Data Modelling & AI Data Structure & Algorithm

Min steps to convert N-digit prime number into another by replacing a digit in each step

23 June 2025

0

Given two N-digit prime numbers A and B, the task is to find the minimum number of steps taken to convert A to B. Condition for the conversion is that only 1 digit of the current prime number can be modified such that the new number formed is also a prime number. If no such conversion is possible, print -1.

Note: The range of N is [1, 5].

Examples:

Input: N = 4, A = 1033, B = 8179
Output: 6
Explanation: The steps of conversion are 1033 -> 1733 -> 3733 -> 3739 -> 3779 -> 8779 -> 8179. While changing numbers from 1033->1733, they differ by only one digit and the same happens for subsequent steps.

Input: N = 4, A = 1373, B = 8179
Output: 7

Input: N = 2, A = 11, B = 37
Output: 2
Explanation: The steps of conversion are 11 -> 17 -> 37

Approach: Using Breadth First Search Algorithm

Find all N digit prime numbers and make a graph with these numbers.
Consider each prime number as a node of a graph and create an edge from one node to another if they differ by a single digit.
Apply BFS traversal and find out the number of edges between A and B.
If no path exists, print -1.
Else print the no. of edges, which is the required solution.

Below is the implementation of the above approach.

C++

// C++ program for the above problem
 
#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define mod 1000000007
#define pb push_back
#define mod 1000000007
#define vi vector<int>
 
// adjacency list for numbers
// till 100001
vi lis[100001];
vi primes;
 
// visited array
int vis[100001];
 
// to store distance of every
// vertex from A
int dis[100001];
 
// function to check if number
// is a prime
bool isPrime(int n)
{
    for (int i = 2;
         i * i <= n; i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// function to check if numbers
// differ by only a single-digit
bool valid(int a, int b)
{
    int c = 0;
    while (a) {
 
        // check the last digit of
        // both numbers and increase
        // count if different
        if ((a % 10) != (b % 10)) {
            c++;
        }
        a = a / 10;
        b = b / 10;
    }
    if (c == 1) {
        return true;
    }
    else {
        return false;
    }
}
 
void makePrimes(int N)
{
    int i, j;
    int L = pow(10, N - 1);
    int R = pow(10, N) - 1;
    // generate all N digit primes
    for (int i = L; i <= R; i++) {
        if (isPrime(i)) {
            primes.pb(i);
        }
    }
    for (i = 0;
         i < primes.size(); i++) {
 
        // for every prime number i
        // check if an edge
        // can be made.
        for (j = i + 1;
             j < primes.size(); j++) {
            int a = primes[i];
            int b = primes[j];
 
            // for every prime number
            // i check if an edge can
            // be made from i to j.
            if (valid(a, b)) {
 
                // if edge is possible then
                // insert in the adjacency
                // list
                lis[a].pb(b);
                lis[b].pb(a);
            }
        }
    }
}
 
// function to count distance
void bfs(int src)
{
    queue<int> q;
    q.push(src);
    vis[src] = 1;
    dis[src] = 0;
    while (!q.empty()) {
        int curr = q.front();
        q.pop();
        for (int x : lis[curr]) {
 
            // if unvisited push onto queue
            // and mark visited as 1 and
            // add the distance of curr+1.
            if (vis[x] == 0) {
                vis[x] = 1;
                q.push(x);
                dis[x] = dis[curr] + 1;
            }
        }
    }
}
 
// Driver code
int main()
{
    int N = 4;
    makePrimes(N);
    int A = 1033, B = 8179;
 
    // Call bfs traversal
    // with root as node A
    bfs(A);
 
    if (dis[B] == -1)
       
        // Indicates not possible
        cout << "-1" << endl;
    else
        cout << dis[B] << endl;
 
    return 0;
}

Java

// Java program for the above problem 
import java.util.*;
public class GFG
{
 
  // Adjacency list for numbers 
  // till 100001
  static Vector<Vector<Integer>> lis = new Vector<Vector<Integer>>();     
  static Vector<Integer> primes = new Vector<Integer>();
 
