Minimize steps required to reach the value N

Given an infinite number line from the range [-INFINITY, +INFINITY] and an integer N, the task is to find the minimum count of moves required to reach the N, starting from 0, by either moving i steps forward or 1 steps backward in every i^th move.

Examples:

Input: N = 18
Output: 6
Explanation:
To reach to the given value of N, perform the operations in the following sequence: 1 – 1 + 3 + 4 + 5 + 6 = 18
Therefore, a total of 6 operations are required.

Input: N = 3
Output: 2
Explanation:
To reach to the given value of N, perform the operations in the following sequence: 1 + 2 = 3
Therefore, a total of 2 operations are required.

Approach: The idea is to initially, keep adding 1, 2, 3 . . . . K, until it is greater than or equal to the required value N. Then, calculate the required number to be subtracted from the current sum. Follow the steps below to solve the problem:

• Initially, increment by K until N is greater than the current value. Now, stop at some position 
  pos = 1 + 2 + …………. + steps = steps ? (steps + 1) / 2 ? N.
  Note: 0 ? pos – N < steps. Otherwise, the last step wasn’t possible.
  • Case 1: If pos = N then, ‘steps’ is the required answer.
  • Case 2: If pos ? N, then replace any iteration of K with -1.
• By replacing any K with -1, the modified value of pos = pos – (K + 1). Since K ? [1, steps], then pos ? [pos – steps – 1, pos – 2].
• It is clear that pos – step < N. If N < pos – 1, then choose the corresponding K = pos – N – 1 and replace K with -1 and get straight to the point N.
• If N + 1 = pos, only one -1 operation is required.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to find the minimum steps
// required to reach N by either moving
// i steps forward or 1 steps backward
int minimumsteps(int N)
{

    // Stores the required count
    int steps = 0;

    // IF total moves required
    // is less than double of N
    while (steps * (steps + 1) < 2 * N) {

        // Update steps
        steps++;
    }

    // Steps required to reach N
    if (steps * (steps + 1) / 2 == N + 1) {

        // Update steps
        steps++;
    }

    cout << steps;
}

// Driver Code
int main()
{

    // Given value of N
    int N = 18;
    minimumsteps(N);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to find the minimum steps
// required to reach N by either moving
// i steps forward or 1 steps backward
static void minimumsteps(int N)
{

    // Stores the required count
    int steps = 0;

    // IF total moves required
    // is less than double of N
    while (steps * (steps + 1) < 2 * N)
    {

        // Update steps
        steps++;
    }

    // Steps required to reach N
    if (steps * (steps + 1) / 2 == N + 1)
    {

        // Update steps
        steps++;
    }

    System.out.println(steps);
}

// Driver code
public static void main(String[] args)
{

    // Given value of N
    int N = 18;
    minimumsteps(N);
}
}

// This code is contributed by code_hunt.

Python3

# Function to find the minimum steps
# required to reach N by either moving
# i steps forward or 1 steps backward

def minimumsteps(N) :

    # Stores the required count
    steps = 0

    # IF total moves required
    # is less than double of N
    while (steps * (steps + 1) < 2 * N) :

        # Update steps
        steps += 1

    # Steps required to reach N
    if (steps * (steps + 1) / 2 == N + 1) :

        # Update steps
        steps += 1
    print(steps)


# Driver code
N = 18;
minimumsteps(N)

# This code is contributed by Dharanendra L V

C#

// C# program for the above approach
using System;

class GFG {

    // Function to find the minimum steps
    // required to reach N by either moving
    // i steps forward or 1 steps backward
    static void minimumsteps(int N)
    {

        // Stores the required count
        int steps = 0;

        // IF total moves required
        // is less than double of N
        while (steps * (steps + 1) < 2 * N) {

            // Update steps
            steps++;
        }

        // Steps required to reach N
        if (steps * (steps + 1) / 2 == N + 1) {

            // Update steps
            steps++;
        }

        Console.WriteLine(steps);
    }

    // Driver code
    static public void Main()
    {

        // Given value of N
        int N = 18;
        minimumsteps(N);
    }
}

// This code is contributed by Dharanendra LV

Javascript

<script>

// JavaScript program for the above approach

// Function to find the minimum steps
// required to reach N by either moving
// i steps forward or 1 steps backward
function minimumsteps(N)
{

    // Stores the required count
    let steps = 0;

    // IF total moves required
    // is less than double of N
    while (steps * (steps + 1) < 2 * N)
    {

        // Update steps
        steps++;
    }

    // Steps required to reach N
    if (steps * Math.floor((steps + 1) / 2) == N + 1)
    {

        // Update steps
        steps++;
    }

    document.write(steps);
}

// Driver Code

// Given value of N
let N = 18;

minimumsteps(N);

// This code is contributed by Surbhi Tyagi.

</script>

Output:

6

Time CompleNity: O(sqrt(N))
AuNiliary Space: O(1)