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Find a number in minimum steps
  • Difficulty Level : Medium
  • Last Updated : 17 Feb, 2021

Given an infinite number line from -INFINITY to +INFINITY and we are on zero. We can move n steps either side at each n’th time. 

Approach 1 : Using Tree 

1st time; we can move only 1 step to both ways, means -1 1;

2nd time we can move 2 steps  from -1 and 1;
-1 :  -3 (-1-2)  1(-1+2)
 1 :  -1 ( 1-2)  3(1+2)

3rd time we can move 3 steps either way from -3, 1, -1, 3 
-3:  -6(-3-3) 0(-3+3)
1:   -2(1-3)   4(1+3)
-1:  -4(-1-3)  2(-1+3)
3:     0(0-3)   6(3+3) 

Find the minimum number of steps to reach a given number n.

Examples: 

Input : n = 10
Output : 4
We can reach 10 in 4 steps,  1, 3, 6, 10 


Input : n = 13
Output : 5
We can reach 10 in 4 steps,  -1, 2, 5, 9, 14

This problem can be modeled as tree. We put initial point 0 at root, 1 and -1 as children of root. Next level contains values at distance 2 and so on.  

              0
            /   \
         -1       1  
        /  \     /  \
       1   -3   -1   3
     /  \  / \  / \  / \

The problem is now to find the closes node to root with value n. The idea is to do Level Order Traversal of tree to find the closest node. Note that using DFS for closest node is never a good idea (we may end up going down many unnecessary levels). 



Below is C++, Python implementation of above idea.

C++




// C++ program to find a number in minimum steps
#include <bits/stdc++.h>
using namespace std;
#define InF 99999
 
// To represent data of a node in tree
struct number {
    int no;
    int level;
 
public:
    number() {}
    number(int n, int l)
        : no(n), level(l)
    {
    }
};
 
// Prints level of node n
void findnthnumber(int n)
{
    // Create a queue and insert root
    queue<number> q;
    struct number r(0, 1);
    q.push(r);
 
    // Do level order traversal
    while (!q.empty()) {
        // Remove a node from queue
        struct number temp = q.front();
        q.pop();
 
        // To avoid infinite loop
        if (temp.no >= InF || temp.no <= -InF)
            break;
 
        // Check if dequeued number is same as n
        if (temp.no == n) {
            cout << "Found number n at level "
                 << temp.level - 1;
            break;
        }
 
        // Insert children of dequeued node to queue
        q.push(number(temp.no + temp.level, temp.level + 1));
        q.push(number(temp.no - temp.level, temp.level + 1));
    }
}
 
// Driver code
int main()
{
    findnthnumber(13);
    return 0;
}


Java




// Java program to find a number in minimum steps
import java.util.*;
class GFG
{
  static final int InF = 99999;
 
  // To represent data of a node in tree
  static class number
  {
    int no;
    int level;
    number() {}
    number(int n, int l)
    {
      this.no = n;
      this.level = l;
    }
  };
 
  // Prints level of node n
  static void findnthnumber(int n)
  {
 
    // Create a queue and insert root
    Queue<number> q = new LinkedList<>();
    number r = new number(0, 1);
    q.add(r);
 
    // Do level order traversal
    while (!q.isEmpty())
    {
 
      // Remove a node from queue
      number temp = q.peek();
      q.remove();
 
      // To astatic void infinite loop
      if (temp.no >= InF || temp.no <= -InF)
        break;
 
      // Check if dequeued number is same as n
      if (temp.no == n)
      {
        System.out.print("Found number n at level "
                         + (temp.level - 1));
        break;
      }
 
      // Insert children of dequeued node to queue
      q.add(new number(temp.no + temp.level, temp.level + 1));
      q.add(new number(temp.no - temp.level, temp.level + 1));
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
    findnthnumber(13);
  }
}
 
// This code is contributed by gauravrajput1


Python3




from collections import deque
 
# Python program to find a number in minimum steps
InF = 99999
 
# To represent data of a node in tree
class number:
    def __init__(self,n,l):
        self.no = n
        self.level = l
 
# Prints level of node n
def findnthnumber(n):
    # Create a queue and insert root
    q = deque()
    r = number(0, 1)
    q.append(r)
 
    # Do level order traversal
    while (len(q) > 0):
        # Remove a node from queue
        temp = q.popleft()
        # q.pop()
 
        # To avoid infinite loop
        if (temp.no >= InF or temp.no <= -InF):
            break
 
        # Check if dequeued number is same as n
        if (temp.no == n):
            print("Found number n at level", temp.level - 1)
            break
 
        # Insert children of dequeued node to queue
        q.append(number(temp.no + temp.level, temp.level + 1))
        q.append(number(temp.no - temp.level, temp.level + 1))
 
# Driver code
if __name__ == '__main__':
    findnthnumber(13)
 
# This code is contributed by mohit kumar 29


C#




// C# program to find a number in minimum steps
using System;
using System.Collections.Generic;
public
class GFG
{
  static readonly int InF = 99999;
 
