Take an integer n (n >= 0) and a digit d (0 <= d <= 9) as an integer.

Square all numbers k (0 <= k <= n) between 0 and n.

Count the numbers of digits d used in the writing of all the k**2.

Implement the function taking n and d as parameters and returning this count.

Examples:

n = 10, d = 1
the k*k are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
We are using the digit 1 in: 1, 16, 81, 100. The total count is then 4.

The function, when given n = 25 and d = 1 as argument, should return 11 since
the k*k that contain the digit 1 are:
1, 16, 81, 100, 121, 144, 169, 196, 361, 441.
So there are 11 digits 1 for the squares of numbers between 0 and 25.

Note that 121 has twice the digit 1.

int nbDig(int n, int d) {
   int count = 0;
  if(d == 0) {
    count++;
  }
  for (int i = 0; i <= n; i++) {
    int square = i * i;
    while (square > 0) {
      if (square % 10 == d) {
        count++;
      }
      square ~/= 10;
    }
  }
  return count;
}

int nbDig(int n, int d) =>
    '$d'.allMatches([for (var i = 0; i <= n; i += 1) i * i].join()).length;

import 'dart:math';

int nbDig(int n, int d) {
  int count1 = 0;
  for (int i = 0; i<=n; i++ ){
    count1= count1 +RegExp('[${d.toString()}]').allMatches(pow(i,2).toString()).length;
  }
  return count1;
}