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2100010006

... well then 2100010006

you have 8th digit as 0, but in your 10 digit number 8 is occurring once.

plzz explain

algos - u have used 9 zeroes so your last or tenth digit should be 8 as in that case only 8 zeroes will left

grumpycat you have used 7 zeroes so your last digit should be 7

Guys please before posting anything think first, posting your comments without thinking is of no use........

The 1st digit describes the no. occurence of 1 - this is 2
The 2nd - no. of occurence of 2 - this is 1.
From 3 to 9 - no occurences
And the last digit is just 1.

There´s no info about zeros in your question.

0123456789 (index)
6210001000 ->

incomplete question. how about the 10th digit ?? should it describe the number of occurrences of the number 10 ?? or 0 ?? ... stupid

9000000000
first digit represent no.of zeros and so on...

0000010006

I took it as this:
13432 => 12110
number has:
1 '4'
2 '3'
1 '2'
1 '1'
0 '0'

public static int totalDigitNumber(int n)
	{
		System.out.print(n);
	    //get lowest sig fig, incrament place, and divid by 10. loop till 0.
	    boolean isNegative = false;
	    if(n<0)
	    { 
	       isNegative = true;
	       n*=-1;
	    }
	    int totalDigit = 0;
	    for(; n>0; n/=10)
	    {
	        int lsd = n%10;
	        totalDigit+=Math.pow(10,lsd);
	    }
	    if(isNegative)
	    	totalDigit*=-1;
	    System.out.println(" => " + totalDigit);
	    
	    return totalDigit;
	}

2100010006

{cannot be formed: because it may lead to infinite loop with repeating (10 digit) number. I n the code below I considered array of size 10 instead of 10 digit number}
void generate(){
int a[]={9,0,0,0,0,0,0,0,0,0};
int k=0;
while(true)
{
int b[]={0,0,0,0,0,0,0,0,0,0};
for(int i=0;i<10;i++)
{
// System.out.println(a[i]+"---"+b[a[i]]);

b[a[i]]=b[a[i]]+1;
}
for(int i=0;i<10;i++)
{
a[i]=b[i];

System.out.print(a[i]+" ");
}
System.out.println();
k++;
}
}

0010000017.....(9876543210 index)

#include<stdio.h>


void toarray(int *p, unsigned long i);
int main(){
unsigned long i;
int a[9];
int count;

for(i = 100000000; i <= 999999999; i++){
int k;
toarray(a, i);
for(k = 0; k < 9; k++){
int j;
count = 0;
for(j = 0; j < 9; j++){
if((k + 1)%9 == a[j]){
count++;
}
}
if(count != a[k]){
break;
}
}
if(k == 9 && count == a[k-1]){
printf("%lu\n", i);
}
}
}

void toarray(int *p, unsigned long i){
int j;
int k = 10;
for(j = 8; j >= 0; j--){
p[j] = i % k;
i = i / k;
}
}
ans : 210010005

Checked it for 4 - 9 digit numbers .. above code is for a 9 digit number .
ps: it is a brute force method.

1000000008

2100000001

1111111111

1111111111