CareerCup

int size;
void dig_gen(int m,int n, char* passwd)
{
int i,j;
if(n==size-1)
{
for(i=m;i<=9;i++)
{
passwd[n]='0'+i;
for(j=0; j<size; j++)
printf("%c",passwd[j]);
printf("\n");
}
}
else
{
for(i=m;i<10-size+n+1;i++)
{
passwd[n]='0'+i;
dig_gen(i+1,n+1,passwd);
}
}
}
void main()
{
char* passwd;
scanf("%d",&size);
passwd=malloc(sizeof(char)*size);
dig_gen(0,0,passwd);
free(passwd);
}

Generate the number first using N. Its probably something like
for(int i = 1; i<10; i+2)
{
for(int j = 1; j<10^N; j+2)
{
generatenumber = i*10^N + j;
something-------

check the digits using another function... best i could come up with...

little bit of comments in the code would make it easier to understand

Please note:
1.strictly ascending
2. N digits

Solution:
if (N<= 0) printf( "No solution\n");
if (N == 10) return printf( "01234456789\n");
for (i1=0; i1< 10-N; i1++)
{
printf("%1d");
if (N<9)
for (i1=1; i1< 10-N; i1++)
{
printf("%1d");
if (N<8)
for (i1=1; i1< 10-N; i1++)
{

... ...
}
}
}
prinf("\n");

My mi
stakes:

Should be:

Solution:
if (N<= 0) printf( "No solution\n");
if (N == 10) return printf( "01234456789\n");
for (i1=0; i1< 10-N; i1++)
{
printf("%1d", i1);
if (N>1)
for (i2=i1+1; i2< 10-N; i2++)
{
printf("%1d", i2);
if (N>2)
for (i3=i2+1; i3< 10-N; i3++)
{
... ...
if (N>8)
for (i9=i8+1; i9< 10-N; i9++)
{
printf("%1d", i9);
}
... ...
}
}
}
prinf("\n");

Sorry, more mistakes:
for (i1=0; i1< 10-N; i1++)
{
printf("%1d", i1);
if (N>1)
for (i2=i1+1; i2< 10-N+1; i2++)
{
printf("%1d", i2);
if (N>2)
for (i3=i2+1; i3< 10-N+2; i3++)
{
... ...
if (N>8)
for (i9=i8+1; i9< 10-N+8; i9++)
{
printf("%1d", i9);
}
... ...
}
}
}
prinf("\n");

int main()
{
int N=5;
int num_array[N];
posnos(N ,-1, num_array);
return 0;
}

int posnos(int n, int val, int* array)
{
int i;
if(n<=0) {
for(i=4; i>=0; i--)
printf("%d", array[i]);
printf("\n");
return;
}

for(i = val+1; i<10; i++) {
array[n-1] = i;
posnos(n-1, i, array);
}
}

#include<stdio.h>
#include<stdlib.h>
int main()
{
int N;
printf("enter the length of the password ");
scanf("%d",&N);
int x,i,j,a;
//since the password should be desired it should always start with (10-N)th
//digit to satisfy the desired condition
for(x=1;x<=(10-N);x++)
{
//start from number greater tham the first digit
for(i=x;i<=(10-N);i++)
{
printf("%d",x);
//go N times adding 1 to the previous digit
for(a=1;a<N;a++)
{

printf("%d ",(i+a));

}

printf("\n");
}
}
}

// This code is written if the number of digits in the pass word are 4.... the code could be modified accordingly for more numbers in the password.
public class DesirableNumber
{

public static void main(String args[]){
int numberOfDigits = 4;

for(int i = 0;i<9;i++){
for(int k =0;k<9;k++){
for(int j =0;j<9;j++){
String number = i+""+j+"" +k;

if(number.length() < numberOfDigits){
FindDesirable(number);
}else{
break;
}
}
}
}

