Minimize operations to reduce N to 0 by replacing N by its divisor at each step

Given a positive integer N. Find the minimum number of operations needed to reduce N to 0 when N can reduced by its divisor at each operation.

Example:

Input: N = 5
Output: 4
Explanation: 
Reduce 5 as 5-1=4.
Reduce 4 as 4-2=2.
Reduce 2 as 2-1=1.
Reduce 1 as 1-1=0.

Input: N = 8
Output: 4
Explanation:
Reduce 8 as 8-4=4.
Reduce 4 as 4-2=2.
Reduce 2 as 2-1=1.
Reduce 1 as 1-1=0.

Naive Approach:

One easy approach is to find highest divisor of N each time until N becomes 0.

Follow the below steps to solve this problem:

• Traverse until N becomes 0.
•  Find the highest divisor of N and subtract from N.
• Count the number of iterations required for N to become 0 this way.
•  Return the count calculated above as the final answer.

C++14

// C++ program to minimize operations
// to reduce N to 0 by replacing
// N by its divisor at each step

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll findElement(ll N)
{
    for (ll i = 2; i * i <= N; i++) {
        if (N % i == 0)
            return N / i;
    }
    return 1;
}

// Function to count minimum number
// of operation
ll minOperations(ll N)
{
    if (N < 0)
        return -1;

    ll count = 0;
    while (N) {
        ll divisor = findElement(N);
        N -= divisor;
        count++;
    }
    return count;
}

// Driver code
int main()
{
    ll N = 5;
    cout << minOperations(N);

    return 0;
}

Java

// Java program to minimize operations
// to reduce N to 0 by replacing
// N by its divisor at each step
import java.io.*;

class GFG {

  static long findElement(long N)
  {
    for (long i = 2; i * i <= N; i++) {
      if (N % i == 0)
        return N / i;
    }
    return 1;
  }

  // Function to count minimum number
  // of operation
  static long minOperations(long N)
  {
    if (N < 0)
      return -1;

    long count = 0;
    while (N > 0) {
      long divisor = findElement(N);
      N -= divisor;
      count++;
    }
    return count;
  }

  // Driver code
  public static void main (String[] args) {
    long N = 5;
    System.out.print(minOperations(N));
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python3 program to minimize operations
# to reduce N to 0 by replacing N by its
# divisor at each step

# function to find the element
def findElement(N):
    i = 2
    while i * i <= N:
        if N % i == 0:
            return int(N / i)
        i += 1
    return 1

# function to count the min number of operations
def minOperations(N):
    if N < 0:
        return -1
    count = 0
    while N:
        divisor = findElement(N)
        N -= divisor
        count += 1
    return count

# Driver Code
N = 5
print(minOperations(N))

# This code is contributed by phasing17

C#

// C# program to minimize operations
// to reduce N to 0 by replacing
// N by its divisor at each step
using System;
class GFG {
  static long findElement(long N)
  {
    for (long i = 2; i * i <= N; i++) {
      if (N % i == 0)
        return N / i;
    }
    return 1;
  }

  // Function to count minimum number
  // of operation
  static long minOperations(long N)
  {
    if (N < 0)
      return -1;

    long count = 0;
    while (N > 0) {
      long divisor = findElement(N);
      N -= divisor;
      count++;
    }
    return count;
  }

  // Driver code
  public static void Main()
  {
    long N = 5;
    Console.WriteLine(minOperations(N));
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript program to minimize operations
    // to reduce N to 0 by replacing
    // N by its divisor at each step
    const findElement = (N) => {
        for (let i = 2; i * i <= N; i++) {
            if (N % i == 0)
                return parseInt(N / i);
        }
        return 1;
    }

    // Function to count minimum number
    // of operation
    const minOperations = (N) => {
        if (N < 0)
            return -1;

        let count = 0;
        while (N) {
            let divisor = findElement(N);
            N -= divisor;
            count++;
        }
        return count;
    }

    // Driver code
    let N = 5;
    document.write(minOperations(N));

// This code is contributed by rakeshsahni

</script>

Output:

4

Time Complexity: O(N^(3/2))
Auxiliary Space: O(1)

Efficient Approach: Above solution can be optimized using pre-computation using Sieve of Eratosthenes for smallest prime factor and modifying it a bit for storing highest factor of N.

Follow the below steps to solve this problem: 

• Precompute sieve using Sieve of Eratosthenes for least prime factor of numbers till N
• Modify above sieve by storing 1 for all zero values else i/sieve[i] for every i-th value
• Traverse until N becomes 0.
• Subtract sieve[N] for every iteration of N.
• Count the number of iterations required for N to become 0 this way.
• Return the count calculated above as the final answer.