  // Visited array
  static int[] vis = new int[100001];
 
  // To store distance of every 
  // vertex from A 
  static int[] dis = new int[100001];
 
  // Function to check if number 
  // is a prime
  static boolean isPrime(int n)
  {
    int i = 2;
    while (i * i <= n)
    {
      if (n % i == 0)
        return false;
      i += 1;
    }
    return true;
  }
 
  // Function to check if numbers 
  // differ by only a single-digit
  static boolean valid(int a, int b)
  {
    int c = 0;        
    while(a > 0)
    {
 
      // Check the last digit of 
      // both numbers and increase 
      // count if different 
      if ((a % 10) != (b % 10))
        c += 1;             
      a = a / 10;
      b = b / 10;
    }     
    if (c == 1)
      return true;
    else
      return false;
  }
 
  static void makePrimes(int N)
  {
 
    // Generate all N digit primes 
    int L = (int)Math.pow(10, N - 1);
    int R = (int)Math.pow(10, N) - 1;
 
    for(int i = L; i < R + 1; i++)
    {
      if (isPrime(i))
        primes.add(i);
    }
 
    for(int i = 0; i < primes.size(); i++)
    {
 
      // For every prime number i 
      // check if an edge 
      // can be made. 
      for(int j = i + 1; j < primes.size(); j++)
      {
        int a = primes.get(i);
        int b = primes.get(j);
 
        // For every prime number 
        // i check if an edge can 
        // be made from i to j. 
        if (valid(a, b))
        {
 
          // If edge is possible then 
          // insert in the adjacency 
          // list
          lis.get(a).add(b);
          lis.get(b).add(a);
        }
      }
    }
  }
 
  // Function to count distance 
  static void bfs(int src)
  {
    Vector<Integer> q = new Vector<Integer>();
    q.add(src);
    vis[src] = 1;
    dis[src] = 0;
    while (q.size() != 0)
    {
      int curr = q.get(0);
      q.remove(0);
      for(int x : lis.get(curr))
      {
        // If unvisited push onto queue 
        // and mark visited as 1 and 
        // add the distance of curr+1.
        if (vis[x] == 0)
        {
          vis[x] = 1;
          q.add(x);
          dis[x] = dis[curr] + 1;
        }
      }
    }
  }
 
  // Driver code
  public static void main(String[] args) 
  {
    for(int i = 0; i < 100001; i++)
    {
      lis.add(new Vector<Integer>());
    }
    int N = 4;
    makePrimes(N); 
    int A = 1033;
    int B = 8179;
 
    // Call bfs traversal 
    // with root as node A 
    bfs(A);
 
    if (dis[B] == -1)
    {
      // Indicates not possible 
      System.out.print(-1);
    }
    else
    {
      System.out.print(dis[B]);
    }
  }
}
 
// This code is contributed by divyesh072019

Python3

# Python3 program for the above problem 
mod = 1000000007
 
# Adjacency list for numbers 
# till 100001 
lis = [[] for i in range(100001)]
primes = []
 
# Visited array
vis = [0 for i in range(100001)]
 
# To store distance of every 
# vertex from A 
dis = [0 for i in range(100001)]
 
# Function to check if number 
# is a prime
def isPrime(n):
     
    i = 2
    while (i * i <= n):
        if (n % i == 0):
            return False
             
        i += 1
         
    return True
 
# Function to check if numbers 
# differ by only a single-digit
def valid(a, b):
     
    c = 0
     
    while(a):
         
        # Check the last digit of 
        # both numbers and increase 
        # count if different 
        if ((a % 10) != (b % 10)):
            c += 1
             
        a = int(a / 10)
        b = int(b / 10)
 
    if (c == 1):
        return True
    else:
        return False
 
def makePrimes(N):
     
    global primes
    global lis
    i = 0
    j = 0
     
    # Generate all N digit primes 
    L = pow(10, N - 1)
    R = pow(10, N) - 1
 
    for i in range(L, R + 1):
        if (isPrime(i)):
            primes.append(i)
             
    for i in range(len(primes)):
         