}

private static void FindDesirable(String number) {

char a = number.charAt(0);
char b = number.charAt(1);
char c = number.charAt(2);
if(a<b && b<c){
System.out.println(number);
}
}
}

AllDesirable ad = new AllDesirable();
			StringBuffer output = new StringBuffer(Integer.parseInt(argv[1]));
			for (int i = 0; i < Integer.parseInt(argv[1]); i++)
				output.append('0');
			ad.printMe(Integer.parseInt(argv[1]), 0, output, 0);

public class AllDesirable {
	public void printMe(int n, int start, StringBuffer output, int level){
		if (n == level){
			System.out.println(output);
			return;
		}
		
		for (int i = start; i <= 9; i++){
			output.setCharAt(level, (char) (i + '0'));
			printMe(n, i + 1, output, level + 1);
		}
	}
}

This comment has been deleted.

int N = 5;
int[] a = new int[N];
getDigit(0, 0, N, a);

public static void getDigit(int digit, int i, int length, int[] a)
{
if(digit==length-1)
for(int j=i;j<=9;j++)
{
a[digit] = j;
for(int num:a)
System.out.print(num);
System.out.println();
}
else
for(int j=i;j<=9;j++)
{
a[digit] = j;
getDigit(digit+1, j+1, length, a);
}
}

Goal: Find all K-combinations (in lexographic order) of 9 digits.

Pseudo Code:

//this function finds the next combination in lexographic order
Nextcomb(comb)
{
while (comb != Lastcomb) do
{
print(comb);
B = comb;
i = 0;
while (i < K) and (B[i+1] == B[i]+1) do
{
B[i] = i; // if there is no gap in elements, and the end has not reached, then B[i]=i;
i++;
}
B[i] = B[i]+1; // otherwise, jump by 1;
Nextcomb(B);
}
}


//Main function
Input = K; // length of password
Lastcomb = 9-K+1,9-K,...,9;
Firstcomb = 0,1,2,3,...,K-1;

Nextcomb(Firstcomb);

List<Integer> PasswordList(int startInt, int N)
{
List<Integer> finalList=new ArrayList<Integer>();
int R = 9-N+1;
if(N==1)
{
for(int t=startInt;t<=9;t++)
{
finalList.add(t);
}
return finalList;
}
for(int i=startInt;i<=R;i++)
{
List<Integer> prevList = PasswordList(i+1,N-1);
for(int j=0;j<prevList.size();j++)
{
Double temp = Math.pow(10,(N-1));
finalList.add(i*temp.intValue() + prevList.get(j));
}
}
return finalList;
}

public class Desire2 {

/**
* @param args
*/
public static void main(String[] args) {
desireNum("0",4);


}

public static void desireNum (String output,int level){
if(level==0){
System.out.print(Integer.parseInt(output)+" ,");
return;
}
for(int i=1;i<10;i++){
if(Integer.parseInt(output.charAt(output.length()-1)+"")<i){
//System.out.println(output.charAt(output.length()-1)+" "+i+" level"+level);
desireNum(output+i,level-1);
}
}

}

}

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<stdlib.h>

void main()
{
int i, n, j, k, flag, lb, ub, x, y, cnt;
char c[10];
clrscr();
printf("\nEnter digits: ");
scanf("%d",&n);
lb=pow(10,n-1);
ub=pow(10,n);
cnt=0;

for(i=lb;i<ub;i++)
{

itoa(i,c,10);

flag=0;
for(j=0;j<n-1;j++)
{
x=c[j];
y=c[j+1];
if(x<y)
flag=flag+1;
}

if(flag==(n-1))
{ printf("\t");
cnt=cnt+1;
for(k=0;k<n;k++)
printf("%c",c[k]);
}

}
printf("\nTotal no. of desirable numbers having %d digits is: %d",n,cnt);

getch();
}

Here's Python:

def getDesirables(N):
    if N < 1:
        return None
    if N is 1:
        return [str(i) for i in range(10)]
    desirables = []
    for s in getDesirables(N-1):
        last_digit = int(s[-1])
        for t in range(last_digit+1,10):
            desirables.append(s+str(t))
    return desirables

