If all decimal numbers are concatenated in a string then we will get a string which looks like string P as shown below. We need to tell Nth character of this string.
P = “12345678910111213141516171819202122232425262728293031….”
N = 10 10th character is 1 N = 11 11th character is 0 N = 50 50th character is 3 N = 190 190th character is 1
We can solve this problem by break-ing the string length-wise. We know that in decimal 9 numbers are of length 1, 90 numbers are of length 2, 900 numbers are of length 3 and so on, so we can skip these numbers according to the given N and can get the desired character.
Processing for N = 190 is explained below, P[184..195] = “979899100101” First getting length of number at N, 190 – 9 = 181 number length is more than 1 181 – 90*2 = 1 number length is more than 2 1 – 900*3 < 0 number length is 3 Now getting actual character at N, 1 character after maximum 2 length number(99) is, 1 Processing for N = 251 is explained below, P[250..255] = “120121” First getting length of number at N, 251 - 9 = 242 number length is more than 1 242 – 90*2 = 62 number length is more than 2 62 – 900*3 < 0 number length is 3 Now getting actual character at N, 62 characters after maximum 2 length number(99) is, Ceil(62/3) = 21, 99 + 21 = 120 120 is the number at N, now getting actual digit, 62%3 = 2, 2nd digit of 120 is 2, so our answer will be 2 only.
Please see below code for better understanding,
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Improved By : Mithun Kumar