + 1); } } } } // Return Answer return (dp[M] == Number.MAX_VALUE) ? -1 : dp[M]; } // Driver Code let N = 4; let M = 576; document.write(minOperationCnt(N, M)); // This code is contributed by Potta Lokesh </script> |
Output
14
Time Complexity: O((M – N)*√(M – N))
Auxiliary Space: O(M)
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