print(getDesirables(0)) # N must be positive, else None
print(getDesirables(1))
print(getDesirables(2))

void ordercombination(int n, char *out, int permlen, int start)
{
if(n == permlen)
{
cout<<out<<endl;
return;
}
for(int i=start;i<10;i++)
{
out[permlen] = i + '0';
ordercombination(n,out,permlen+1,i+1);
}
}

package b;

public class DesirableNum {
	public static void main(String[] args){
		int N = 3;			
		printDesirable("", 0, N);
	}
	
	public static void printDesirable(String op, int start, int N){
		if(N == 0){
			System.out.println(op);
			return;
		}
					
			for(int i = start; i < 10; i++){
				String initial = op;
				op += i;
				printDesirable(op, i+1, N-1);				
				op = initial;				
			}	
	}
}

{{
main(N)
{
// form smallest N digit no
int num = pow(10,N-1);
while(num<pow(10,N))
{
status = checkAscending(num,N);
if status == -1
num++;
else
print(num)
}
}
checkAcending(num,N)
{
char *n = itoa(num);
for(i=0;i<N-1;i++)
{
if n[i]>n[i+1]
return(-1);
}
return(1);

}

<pre lang="c#" line="1" title="CodeMonkey35355" class="run-this"> class Combination
{
static void Main(string[] args)
{
Console.Write("Enter Length: ");
int length = Convert.ToInt32(Console.ReadLine());
Combinations(length, "", 0);
Console.ReadLine();
}

private static void Combinations(int length, string combination, int intIndex)
{
if (combination.Length == length)
{
Console.WriteLine(combination);
}
else if (combination.Length < length)
{
for (int i = intIndex; i < 10; i++)
{
string newCombination = combination + i.ToString();
Combinations(length, newCombination, ++intIndex);
}
}
}
}</pre>

// Good question for practice
// Here is pesudo code , very simple
str = string of n length
For(i = 1;i<9-N;i++)
{
str[0] = ‘0’+i;
Print_desire(str,1,N,i);
}

Print_desire(char *str,int i,int n,int k)
{
If(i>=N)
{
Return;
}
If(i==N-1)
{
Print str;
}
For(j = 2;j<9;j++)
If(j>i)
{
Str[i] = ‘0’+j;
Print_desire(str,i+1,n,j);
}




}


}



// let me know if it is wrong !

//sort the numbers . Since it would n digits only, so we could use count sort,
    //then use the permutation and avoid if the number is repeating. 
    class Program
    {
        static void Main(string[] args)
        {
            int[] digits = { 1, 2, 4, 5, 7, 8};
            bool[] used = new bool[digits.Length];
            Permutation(digits, used, 0, digits.Length-2, "",-10);
            Console.Read();
        }
        public static void Permutation(int[] digits, bool[] used, int count, int length, string result,int pre)
        {
            if (count == length && result != " ")
            {
                Console.WriteLine(result);               
                return;
            }
            for (int i = 0; i < digits.Length; ++i)
            {
                if (used[i]||pre > digits[i])
                    continue;
               
                pre = digits[i];
                used[i] = true;
                Permutation(digits, used, count + 1, length, result +" "+ digits[i],pre);           
                used[i] = false;
            }

