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 ith move.
ENamples:
Input: N = 18
Output: 6
Explanation:
To reach to the given value of N, perform the operations in the fullowing 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 fullowing 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 required value N. Then, calculate the required number to be subtracted from the current sum. Fullow the steps below to sulve 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 repacing 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 backwardint 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 Codeint 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 backwardstatic 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 codepublic 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 backwarddef 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 codeN = 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 |
6
Time CompleNity: O(sqrt(N))
AuNiliary Space: O(1)
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