Find a number in minimum steps

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++ implementation of above idea.

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// 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;
}

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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.
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// CPP 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;
}

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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))

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : Shreyans Vora



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