Given an integer N, the task is to find the minimum number of steps to obtain N from M (M = 1 initially). In each step, M/X can be added to M where X is any positive integer.
Examples:
Input: N = 5
Output: 3
Explanation: Initially the number is 1. 1/1 is added to make it 2.
In next step adding 2/1 = 2 it becomes 4. At last add 4/4 = 1 to get the 5.
This is the minimum steps required to convert 1 to 5Input: N = 7
Output: 4
Explanation: Initially the number is 1.
Now 1/1 is added to it and it becomes 2.
After adding 2/1 = 2 it becomes 4. In the third 4/2 = 2 is added and it becomes 6.
At the final step add 6/6 = 1 and it becomes 7.
Approach: The approach of this question is using Dynamic Programming. For each integer, there are many possible moves. Store the minimum steps required to reach every number in the dp[] array and use this for the next numbers. Follow the steps mentioned below.
- Start iterating from 2 to N.
- For each number i, do the following:
- Check from which numbers (less than i), we can reach i.
- Now for those numbers find the minimum steps required to reach i by using the relation dp[i] = min(dp[i], dp[j]+1) where j is such a number from where i can be reached.
- Store that minimum value in dp[i] array.
- After iteration is done for all elements up to N return value of dp[N].
Below is the implementation of the above approach:
C++
// C++ code to implement above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum steps requiredint minSteps(int N){ vector<int> dp(N + 1, INT_MAX); dp[1] = 0; // Loop to find the minimum steps to // reach N from 1 for (int i = 2; i <= N; ++i) { for (int j = 1; j <= i; ++j) { // Finding the distance // between two numbers int distance = i - j; if (distance == 0) { continue; } // Divide the number int divide = j / distance; if (divide != 0) { // Checking if the number // can be reached or not if (j / divide == distance) { dp[i] = min(dp[j] + 1, dp[i]); } } } } return dp[N];}// Driver codeint main(){ int N = 7; int ans = minSteps(N); cout << (ans); return 0;}// This code is contributed by rakeshsahni |
Java
// Java code to implement above approachimport java.io.*;import java.util.*;class GFG { // Function to find the minimum steps required static int minSteps(int N) { int dp[] = new int[N + 1]; Arrays.fill(dp, Integer.MAX_VALUE); dp[1] = 0; // Loop to find the minimum steps to // reach N from 1 for (int i = 2; i <= N; ++i) { for (int j = 1; j <= i; ++j) { // Finding the distance // between two numbers int distance = i - j; if (distance == 0) { continue; } // Divide the number int divide = j / distance; if (divide != 0) { // Checking if the number // can be reached or not if (j / divide == distance) { dp[i] = Math.min(dp[j] + 1, dp[i]); } } } } return dp[N]; } // Driver code public static void main(String[] args) { int N = 7; int ans = minSteps(N); System.out.println(ans); }} |
Python
# Python] code to implement above approachimport sys# Function to find the minimum steps requireddef minSteps(N): dp = [] dp = [sys.maxsize for i in range(N + 1)] dp[1] = 0; # Loop to find the minimum steps to # reach N from 1 for i in range(2, N + 1): for j in range(1, i + 1): # Finding the distance # between two numbers distance = i - j if (distance == 0): continue # Divide the number divide = j // distance; if (divide != 0): # Checking if the number # can be reached or not if (j // divide == distance): dp[i] = min(dp[j] + 1, dp[i]) return dp[N]# Driver codeN = 7ans = minSteps(N);print(ans)# This code is contributed by Samim Hossain Mondal. |
C#
// C# program for the above approachusing System;public class GFG{ // Function to find the minimum steps required static int minSteps(int N) { int[] dp = new int[N + 1]; for(int i = 0; i < N + 1; i++) dp[i] = Int32.MaxValue; dp[1] = 0; // Loop to find the minimum steps to // reach N from 1 for (int i = 2; i <= N; ++i) { for (int j = 1; j <= i; ++j) { // Finding the distance // between two numbers int distance = i - j; if (distance == 0) { continue; } // Divide the number int divide = j / distance; if (divide != 0) { // Checking if the number // can be reached or not if (j / divide == distance) { dp[i] = Math.Min(dp[j] + 1, dp[i]); } } } } return dp[N]; } // Driver code static public void Main (){ int N = 7; int ans = minSteps(N); Console.Write(ans); }}// This code is contributed by hrithikgarg03188. |
Javascript
<script>// JavaScript code for the above approach// Function to find the minimum steps requiredfunction minSteps( N){ let dp = new Array(N + 1).fill(Number.MAX_VALUE); dp[1] = 0; // Loop to find the minimum steps to // reach N from 1 for (let i = 2; i <= N; ++i) { for (let j = 1; j <= i; ++j) { // Finding the distance // between two numbers let distance = i - j; if (distance == 0) { continue; } // Divide the number let divide =Math.floor(j / distance); if (divide != 0) { // Checking if the number // can be reached or not if (j / divide == distance) { dp[i] = Math.min(dp[j] + 1, dp[i]); } } } } return dp[N];}// Driver code let N = 7; let ans = minSteps(N); document.write((ans));// This code is contributed by Potta Lokesh </script> |
Output
4
Time Complexity: O(N*N)
Auxiliary Space: O(N)
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