Create 040.js

Author: Walker30263

040.js

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+/*
+This problem can also be solved using math...but I made a bashy program solution anyways since it's like 5 lines of code
+Also made a less bashy program solution using the following algorithm because I'm bored
+
+By concatenating all x-digit long natural numbers, we will obtain a string x * 9 * 10^(n-1) characters long
+
+Let c(x) be a function that returns the length of the string obtained by concatenating all x-digit long natural numbers:
+
+c(0): 0-character long string (duh)
+c(1): 9-character long string
+c(2): 180-character long string
+c(3): 2700-character long string
+c(4): 36000-character long string
+c(5): 450000-character long string
+c(6): 5400000-character long string
+
+Thus, to find the nth digit in C_10 (Base 10 Champernowne's Constant), we can use the following algorithm:
+Starting at k = 0, subtract c(k) from n until n' <= c(k+1). Then, nth digit is contained in the ceil(n' / (k+1))-th k+1-digit number.
+Index of nth digit in the number will be equal to n' % k+1. (If 0 then (n' % k+1)-th position)
+
+1st digit: 1 - c(0) = 1 = n' <= c(1). 1st digit is contained in the ceil(1 / 1) = 1st 1-digit number = 10^(1-1) + 1 - 1 = "1". Index of digit in "1" = 1 % 1 = 0, so 1st position. 
+Therefore, 1st digit = 1.
+10th digit: 10 - c(0) - c(1) = 1 = n' <= c(2). 10th digit is contained in the ceil(1 / 2) = 1st 2-digit number = 10^(2-1) + 1 - 1 = "10". Index of digit in "10" = 1 % 2 = 1, so 1st position.
+Therefore, 10th digit = 1.
+100th digit: 100 - c(0) - c(1) = 91 = n' <= c(3). 100th digit is contained in the ceil(91 / 2) = 46th 2-digit number = 10^(2-1) + 46 - 1 = "55". Index of digit in "55" = 91 % 2 = 1, so 1st position.
+Therefore, 100th digit = 5.
+1000th digit: 1000 - c(0) - c(1) - c(2) = 811 = n' <= c(3). 1000th digit is contained in the ceil(811 / 3) = 271th 3-digit number = 10^(3-1) + 271 - 1 = "370". Index of digit in "370" = 811 % 3 = 1, so 1st position.
+Therefore, 1000th digit = 3.
+10000th digit: 10000 - c(0) - c(1) - c(2) - c(3) = 7111 = n' <= c(4). 10000th digit is contained in the ceil(7111 / 4) = 1778th 4-digit number = 10^(4-1) + 1778 - 1 = "2777". Index of digit in "2777" = 7111 % 4 = 3, so 3rd position.
+Therefore, 10000th digit = 7.
+100000th digit: 100000 - c(0) - c(1) - c(2) - c(3) - c(4) = 61111 = n' <= c(5). 100000th digit is contained in the ceil(61111 / 5) = 12223th 5-digit number = 10^(5-1) + 12223 - 1 = "22222". Index of digit in "22222" = 61111 % 5 = 1, so 1st position.
+Therefore, 100000th digit = 2.
+1000000th digit: 1000000 - c(0) - c(1) - c(2) - c(3) - c(4) - c(5) = 511111 = n' <= c(6). 1000000th digit is contained in the ceil(511111 / 6) = 85186th 6-digit number = 10^(6-1) + 85186 - 1 = 185185. Index of digit in "185185" = 511111 % 6 = 1, so 1st position.
+Therefore, 1000000th digit = 1.
+
+Thus, the answer is 1 * 1 * 5 * 3 * 7 * 2 * 1 = 210.
+*/
+
+
+
+//Bashy solution:
+
+let champernownesConstant = "";
+
+for (let i = 1; i <= 500000; i++) { //string should definitely be at least a million characters long now 
+    champernownesConstant += i;
+}
+
+let answer = parseInt(champernownesConstant.charAt(1 - 1)) * parseInt(champernownesConstant.charAt(10 - 1)) * parseInt(champernownesConstant.charAt(100 - 1)) * parseInt(champernownesConstant.charAt(1000 - 1)) * parseInt(champernownesConstant.charAt(10000 - 1)) * parseInt(champernownesConstant.charAt(100000 - 1)) * parseInt(champernownesConstant.charAt(1000000 - 1));
+console.log(answer);
+
+
+
+
+
+//Algorithmic solution:
+
+//function that returns the nth digit in C (Champernowne's Constant in Base 10) using the algorithm described at the top of the code
+function nthDigitInC(n) {
+    let k = 0;
+    let m = n;
+
+    // "Starting at k = 0, subtract c(k) from n until n' <= c(k+1)." (here we're just calling n' "m")
+
+    while (m > lengthOfStringOfAllXDigitNumbersConcatenated(k+1)) {
+        m -= lengthOfStringOfAllXDigitNumbersConcatenated(k+1);
+        k++;
+    }
+
+    
+
+    // "Then, nth digit is contained in the ceil(n' / (k+1))-th k+1-digit number."
+
+    let numberDigitIsContainedIn = (10**(k)) + Math.ceil(m / (k + 1)) - 1;
+
+    // "Index of nth digit in the number will be equal to n' % k+1. (If 0 then (n' % k+1)-th position)"
+
+    let index = m % (k+1);
+
+    if (index === 0) {
+        return numberDigitIsContainedIn.toString().charAt(index);
+    } else {
+        return numberDigitIsContainedIn.toString().charAt(index - 1);
+    }
+}
+
+//helper function that returns the returns the length of the string obtained by concatenating all x-digit long natural numbers
+function lengthOfStringOfAllXDigitNumbersConcatenated(x) {
+    return (x * 9 * (10 ** (x-1)))
+}
+
+let answer2 = 1; //accumulator kinda
+
+for (let i = 0; i <= 6; i++) {
+    answer2 *= nthDigitInC(10 ** i);
+}
+
+console.log(answer2);