Minimum number of operations required to reduce N to 0

19 June 2025

0

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 n
int 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 Code
int main()
{
    int n = 4;
    cout << downToZero(n);
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to count the minimum
// steps required to reduce n
static 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 Code
public 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 approach
import math
import sys
 
# Function to count the minimum
# steps required to reduce n
def 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 Code
if __name__ == "__main__":
     
    n = 4
 
    print(downToZero(n))
 
# This code is contributed by chitranayal    

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to count the minimum
// steps required to reduce n
static 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 Code
public 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:

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 n
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 Code
int main()
{
    int n = 4;
    cout << downToZero(n);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
class GFG{
  
// Function to find the minimum
// steps required to reduce n
static 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 Code
public 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 n
def 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 Code
if __name__ == '__main__':
    n = 4;
    print(downToZero(n));
     
# This code is contributed by Rohit_ranjan

C#

// C# Program to implement
// the above approach
using System;
class GFG{
  
// Function to find the minimum
// steps required to reduce n
static 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 Code
public 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:

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

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