Given an integer N, the task is to count the minimum steps required to reduce the value of N to 0 by performing the following two operations:
- Consider integers A and B where N = A * B (A != 1 and B != 1), reduce N to min(A, B)
- Decrease the value of N by 1
Examples :
Input: N = 3
Output: 3
Explanation:
Steps involved are 3 -> 2 -> 1 -> 0
Therefore, the minimum steps required is 3.
Input: N = 4
Output: 3
Explanation:
Steps involved are 4->2->1->0.
Therefore, the minimum steps required is 3.
Naive Approach: The idea is to use the concept of Dynamic Programming. Follow the steps below to solve the problem:
- The simple solution to this problem is to replace N with each possible value until it is not 0.
- When N reaches 0, compare the count of moves with the minimum obtained so far to obtain the optimal answer.
- Finally, print the minimum steps calculated.
Illustration:
N = 4,
- On applying the first rule, factors of 4 are [ 1, 2, 4 ].
Therefore, all possible pairs (a, b) are (1 * 4), (2 * 2), (4 * 1).
Only pair satisfying the condition (a!=1 and b!=1) is (2, 2) . Therefore, reduce 4 to 2.
Finally, reduce N to 0, in 3 steps(4 -> 2 -> 1 -> 0)- On applying the second rule, steps required is 4, (4 -> 3 -> 2 -> 1 -> 0).
Recursive tree for N = 4 is 4 / \ 3 2(2*2) | | 2 1 | | 1 0 | 0
- Therefore, minimum steps required to reduce N to 0 is 3.
Therefore, the relation is:
f(N) = 1 + min( f(N-1), min(f(x)) ), where N % x == 0 and x is in range [2, K] where K = sqrt(N)
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to count the minimum// steps required to reduce nint downToZero(int n){ // Base case if (n <= 3) return n; // Allocate memory for storing // intermediate results vector<int> dp(n + 1, -1); // Store base values dp[0] = 0; dp[1] = 1; dp[2] = 2; dp[3] = 3; // Stores square root // of each number int sqr; for (int i = 4; i <= n; i++) { // Compute square root sqr = sqrt(i); int best = INT_MAX; // Use rule 1 to find optimized // answer while (sqr > 1) { // Check if it perfectly divides n if (i % sqr == 0) { best = min(best, 1 + dp[sqr]); } sqr--; } // Use of rule 2 to find // the optimized answer best = min(best, 1 + dp[i - 1]); // Store computed value dp[i] = best; } // Return answer return dp[n];}// Driver Codeint main(){ int n = 4; cout << downToZero(n); return 0;} |
Java
// Java program to implement// the above approachclass GFG{// Function to count the minimum// steps required to reduce nstatic int downToZero(int n){ // Base case if (n <= 3) return n; // Allocate memory for storing // intermediate results int []dp = new int[n + 1]; for(int i = 0; i < n + 1; i++) dp[i] = -1; // Store base values dp[0] = 0; dp[1] = 1; dp[2] = 2; dp[3] = 3; // Stores square root // of each number int sqr; for(int i = 4; i <= n; i++) { // Compute square root sqr = (int)Math.sqrt(i); int best = Integer.MAX_VALUE; // Use rule 1 to find optimized // answer while (sqr > 1) { // Check if it perfectly divides n if (i % sqr == 0) { best = Math.min(best, 1 + dp[sqr]); } sqr--; } // Use of rule 2 to find // the optimized answer best = Math.min(best, 1 + dp[i - 1]); // Store computed value dp[i] = best; } // Return answer return dp[n];}// Driver Codepublic static void main(String[] args){ int n = 4; System.out.print(downToZero(n));}}// This code is contributed by amal kumar choubey |
Python3
# Python3 program to implement# the above approachimport mathimport sys# Function to count the minimum# steps required to reduce ndef downToZero(n): # Base case if (n <= 3): return n # Allocate memory for storing # intermediate results dp = [-1] * (n + 1) # Store base values dp[0] = 0 dp[1] = 1 dp[2] = 2 dp[3] = 3 # Stores square root # of each number for i in range(4, n + 1): # Compute square root sqr = (int)(math.sqrt(i)) best = sys.maxsize # Use rule 1 to find optimized # answer while (sqr > 1): # Check if it perfectly divides n if (i % sqr == 0): best = min(best, 1 + dp[sqr]) sqr -= 1 # Use of rule 2 to find # the optimized answer best = min(best, 1 + dp[i - 1]) # Store computed value dp[i] = best # Return answer return dp[n]# Driver Codeif __name__ == "__main__": n = 4 print(downToZero(n))# This code is contributed by chitranayal |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to count the