  // To represent data of a node in tree
  public
 class number
  {
    public
 int no;
    public
 int level;
    public
 number() {}
    public
 number(int n, int l)
    {
      this.no = n;
      this.level = l;
    }
  };
 
  // Prints level of node n
  static void findnthnumber(int n)
  {
 
    // Create a queue and insert root
    Queue<number> q = new Queue<number>();
    number r = new number(0, 1);
    q.Enqueue(r);
 
    // Do level order traversal
    while (q.Count != 0)
    {
 
      // Remove a node from queue
      number temp = q.Peek();
      q.Dequeue();
 
      // To astatic void infinite loop
      if (temp.no >= InF || temp.no <= -InF)
        break;
 
      // Check if dequeued number is same as n
      if (temp.no == n)
      {
        Console.Write("Found number n at level "
                         + (temp.level - 1));
        break;
      }
 
      // Insert children of dequeued node to queue
      q.Enqueue(new number(temp.no + temp.level, temp.level + 1));
      q.Enqueue(new number(temp.no - temp.level, temp.level + 1));
    }
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    findnthnumber(13);
  }
}
 
// This code is contributed by gauravrajput1


Output : 

Found number n at level 5

The above solution is contributed by Mu Ven
 

Approach 2 : Using Vector

The above solution uses binary tree for nth time instance i.e. -n and n. But as the level of tree increases this becomes inefficient. For values like abs(200) or more above program gives segmentation fault.
Below solution does not make a tree and takes complexity equal to exact number of steps required. Also the steps required are printed in the array which equals the exact sum required.

Main Idea:  

  • Distance between +n and -n is 2*n. So if you negate a number from +ve to -ve it will create difference of 2*n from previous sum.
  • If a number lies between n(n+1)/2 and (n+1)(n+2)/2 for any n then we go to step (n+1)(n+2)/2 and try to decrease the sum to the difference using idea discussed above.
  • If we go to n(n+1)/2 and then try to increase than it will ultimately lead you to same number of steps. 
    And since you cannot negate any number (as sum is already less than required sum) from n(n+1)/2 this proves that it takes minimum number of steps.

C++




// C++ program to Find a
// number in minimum steps
#include <bits/stdc++.h>
using namespace std;
 
vector<int> find(int n)
{
    // Steps sequence
    vector<int> ans;
 
    // Current sum
    int sum = 0;
    int i;
 
    // Sign of the number
    int sign = (n >= 0 ? 1 : -1);
    n = abs(n);
 
    // Basic steps required to get
    // sum >= required value.
    for (i = 1; sum < n; i++) {
        ans.push_back(sign * i);
        sum += i;
    }
 
    // If we have reached ahead to destination.
    if (sum > sign * n) {
        /*If the last step was an odd number,
         then it has following mechanism for
         negating a particular number and
         decreasing the sum to required number
         Also note that it may require
         1 more step in order to reach the sum. */
        if (i % 2) {
            sum -= n;
            if (sum % 2) {
                ans.push_back(sign * i);
                sum += i++;
            }
            ans[(sum / 2) - 1] *= -1;
        }
        else {
            /* If the current time instance is even
            and sum is odd than it takes
            2 more steps and few
            negations in previous elements
            to reach there. */
            sum -= n;
            if (sum % 2) {
                sum--;
                ans.push_back(sign * i);
                ans.push_back(sign * -1 * (i + 1));
            }
            ans[(sum / 2) - 1] *= -1;
        }
    }
    return ans;
}
 
// Driver Program
int main()
{
    int n = 20;
    if (n == 0)
        cout << "Minimum number of Steps: 0\nStep sequence:\n0";
    else {
        vector<int> a = find(n);
        cout << "Minimum number of Steps: " << a.size() << "\nStep sequence:\n";
        for (int i : a)
            cout << i << " ";
    }
    return 0;
}


Java




// Java program to Find a
// number in minimum steps
import java.util.*;
class GFG
{
  static Vector<Integer> find(int n)
  {
 
    // Steps sequence
    Vector<Integer> ans = new Vector<>();
 
    // Current sum
    int sum = 0;
    int i = 1;
 
    // Sign of the number
    int sign = (n >= 0 ? 1 : -1);
    n = Math.abs(n);
 
    // Basic steps required to get
    // sum >= required value.
    for (; sum < n;)
    {
      ans.add(sign * i);
      sum += i;
      i++;
    }
 
    // If we have reached ahead to destination.
    if (sum > sign * n)
    {
 
      /*If the last step was an odd number,
         then it has following mechanism for
         negating a particular number and
         decreasing the sum to required number
         Also note that it may require
         1 more step in order to reach the sum. */
      if (i % 2 != 0)
      {
        sum -= n;
        if (sum % 2 != 0)
        {
          ans.add(sign * i);
          sum += i;
          i++;
        }
        int a = ans.get((sum / 2) - 1);
        ans.remove((sum / 2) - 1);
        ans.add(((sum / 2) - 1), a*(-1));
      }
      else
      {
 
