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Minimum number of moves to reach N starting from (1, 1)

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  • Difficulty Level : Medium
  • Last Updated : 07 Mar, 2022

Given an integer N and an infinite table where ith row and jth column contains the value i *j. The task is to find the minimum number of moves to reach the cell containing N starting from the cell (1, 1)
Note: From (i, j) only valid moves are (i + 1, j) and (i, j + 1) 
Examples: 
 

Input: N = 10 
Output:
(1, 1) -> (2, 1) -> (2, 2) -> (2, 3) -> (2, 4) -> (2, 5)
Input: N = 7 
Output:
 

 

Approach: Note that any cell (i, j) can be reached in i + j – 2 steps. Thus, only the pair (i, j) is required with i * j = N that minimizes i + j. It can be found out by finding all the possible pairs (i, j) and check them in O(√N). To do this, without loss of generality, it can be assumed that i ≤ j and i ≤ √N since N = i * j ≥ i2. So √N ≥ i2 i.e. √N ≥ i
Thus, iterate over all the possible values of i from 1 to √N and, among all the possible pairs (i, j), pick the lowest value of i + j – 2 and that is the required answer.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum number
// of moves required to reach the cell
// containing N starting from (1, 1)
int min_moves(int n)
{
    // To store the required answer
    int ans = INT_MAX;
 
    // For all possible values of divisors
    for (int i = 1; i * i <= n; i++) {
 
        // If i is a divisor of n
        if (n % i == 0) {
 
            // Get the moves to reach n
            ans = min(ans, i + n / i - 2);
        }
    }
 
    // Return the required answer
    return ans;
}
 
// Driver code
int main()
{
    int n = 10;
 
    cout << min_moves(n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the minimum number
// of moves required to reach the cell
// containing N starting from (1, 1)
static int min_moves(int n)
{
    // To store the required answer
    int ans = Integer.MAX_VALUE;
 
    // For all possible values of divisors
    for (int i = 1; i * i <= n; i++)
    {
 
        // If i is a divisor of n
        if (n % i == 0)
        {
 
            // Get the moves to reach n
            ans = Math.min(ans, i + n / i - 2);
        }
    }
 
    // Return the required answer
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 10;
 
    System.out.println(min_moves(n));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python3 implementation of the approach
import sys
 
from math import sqrt
 
# Function to return the minimum number
# of moves required to reach the cell
# containing N starting from (1, 1)
def min_moves(n) :
 
    # To store the required answer
    ans = sys.maxsize;
 
    # For all possible values of divisors
    for i in range(1, int(sqrt(n)) + 1) :
 
        # If i is a divisor of n
        if (n % i == 0) :
 
            # Get the moves to reach n
            ans = min(ans, i + n // i - 2);
 
    # Return the required answer
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    n = 10;
 
    print(min_moves(n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
     
class GFG
{
     
// Function to return the minimum number
// of moves required to reach the cell
// containing N starting from (1, 1)
static int min_moves(int n)
{
    // To store the required answer
    int ans = int.MaxValue;
 
    // For all possible values of divisors
    for (int i = 1; i * i <= n; i++)
    {
 
        // If i is a divisor of n
        if (n % i == 0)
        {
 
            // Get the moves to reach n
            ans = Math.Min(ans, i + n / i - 2);
        }
    }
 
    // Return the required answer
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 10;
 
    Console.WriteLine(min_moves(n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // Function to return the minimum number
    // of moves required to reach the cell
    // containing N starting from (1, 1)
    function min_moves(n)
    {
        // To store the required answer
        let ans = Number.MAX_VALUE;
 
        // For all possible values of divisors
        for (let i = 1; i * i <= n; i++)
        {
 
            // If i is a divisor of n
            if (n % i == 0)
            {
 
                // Get the moves to reach n
                ans = Math.min(ans, i + parseInt(n / i, 10) - 2);
            }
        }
 
        // Return the required answer
        return ans;
    }
     
    let n = 10;
   
    document.write(min_moves(n));
             
</script>

Output: 

5

 

Time Complexity: O(sqrt(n))

Auxiliary Space: O(1)


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