        # For every prime number i 
        # check if an edge 
        # can be made. 
        for j in range(i + 1, len(primes)):
            a = primes[i]
            b = primes[j]
 
            # For every prime number 
            # i check if an edge can 
            # be made from i to j. 
            if (valid(a, b)):
                 
                # If edge is possible then 
                # insert in the adjacency 
                # list
                lis[a].append(b)
                lis[b].append(a)
 
# Function to count distance 
def bfs(src):
     
    global vis
    global dis
    q = []
    q.append(src)
    vis[src] = 1
    dis[src] = 0
 
    while (len(q) != 0):
        curr = q[0]
        q.pop(0)
         
        for x in lis[curr]:
             
            # If unvisited push onto queue 
            # and mark visited as 1 and 
            # add the distance of curr+1.
            if (vis[x] == 0):
                vis[x] = 1
                q.append(x)
                dis[x] = dis[curr] + 1
 
# Driver code 
N = 4
makePrimes(N) 
A = 1033
B = 8179
 
# Call bfs traversal 
# with root as node A 
bfs(A)
 
if (dis[B] == -1):
     
    # Indicates not possible 
    print(-1)
else:
    print(dis[B])
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program for the above problem 
using System;
using System.Collections.Generic;
class GFG {
  
    // Adjacency list for numbers 
    // till 100001
    static List<List<int>> lis = new List<List<int>>();
     
    static List<int> primes = new List<int>();
      
    // Visited array
    static int[] vis = new int[100001];
      
    // To store distance of every 
    // vertex from A 
    static int[] dis = new int[100001];
     
    // Function to check if number 
    // is a prime
    static bool isPrime(int n)
    {
        int i = 2;
        while (i * i <= n)
        {
            if (n % i == 0)
                return false;
            i += 1;
        }
        return true;
    }
     
    // Function to check if numbers 
    // differ by only a single-digit
    static bool valid(int a, int b)
    {
        int c = 0;
          
        while(a > 0)
        {
            // Check the last digit of 
            // both numbers and increase 
            // count if different 
            if ((a % 10) != (b % 10))
                c += 1;
                  
            a = a / 10;
            b = b / 10;
        }
      
        if (c == 1)
            return true;
        else
            return false;
    }
     
    static void makePrimes(int N)
    {
       
        // Generate all N digit primes 
        int L = (int)Math.Pow(10, N - 1);
        int R = (int)Math.Pow(10, N) - 1;
      
        for(int i = L; i < R + 1; i++)
        {
            if (isPrime(i))
                primes.Add(i);
        }
                  
        for(int i = 0; i < primes.Count; i++)
        {
           
            // For every prime number i 
            // check if an edge 
            // can be made. 
            for(int j = i + 1; j < primes.Count; j++)
            {
                int a = primes[i];
                int b = primes[j];
      
                // For every prime number 
                // i check if an edge can 
                // be made from i to j. 
                if (valid(a, b))
                {
                    // If edge is possible then 
                    // insert in the adjacency 
                    // list
                    lis[a].Add(b);
                    lis[b].Add(a);
                }
            }
        }
    }
     
    // Function to count distance 
    static void bfs(int src)
    {
        List<int> q = new List<int>();
        q.Add(src);
        vis[src] = 1;
        dis[src] = 0;
      
        while (q.Count != 0)
        {
            int curr = q[0];
            q.RemoveAt(0);
              
            foreach(int x in lis[curr])
            {
                // If unvisited push onto queue 
                // and mark visited as 1 and 
                // add the distance of curr+1.
                if (vis[x] == 0)
                {
                    vis[x] = 1;
                    q.Add(x);
                    dis[x] = dis[curr] + 1;
                }
            }
        }
    }
 
  // Driver code
  static void Main() 
  {
    for(int i = 0; i < 100001; i++)
    {
        lis.Add(new List<int>());
    }
    int N = 4;
    makePrimes(N); 
    int A = 1033;
    int B = 8179;
      