        }
    }

#include<iostream>
using namespace std;
void listAllOrderedNumbers(int str[],int length,bool mask[]);
int main()
{
int str[5];
str[0]=-1;
int length=1;
bool mask[10];
for(int i=1;i<=9;i++)
	mask[i]=false;
listAllOrderedNumbers(str,1,mask);
}
void listAllOrderedNumbers(int str[],int length,bool mask[])
{	
	if(length==5)
	{
		for(int j=1;j<5;j++)
			cout<<str[j]<<" ";
		cout<<endl;		
	}
	else
	{	
		for(int i=1;i<=9;i++)
		{
				if(mask[i]==true)
					continue;
				if(str[length-1]< i)
				{
					str[length]=i;
					mask[i]=true;
					listAllOrderedNumbers(str,length+1,mask);
					mask[i]=false;
				}
		}	
	}	
}

public void printNum(int digits)
{
genNums(0,digits,-1,"");
}

public void genNums(int idx, int digits, int lastNum, String temp)
{
if(temp.length() == digits)
{System.out.println(temp); return;};

for(int n=lastNum+1;n<=10-digits+idx;n++)
{
genNums(idx+1,digits,n,temp+String.valueOf(n));
}
}

public void printNum(int digits)
  {
    genNums(0,digits,-1,"");
  }
  
  public void genNums(int idx, int digits, int lastNum, String temp)
  {
    if(temp.length() == digits) 
    {System.out.println(temp); return;};
      
    for(int n=lastNum+1;n<=10-digits+idx;n++)
    {
        genNums(idx+1,digits,n,temp+String.valueOf(n));
    }
  }

/*
 * password.cpp
 *
 *  Created on: Aug 13, 2012
 *    
 */

#include <stack>
#include <stdio.h>
using namespace std;
void pass(stack <int> a,int n)
{
 if(n==0)
 {
	 while(!a.empty())
	 {
		 printf("%d ",a.top());
		 a.pop();

	 }
	 printf("\n");
 }

 else{

	  for(int i = 0;i<a.top();i++)
	  {
		  a.push(i);
		  pass(a,n-1);
		  a.pop();

	  }

 }

}
int main()
{
int n;
scanf("%d",&n);
  stack <int> a;
  a.push(9);
  pass(a,n-1);
}

Like most of the other code, I am using tail recursion. When numDigits is 0, I will just print what I have so far in the "curr" argument. Otherwise I will construct the new "curr" argument by multiplying it by 10 and adding the next digit to it. My only "innovation" is the use of the ones-digit to determine the minimum digit I should start iterating from.

static void printAll(int numDigits, int curr) {
	if(numDigits == 0) {
		System.out.println(curr);
		return;
	}
	int min = 1 + (curr > 0 ? curr%10 : 0);
	for(int d=min; d<=9; d++) {
		printAll(numDigits-1, curr*10 + d);
	}
}

int permutation_without_repetition(int start,int size,int index,char* password)
{
   int num_of_instance=0;
   if(size == 0) 
   {
       cout<<password<<endl;
       return 1;
   }
   for(int i=start;i<=10-size;i++)
   {
       password[index]='0'+i;
       num_of_instance += permutation_without_repetition(i+1,size-1,index+1,password);
   }
    return num_of_instance;
}

void password_prob(int size)
{
    char* password;
    
    if(size>0){
        password = new char[size];
        int count = permutation_without_repetition(0,size,0,password);
        cout<<"No. of possible passwords: "<<count<<endl;
    }
    else
    {
        cout<<"Invalid input !";
    }
}

According to the problem, I rarely find any of above solutions which gives correct output. can you please look into this? Comments and feedbacks are highly appreciatable. Thanks.

/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
/**
*
* @author dell
*/
public class Main2 {

/**
* @param args the command line arguments
*/
public static void main(String[] args) {
new Main2().substringCombination("", 3, 0);
}

public void substringCombination(String prefix, int n, int k) {
if (n == 0) {
System.out.println(prefix);
return;
}
for (int i = k+1; i <=9; i++) {
substringCombination(prefix +i , n-1,i);
}
}
}

#include<stdio.h>

void print(int dig,int n,char *str,int ind)
{
    if(dig<=0)
    {
        str[ind]='\0';
        printf("%s\n",str);
        return;
    }
    else
    {
        for(int i=n;i<=9;i++)
        {
            str[ind]=i+48;
            print(dig-1,i+1,str,ind+1);
        }
    }
}

int main()
{
    int n;
    char str[3];
    print(3,0,str,0);
}

#!/usr/bin/python

##print password recursively from leftmost digit
##s is current string, minv is min value of leftmost digit, m is number of digits
def printPW(s, minv, m):
    for v in range(minv, 11 - m):
        if m==1:
            print s + str(v)
        else:
            printPW(s + str(v), v + 1, m - 1)

printPW('', 6, 3)

