Given a positive value N, we need to find the count of numbers smaller than N such that the difference between the number and sum of its digits is greater than or equal to given specific value diff.
Input : N = 13, diff = 2 Output : 4 Then 10, 11, 12 and 13 satisfy the given condition shown below, 10 – sumofdigit(10) = 9 >= 2 11 – sumofdigit(11) = 9 >= 2 12 – sumofdigit(12) = 9 >= 2 13 – sumofdigit(13) = 9 >= 2 Whereas no number from 1 to 9 satisfies above equation so final result will be 4
We can solve this problem by observing a fact that for a number k less than N,
if k – sumofdigit(k) >= diff then above equation will be true for (k+1) also because we know that sumofdigit(k+1) is not greater than sumofdigit(k) + 1 so, k + 1 - sumofdigit(k + 1) >= k - sumofdigit(k) but we know that right side of above inequality is greater than diff, so left side will also be greater than diff.
So, finally we can say that if a number k satisfies the difference condition then (k + 1) will also satisfy same equation so our job is to find the smallest number which satisfies the difference condition then all numbers greater than this and up to N will satisfy the condition so our answer will be N – smallest number we found.
We can find the smallest number satisfying this condition using binary search so total time complexity of solution will be O(log N)
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Improved By : nitin mittal