    // Call bfs traversal 
    // with root as node A 
    bfs(A);
      
    if (dis[B] == -1)
    {
        // Indicates not possible 
        Console.Write(-1);
    }
    else
    {
        Console.Write(dis[B]);
    }
  }
}

Javascript

// JS program for the above problem
let mod = 1000000007
 
// adjacency list for numbers
// till 100001
let lis = new Array(100001);
for (var i = 0; i < 100001; i++)
    lis[i] = new Array();
     
let primes = [];
 
// visited array
let vis = new Array(100001).fill(0);
 
// to store distance of every
// vertex from A
let dis = new Array(100001).fill(0);
 
// function to check if number
// is a prime
function isPrime(n)
{
    for (let i = 2;
         i * i <= n; i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// function to check if numbers
// differ by only a single-digit
function valid(a, b)
{
    let c = 0;
    while (a > 0) {
 
        // check the last digit of
        // both numbers and increase
        // count if different
        if ((a % 10) != (b % 10)) {
            c++;
        }
        a = Math.floor(a / 10);
        b = Math.floor(b / 10);
    }
    if (c == 1) {
        return true;
    }
    else {
        return false;
    }
}
 
function makePrimes( N)
{
    let i, j;
    let L = 10 ** (N - 1);
    let R = (10 ** N) - 1;
     
    // generate all N digit primes
    for (i = L; i <= R; i++) {
        if (isPrime(i)) {
            primes.push(i);
        }
    }
     
    for (i = 0; i < primes.length; i++) {
 
        // for every prime number i
        // check if an edge
        // can be made.
        for (j = i + 1; j < primes.length; j++) {
            let a = primes[i];
            let b = primes[j];
 
            // for every prime number
            // i check if an edge can
            // be made from i to j.
            if (valid(a, b)) {
                 
                //console.log(lis[a], lis[b])
                // if edge is possible then
                // insert in the adjacency
                // list
                let l1 = lis[a]
                let l2 = lis[b]
                l1.push(b)
                l2.push(a)
                lis[a] = l1
                lis[b] = l2;
            }
        }
    }
}
 
// function to count distance
function bfs(src)
{
    let q = [];
    q.push(src);
    vis[src] = 1;
    dis[src] = 0;
    while ( q.length > 0) {
        let curr = q[0];
        q.shift();
        for (let x of lis[curr]) {
 
            // if unvisited push onto queue
            // and mark visited as 1 and
            // add the distance of curr+1.
            if (vis[x] == 0) {
                vis[x] = 1;
                q.push(x);
                dis[x] = dis[curr] + 1;
            }
        }
    }
}
 
// Driver code
let N = 4;
makePrimes(N);
let A = 1033, B = 8179;
 
// Call bfs traversal
// with root as node A
bfs(A);
 
if (dis[B] == -1)
    // Indicates not possible
    console.log(-1)
else
    console.log(dis[B]);
 
// This code is contributed by phasing17.

Output:

Time Complexity: O(10^2N)
Auxiliary Space Complexity: O(10⁵)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Min steps to convert N-digit prime number into another by replacing a digit in each step

C++

Java

Python3

C#

Javascript

Nano Banana + Milvus: Turning Hype into Enterprise-Ready Multimodal RAG

LangExtract + Milvus: A Practical Guide to Building a Hybrid Document Processing and Search System

Stop Your AI Assistant from Writing Outdated Code with Milvus SDK Code Helper

LEAVE A REPLY Cancel reply

Most Popular

Android’s most beloved launcher may be done for good

Samsung One UI Home: Everything you need to know

Google Play Store will call out watch faces that are hell on your battery

Nano Banana + Milvus: Turning Hype into Enterprise-Ready Multimodal RAG

EDITOR PICKS

Android’s most beloved launcher may be done for good

Samsung One UI Home: Everything you need to know

Google Play Store will call out watch faces that are hell on your battery

POPULAR POSTS

Android’s most beloved launcher may be done for good

Samsung One UI Home: Everything you need to know

Google Play Store will call out watch faces that are hell on your battery

POPULAR CATEGORY

ABOUT US

FOLLOW US