Minimum number of moves to reach N starting from (1, 1)
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: 5
(1, 1) -> (2, 1) -> (2, 2) -> (2, 3) -> (2, 4) -> (2, 5)
Input: N = 7
Output: 6
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> |
5
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1), since no extra space has been taken.
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