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++ 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 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 |
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# 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 |
<script> // JavaScript program to find a number in minimum steps var InF = 99999;
// To represent data of a node in tree
class number
{
constructor(n, l)
{
this .no = n;
this .level = l;
}
};
// Prints level of node n
function findnthnumber(n)
{
// Create a queue and insert root
var q = [];
var r = new number(0, 1);
q.push(r);
// Do level order traversal
while (q.length != 0)
{
// Remove a node from queue
var temp = q[0];
q.shift();
// 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)
{
document.write( "Found number n at level "
+ (temp.level - 1));
break ;
}
// Insert children of dequeued node to queue
q.push( new number(temp.no + temp.level, temp.level + 1));
q.push( new number(temp.no - temp.level, temp.level + 1));
}
}
// Driver code
findnthnumber(13);
</script> |
Found number n at level 5
Time Complexity: O(n)
Space Complexity: O(n)
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++ 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 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 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# 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 |
<script> // JavaScript program to Find a
// number in minimum steps
function find(n)
{
// Steps sequence
let ans = [];
// Current sum
let sum = 0;
let i = 1;
// Sign of the number
let sign = (n >= 0 ? 1 : -1);
n = Math.abs(n);
// Basic steps required to get
// sum >= required value.
for (; sum < n;)
{
ans.push(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.push(sign * i);
sum += i;
i++;
}
ans[parseInt(sum / 2, 10) - 1] =
ans[parseInt(sum / 2, 10) - 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--;
ans.push(sign * i);
ans.push(sign * -1 * (i + 1));
}
ans[parseInt(sum / 2, 10) - 1] =
ans[parseInt(sum / 2, 10) - 1] * -1;
}
}
return ans;
}
let n = 20;
if (n == 0)
document.write( "Minimum number of Steps: 0" +
"</br>" + "Step sequence:" + "</br>" + "0" );
else
{
let a = find(n);
document.write( "Minimum number of Steps: " +
a.length + "</br>" + "Step sequence:" +
"</br>" );
for (let i = 0; i < a.length; i++)
document.write(a[i] + " " );
}
</script> |
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))