Minimum step to reach one - GeeksforGeeks

13 Jul, 2021
Given a positive number N, we need to reach to 1 in minimum number of steps where a step is defined as converting N to (N-1) or converting N to its one of the bigger divisor. 
Formally, if we are at N, then in 1 step we can reach to (N – 1) or if N = u*v then we can reach to max(u, v) where u > 1 and v > 1. 
Examples:
Input : N = 17
Output : 4
We can reach to 1 in 4 steps as shown below,
17 -> 16(from 17 - 1) -> 4(from 4 * 4) -> 
2(from 2 * 2) -> 1(from 2 - 1)

Input : N = 50
Output : 5
We can reach to 1 in 5 steps as shown below,
50 -> 10(from 5 * 10) -> 5(from 2 * 5) -> 
4(from 5 - 1) -> 2(from 2 *2) -> 1(from 2 - 1)
We can solve this problem using breadth first search because it works level by level so we will reach to 1 in minimum number of steps where next level for N contains (N – 1) and bigger proper factors of N. Complete BFS procedure will be as follows, First we will push N with steps 0 into the data queue then at each level we will push their next level elements with 1 step more than its previous level elements. In this way when 1 will be popped out from queue, it will contain minimum number of steps with it, which will be our final result. In below code a queue of a structure of ‘data’ type is used which stores value and steps from N in it, another set of integer type is used to save ourselves from pushing the same element more than once which can lead to an infinite loop. So at each step, we push the value into set after pushing that into the queue so that the value won’t be visited more than once. 
Please see below code for better understanding,  
C++
Java
C#
Javascript
//  C++ program to get minimum step to reach 1// under given constraints#include <bits/stdc++.h>using namespace std; //  structure represent one node in queuestruct data{    int val;    int steps;    data(int val, int steps) : val(val), steps(steps)    {}}; //  method returns minimum step to reach oneint minStepToReachOne(int N){    queue<data> q;    q.push(data(N, 0));     // set is used to visit numbers so that they    // won't be pushed in queue again    set<int> st;     //  loop until we reach to 1    while (!q.empty())    {        data t = q.front();     q.pop();                 // if current data value is 1, return its        // steps from N        if (t.val == 1)            return t.steps;         //  check curr - 1, only if it not visited yet        if (st.find(t.val - 1) == st.end())        {            q.push(data(t.val - 1, t.steps + 1));            st.insert(t.val - 1);        }         //  loop from 2 to sqrt(value) for its divisors        for (int i = 2; i*i <= t.val; i++)        {             // check divisor, only if it is not visited yet            // if i is divisor of val, then val / i will            // be its bigger divisor            if (t.val % i == 0 && st.find(t.val / i) == st.end())            {                q.push(data(t.val / i, t.steps + 1));                st.insert(t.val / i);            }        }    }} //  Driver code to test above methodsint main(){    int N = 17;    cout << minStepToReachOne(N) << endl;}
// Java program to get minimum step to reach 1// under given constraintsimport java.util.*; class GFG{ // structure represent one node in queuestatic class data{    int val;    int steps;    public data(int val, int steps)    {        this.val = val;        this.steps = steps;    }     }; // method returns minimum step to reach onestatic int minStepToReachOne(int N){    Queue<data> q = new LinkedList<>();    q.add(new data(N, 0));     // set is used to visit numbers so that they    // won't be pushed in queue again    HashSet<Integer> st = new HashSet<Integer>();     // loop until we reach to 1    while (!q.isEmpty())    {        data t = q.peek(); q.remove();                 // if current data value is 1, return its        // steps from N        if (t.val == 1)            return t.steps;         // check curr - 1, only if it not visited yet        if (!st.contains(t.val - 1))        {            q.add(new data(t.val - 1, t.steps + 1));            st.add(t.val - 1);        }         // loop from 2 to Math.sqrt(value) for its divisors        for (int i = 2; i*i <= t.val; i++)        {             // check divisor, only if it is not visited yet            // if i is divisor of val, then val / i will            // be its bigger divisor            if (t.val % i == 0 && !st.contains(t.val / i) )            {                q.add(new data(t.val / i, t.steps + 1));                st.add(t.val / i);            }        }    }    return -1;} // Driver code public static void main(String[] args){    int N = 17;    System.out.print(minStepToReachOne(N) +"\n");}} // This code is contributed by 29AjayKumar
// C# program to get minimum step to reach 1// under given constraintsusing System;using System.Collections.Generic; class GFG{ // structure represent one node in queueclass data{    public int val;    public int steps;    public data(int val, int steps)    {        this.val = val;        this.steps = steps;    }}; // method returns minimum step to reach onestatic int minStepToReachOne(int N){    Queue<data> q = new Queue<data>();    q.Enqueue(new data(N, 0));     // set is used to visit numbers so that they    // won't be pushed in queue again    HashSet<int> st = new HashSet<int>();     // loop until we reach to 1    while (q.Count != 0)    {        data t = q.Peek(); q.Dequeue();                 // if current data value is 1, return its        // steps from N        if (t.val == 1)            return t.steps;         // check curr - 1, only if it not visited yet        if (!st.Contains(t.val - 1))        {            q.Enqueue(new data(t.val - 1, t.steps + 1));            st.Add(t.val - 1);        }         // loop from 2 to Math.Sqrt(value) for its divisors        for (int i = 2; i*i <= t.val; i++)        {             // check divisor, only if it is not visited yet            // if i is divisor of val, then val / i will            // be its bigger divisor            if (t.val % i == 0 && !st.Contains(t.val / i) )            {                q.Enqueue(new data(t.val / i, t.steps + 1));                st.Add(t.val / i);            }        }    }    return -1;} // Driver codepublic static void Main(String[] args){    int N = 17;    Console.Write(minStepToReachOne(N) +"\n");}} // This code is contributed by 29AjayKumar
<script> // Javascript program to get minimum step// to reach 1 under given constraints // Structure represent one node in queueclass data{    constructor(val, steps)    {        this.val = val;        this.steps = steps;    }} // Method returns minimum step to reach onefunction minStepToReachOne(N){    let q = [];    q.push(new data(N, 0));       // Set is used to visit numbers    // so that they won't be pushed    // in queue again    let st = new Set();       // Loop until we reach to 1    while (q.length != 0)    {        let t = q.shift();                   // If current data value is 1,        // return its steps from N        if (t.val == 1)            return t.steps;           // Check curr - 1, only if        // it not visited yet        if (!st.has(t.val - 1))        {            q.push(new data(t.val - 1,                          t.steps + 1));            st.add(t.val - 1);        }           // Loop from 2 to Math.sqrt(value)        // for its divisors        for(let i = 2; i*i <= t.val; i++)        {                         // Check divisor, only if it is not            // visited yet if i is divisor of val,            // then val / i will be its bigger divisor            if (t.val % i == 0 && !st.has(t.val / i))            {                q.push(new data(t.val / i,                              t.steps + 1));                st.add(t.val / i);            }        }    }    return -1;} // Driver code let N = 17;document.write(minStepToReachOne(N) + "<br>"); // This code is contributed by rag2127 </script>
Output: 
4
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