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facebook · Dec 11, 2014 10:17 AM

N(N+1)/2=1095.. Solve for N. that's the answer

sajhamitra · Dec 11, 2014 10:28 AM

From 1 to 9 ====> 9 digits 10 to 99 =====> 180 digits .                           + ---------- Total of 1 digit and 2 digit no.s  = 189 digits Now rest of the pages are the three numbered pages: N - 189 = 1095 - 189 = 906 906/3 = 302 So the total pages: 9 + 90 + 302 = 401 Last edited: 11-Dec-14 11:34 AM

Slackdemic · Dec 11, 2014 10:41 AM

1 to N = 1 to 129

sara_sota11 · Dec 11, 2014 11:19 AM

Slackdemic got it right. Answer is 401. Todays score: Slackdemic +1 Sajhamitra -1 sara_solta11 -1

helpjava11 · Dec 11, 2014 1:13 PM

Here is my tricky question. You have range of contiguous numbers in array from 1 to 1 million. Among of them one of the numbers is missing. Find out that missing number. You can use any data structure or use a simple math.

virusno1 · Dec 11, 2014 1:35 PM

अहिलेको जमानामा यसो हल्का गुगल गर्नु पर्छ क्या, गुला खेलाउदै दिमाग चलाउन खोजेर मात्र हुदैन। http://codepad.org/Hrj8BKZ7 int getMissingNo (int a[], int n) { int i, total; total = (n+1)*(n+2)/2; for ( i = 0; i< n; i++) total -= a[i]; return total; } /*program to test above function */ int main() { int a[] = {1,2,4,5,6}; int miss = getMissingNo(a,5); printf("%d", miss); /*getchar();*/ } Last edited: 11-Dec-14 03:05 PM

aaitey · Dec 11, 2014 3:04 PM

Do it with a datastructure.

sajhamitra · Dec 11, 2014 3:40 PM

if the numbers are in sequence, this function works fine. if its not in sequence its different story. main method(){ int[] arrayofMillionNumbers = new int[100000]; for(int i = 1 ; i 1) { missingNumber = num[i] + 1; break; } } return missingNumber; }

helpjava11 · Dec 12, 2014 10:22 AM

if the numbers are in sequence, this function works fine. if its not in sequence its different story. m = missing number x = 1; for i = 1 to million { if x is not equal to i { m = x = missing number; stop iteration;} x = x + 1; } output m;

neurologist · Dec 12, 2014 10:31 AM

those numbers are continuous but distributed randomly. like 1,3,2,5,4.....etcyour solution works only if numbers are continuous but there should be some efficient way to solve if sequential numbers are randomly distributed. Last edited: 12-Dec-14 10:44 AM

sajhamitra · Dec 12, 2014 10:42 AM

KaliKoPoi · Dec 12, 2014 10:46 AM

Add all the numbers in the array. Then find n! ( that is 1,000,000!) Subtract them, that would be number that is missing. This is would be complete in O(n) which is better than sorting the number and then using one of above algorithm to find the missing one.

Mr. D · Dec 12, 2014 11:38 AM

Yes that's true but there is another solution too. Hint: You can use an array.

virusno1 · Dec 12, 2014 12:43 PM