        /* If the current time instance is even
            and sum is odd than it takes
            2 more steps and few
            negations in previous elements
            to reach there. */
        sum -= n;
        if (sum % 2 != 0)
        {
          sum--;
          ans.add(sign * i);
          ans.add(sign * -1 * (i + 1));
        }
        ans.add((sum / 2) - 1, ans.get((sum / 2) - 1) * -1);
      }
    }
    return ans;
  }
 
  // Driver Program
  public static void main(String[] args)
  {
    int n = 20;
    if (n == 0)
      System.out.print("Minimum number of Steps: 0\nStep sequence:\n0");
    else
    {
      Vector<Integer> a = find(n);
      System.out.print("Minimum number of Steps: "
                       a.size()+ "\nStep sequence:\n");
      for (int i : a)
        System.out.print(i + " ");
    }
  }
}
 
// This code is contributed by aashish1995


Python3




# Python3 program to Find a
# number in minimum steps
def find(n):
     
    # Steps sequence
    ans = []
     
    # Current sum
    Sum = 0
    i = 0
     
    # Sign of the number
    sign = 0
     
    if (n >= 0):
        sign = 1
    else:
        sign = -1
 
    n = abs(n)
    i = 1
     
    # Basic steps required to get
    # sum >= required value.
    while (Sum < n):
        ans.append(sign * i)
        Sum += i
        i += 1
 
    # If we have reached ahead to destination.
    if (Sum > sign * n):
         
        # If the last step was an odd number,
        # then it has following mechanism for
        # negating a particular number and
        # decreasing the sum to required number
        # Also note that it may require
        # 1 more step in order to reach the sum.
        if (i % 2!=0):
            Sum -= n
             
            if (Sum % 2 != 0):
                ans.append(sign * i)
                Sum += i
                i += 1
                 
            ans[int(Sum / 2) - 1] *= -1
        else:
             
            # If the current time instance is even
            # and sum is odd than it takes
            # 2 more steps and few
            # negations in previous elements
            # to reach there.
            Sum -= n
             
            if (Sum % 2 != 0):
                Sum -= 1
                ans.append(sign * i)
                ans.append(sign * -1 * (i + 1))
                 
            ans[int((sum / 2)) - 1] *= -1
     
    return ans
 
# Driver code
n = 20
 
if (n == 0):
    print("Minimum number of Steps: 0\nStep sequence:\n0")
else:
    a = find(n)
    print("Minimum number of Steps:", len(a))
    print("Step sequence:")
    print(*a, sep = " ")
 
# This code is contributed by rag2127


C#




// C# program to Find a
// number in minimum steps
using System;
using System.Collections.Generic;
public class GFG
{
  static List<int> find(int n)
  {
 
    // Steps sequence
    List<int> ans = new List<int>();
 
    // Current sum
    int sum = 0;
    int i = 1;
 
    // Sign of the number
    int sign = (n >= 0 ? 1 : -1);
    n = Math.Abs(n);
 
    // Basic steps required to get
    // sum >= required value.
    for (; sum < n;)
    {
      ans.Add(sign * i);
      sum += i;
      i++;
    }
 
    // If we have reached ahead to destination.
    if (sum > sign * n)
    {
 
      /*If the last step was an odd number,
         then it has following mechanism for
         negating a particular number and
         decreasing the sum to required number
         Also note that it may require
         1 more step in order to reach the sum. */
      if (i % 2 != 0)
      {
        sum -= n;
        if (sum % 2 != 0)
        {
          ans.Add(sign * i);
          sum += i;
          i++;
        }
        int a = ans[((sum / 2) - 1)];
        ans.RemoveAt((sum / 2) - 1);
        ans.Insert(((sum / 2) - 1), a*(-1));
      }
      else
      {
 
        /* If the current time instance is even
            and sum is odd than it takes
            2 more steps and few
            negations in previous elements
            to reach there. */
        sum -= n;
        if (sum % 2 != 0)
        {
          sum--;
          ans.Add(sign * i);
          ans.Add(sign * -1 * (i + 1));
        }
        ans.Insert((sum / 2) - 1, ans[(sum / 2) - 1] * -1);
      }
    }
    return ans;
  }
 
  // Driver Program
  public static void Main(String[] args)
  {
    int n = 20;
    if (n == 0)
      Console.Write("Minimum number of Steps: 0\nStep sequence:\n0");
    else
    {
      List<int> a = find(n);
      Console.Write("Minimum number of Steps: "
                    a.Count+ "\nStep sequence:\n");
      foreach (int i in a)
        Console.Write(i + " ");
    }
  }
}
 
// This code is contributed by Rajput-Ji


Output : 

 
Minimum number of Steps: 7
Step sequence:
1 2 3 -4 5 6 7

If n is the sum that it is required and s is the minimum steps then: 
n = (s+1)*(s+2)/2 + 1 (or +2) 
Hence n = O(s*s)
Therefore s = O(sqrt(n)) 
Space Complexity : O(sqrt(n)) 
Time complexity : O(sqrt(n)) 
https://youtu.be/GcrapHAFnLg
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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