C++14

// C++ program to minimize operations
// to reduce N to 0 by replacing
// N by its divisor at each step

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAX 10000001
ll sieve[MAX];

// Function to calculate sieve
void makeSieve()
{
    ll i, j;
    sieve[0] = 0;
    sieve[1] = 1;

    for (i = 2; i < MAX; i++)
        sieve[i] = 0;

    for (i = 2; i * i <= MAX; i++) {
        if (!sieve[i]) {
            sieve[i] = i;
            for (j = i * i; j < MAX; j += i)
                if (!sieve[j])
                    sieve[j] = i;
        }
    }
    for (i = 2; i < MAX; i++) {
        if (!sieve[i])
            sieve[i] = 1;
        else
            sieve[i] = i / sieve[i];
    }
}

// Function to count minimum operations
ll minOperations(ll& N)
{
    if (N < 0)
        return -1;

    ll count = 0;
    makeSieve();
    while (N) {
        N -= sieve[N];
        count++;
    }
    return count;
}

// Driver Code
int main()
{
    ll N = 8;
    cout << minOperations(N);

    return 0;
}

Java

// JAVA code to implement the above approach
import java.util.*;
class GFG {

  static int MAX= 10000001;
  static int[] sieve = new int[MAX];

  // Function to calculate sieve
  static void makeSieve()
  {
    int i, j;
    sieve[0] = 0;
    sieve[1] = 1;

    for (i = 2; i < MAX; i++)
      sieve[i] = 0;

    for (i = 2; i * i <= MAX; i++) {
      if (sieve[i]==0) {
        sieve[i] = i;
        for (j = i * i; j < MAX; j += i)
          if (sieve[j]==0)
            sieve[j] = i;
      }
    }
    for (i = 2; i < MAX; i++) {
      if (sieve[i]==0)
        sieve[i] = 1;
      else
        sieve[i] = i / sieve[i];
    }
  }

  // Function to count minimum operations
  static int minOperations(int N)
  {
    if (N < 0)
      return -1;

    int count = 0;
    makeSieve();
    while (N != 0) {
      N -= sieve[N];
      count++;
    }
    return count;
  }

  // Driver code
  public static void main(String[] args)
  {
    int N = 8;
    System.out.print(minOperations(N));
  }
}

// This code is contributed by sanjoy_62.

Python3

# Python3 code to implement the above approach

MAX = 10000001


def makeSieve():
    sieve = [0] * MAX
    sieve[1] = 1
    for i in range(2, 1 + int(MAX ** 0.5)):
        if not sieve[i]:
            sieve[i] = i
            for j in range(i ** 2, MAX):
                if not sieve[j]:
                    sieve[j] = i
    for i in range(2, MAX):
        if not sieve[i]:
            sieve[i] = 1
        else:
            sieve[i] = (i // sieve[i])
    return sieve


def minOperations(N):
    if N < 0:
        return -1
    count = 0
    sieve = makeSieve()
    while N > 0:
        N -= (sieve[N])
        count += 1
    return count


# Driver Code
N = 8
print(minOperations(N))

C#

// C# implementation of above approach
using System;

class GFG{

  static int MAX= 10000001;
  static int[] sieve = new int[MAX];

  // Function to calculate sieve
  static void makeSieve()
  {
    int i, j;
    sieve[0] = 0;
    sieve[1] = 1;

    for (i = 2; i < MAX; i++)
      sieve[i] = 0;

    for (i = 2; i * i <= MAX; i++) {
      if (sieve[i]==0) {
        sieve[i] = i;
        for (j = i * i; j < MAX; j += i)
          if (sieve[j]==0)
            sieve[j] = i;
      }
    }
    for (i = 2; i < MAX; i++) {
      if (sieve[i]==0)
        sieve[i] = 1;
      else
        sieve[i] = i / sieve[i];
    }
  }

  // Function to count minimum operations
  static int minOperations(int N)
  {
    if (N < 0)
      return -1;

    int count = 0;
    makeSieve();
    while (N != 0) {
      N -= sieve[N];
      count++;
    }
    return count;
  }

  // Driver Code
  static public void Main (){

    int N = 8;
    Console.Write(minOperations(N));
  }
}

// This code is contributed by code_hunt.

Javascript

<script>
// JavaScript program to minimize operations
// to reduce N to 0 by replacing
// N by its divisor at each step

let MAX = 10000001
let sieve= new Array(MAX);

// Function to calculate sieve
function makeSieve()
{
    let i, j;
    sieve[0] = 0;
    sieve[1] = 1;

    for (i = 2; i < MAX; i++)
        sieve[i] = 0;

    for (i = 2; i * i <= MAX; i++) {
        if (!sieve[i]) {
            sieve[i] = i;
            for (j = i * i; j < MAX; j += i)
                if (!sieve[j])
                    sieve[j] = i;
        }
    }
    for (i = 2; i < MAX; i++) {
        if (!sieve[i])
            sieve[i] = 1;
        else
            sieve[i] = i / sieve[i];
    }
}

// Function to count minimum operations
function minOperations(N)
{
    if (N < 0)
        return -1;

    let count = 0;
    makeSieve();
    while (N) {
        N -= sieve[N];
        count++;
    }
    return count;
}


// Driver code
    let N = 8;

// Function call
    document.write(minOperations(N));

    // This code is contributed by jana_sayantan.
</script>

Output:

4

Time Complexity: O(N * log(log N)
Auxiliary Space: O(MAX), where MAX is the limit for sieve (here MAX = 10000001).