minimum// steps required to reduce nstatic int downToZero(int n){ // Base case if (n <= 3) return n; // Allocate memory for storing // intermediate results int []dp = new int[n + 1]; for(int i = 0; i < n + 1; i++) dp[i] = -1; // Store base values dp[0] = 0; dp[1] = 1; dp[2] = 2; dp[3] = 3; // Stores square root // of each number int sqr; for(int i = 4; i <= n; i++) { // Compute square root sqr = (int)Math.Sqrt(i); int best = int.MaxValue; // Use rule 1 to find optimized // answer while (sqr > 1) { // Check if it perfectly divides n if (i % sqr == 0) { best = Math.Min(best, 1 + dp[sqr]); } sqr--; } // Use of rule 2 to find // the optimized answer best = Math.Min(best, 1 + dp[i - 1]); // Store computed value dp[i] = best; } // Return answer return dp[n];}// Driver Codepublic static void Main(String[] args){ int n = 4; Console.Write(downToZero(n));}}// This code is contributed by amal kumar choubey |
Javascript
<script> // Javascript Program to implement // the above approach // Function to count the minimum // steps required to reduce n function downToZero(n) { // Base case if (n <= 3) return n; // Allocate memory for storing // intermediate results let dp = new Array(n + 1) dp.fill(-1); // Store base values dp[0] = 0; dp[1] = 1; dp[2] = 2; dp[3] = 3; // Stores square root // of each number let sqr; for (let i = 4; i <= n; i++) { // Compute square root sqr = Math.sqrt(i); let best = Number.MAX_VALUE; // Use rule 1 to find optimized // answer while (sqr > 1) { // Check if it perfectly divides n if (i % sqr == 0) { best = Math.min(best, 1 + dp[sqr]); } sqr--; } // Use of rule 2 to find // the optimized answer best = Math.min(best, 1 + dp[i - 1]); // Store computed value dp[i] = best; } // Return answer return dp[n]; } let n = 4; document.write(downToZero(n)); // This code is contributed by divyesh072019.</script> |
Output:
3
Time complexity: O(N * sqrt(n))
Auxiliary Space: O(N)
Efficient Approach: The idea is to observe that it is possible to replace N by N’ where N’ = min(a, b) (N = a * b) (a != 1 and b != 1).
- If N is even, then the smallest value that divides N is 2. Therefore, directly calculate f(N) = 1 + f(2) = 3.
- If N is odd, then reduce N by 1 from it i.e N = N – 1. Apply the same logic as used for even numbers. Therefore, for odd numbers, the minimum steps required is 4.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum// steps required to reduce nint downToZero(int n){ // Base case if (n <= 3) return n; // Return answer based on // parity of n return n % 2 == 0 ? 3 : 4;}// Driver Codeint main(){ int n = 4; cout << downToZero(n); return 0;} |
Java
// Java Program to implement// the above approachclass GFG{ // Function to find the minimum// steps required to reduce nstatic int downToZero(int n){ // Base case if (n <= 3) return n; // Return answer based on // parity of n return n % 2 == 0 ? 3 : 4;} // Driver Codepublic static void main(String[] args){ int n = 4; System.out.println(downToZero(n));}}// This code is contributed by rock_cool |
Python3
# Python3 Program to implement# the above approach# Function to find the minimum# steps required to reduce ndef downToZero(n): # Base case if (n <= 3): return n; # Return answer based on # parity of n if(n % 2 == 0): return 3; else: return 4;# Driver Codeif __name__ == '__main__': n = 4; print(downToZero(n)); # This code is contributed by Rohit_ranjan |
C#
// C# Program to implement// the above approachusing System;class GFG{ // Function to find the minimum// steps required to reduce nstatic int downToZero(int n){ // Base case if (n <= 3) return n; // Return answer based on // parity of n return n % 2 == 0 ? 3 : 4;} // Driver Codepublic static void Main(String[] args){ int n = 4; Console.WriteLine(downToZero(n));}}// This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript Program to implement // the above approach // Function to find the minimum // steps required to reduce n function downToZero(n) { // Base case if (n <= 3) return n; // Return answer based on // parity of n return n % 2 == 0 ? 3 : 4; } let n = 4; document.write(downToZero(n)); // This code is contributed by divyeshrabadiya07.</script> |
Output:
3
Time complexity: O(1)
Auxiliary Space: O(1)
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