#include <stdio.h>

void print(int arr[], int n)
{
int i;
for(i=0;i<n;i++)
{
printf("%d->",arr[i]);
}
printf("\n");
}

void generatePass(int arr[],int n, int start, int index)
{
int i;

// base case:
if(index == n)
{
print(arr,n);
}

// processing
for(i=start;i<=9;i++)
{
arr[index] = i;
generatePass(arr,n,i+1,index+1);
}
}

int main(void) {
// your code goes here
int arr[3];
int n=3;
int start =1;
int index = 0;
generatePass(arr,n,start,index);
return 0;
}

#include <stdio.h>
#include <stdlib.h>

int valid(int n)
{
        int temp = n / 100, first = n % 10, second = (n/10) % 10;
        if(first <= second)
                return 0;
        while(temp > 0)
        {
                first = second;
                second = temp % 10;
                if(first <= second)
                        return 0;
                temp = temp / 10;
        }
        return 1;
}

void ascending(int s, int e, int size)
{
        int i =0;
        for(i=s;i<=e;i++)
        {
                if(valid(i))
                        printf("%d\n",i);
        }
}

int main()
{
        int start = 1, end =1, i=0, size = 2;
        for(i=1;i<size;i++)
                start = start * 10;
        end = start * 10 - 1;
        ascending(start, end, size);
        return 0;
}

public static void main(String[] args) 
	{
		String s = "";
		int n = 5;
		func(s,n,0);
	}
	
	static void func(String s, int n, int j)
	{
		if (s.length() == n)
		{
			System.out.println(s);
			return;
		}
		else
		{
			for (int i = j+1 ; i < 10 ; i++)
			{
				func(s+i,n, i);
			}
		}
	}

void dig_gen(int begin,int level, int sz , const int max, char* passwd) 
{ 
    int i=begin;
    int j=0;
    //printf("B=%d,L=%d,sz=%d\n",i,level,sz);
    if (level >= max)
    {
        for ( j=0;j<max;++j)
           printf("%d",passwd[j]);
        printf(",");   
        return ;
    }
    
    for (i = begin; i<= 10-sz;++i)
    {
        passwd[level]=i;
        //printf("%d",i);
        dig_gen(i+1,level+1,sz-1,max,passwd);
    }
} 

void main() 
{ 
char* passwd; 
//scanf("%d",&size); 
int size=3;
printf("start\n");
passwd=malloc(sizeof(char)*size); 
dig_gen(0,0,size,size,passwd); 
free(passwd);

}

#include <bits/stdc++.h>
using namespace std;

void findDesirable(int start, string passwd, int size)
{
    if(size == 0)
        cout << passwd << endl;
    else
    {
        for(int i = start; i <= 9; i++)
        {
            string str = passwd + to_string(i);
            findDesirable(i + 1, str , size - 1);
        }
    }
}

int main()
{
    string str;
    int size = 3;

    findDesirable(0, str, size);
}

9

This is basically the problem of selecting M elements from N. Here is N <= 10 (i.e., 0 -9). There are a lot of algorithms designed for this problem.

this is similar to permutations with only one loop starting with 1 to maintain the ascending order

for(i = 0; i < 10; i++)
{
for(j = i; j < 10; j++)
{
for(k = j; k < 10; k++)
{
printf(" %d %d %d ", i,j,k );
}
}
}

for(i = 0; i < 10; i++)
{
for(j = i+1; j < 10; j++)
{
for(k = j+1; k < 10; k++)
{
printf(" %d %d %d ", i,j,k );
}
}
}