Given a Natural number N and a whole number L, the task is to find the count of numbers, smaller than or equal to N, such that the difference between the number and sum of its digits is not less than L.
Input: N = 1500, L = 30 Output: 1461 Input: N = 1546300, L = 30651 Output: 1515631
Our solution depends on a simple observation that if a number, say X, is such that the difference of X and sumOfDigits(X) is less than or equal to L, then X+1 is also a valid number. Hence, we have to find a minimum such X using binary search.
Time Complexity: O(log N)
Auxiliary Space: O(1)
- Count Numbers with N digits which consists of odd number of 0's
- Count Numbers with N digits which consists of even number of 0’s
- Count numbers with difference between number and its digit sum greater than specific value
- Count numbers < = N whose difference with the count of primes upto them is > = K
- Count of numbers from range [L, R] whose sum of digits is Y
- Number of digits in the product of two numbers
- Number of digits before the decimal point in the division of two numbers
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- Program to find count of numbers having odd number of divisors in given range
- Minimum number of digits to be removed so that no two consecutive digits are same
- Print numbers with digits 0 and 1 only such that their sum is N
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- Minimum sum of two numbers formed from digits of an array in O(n)
- Recursive program to print all numbers less than N which consist of digits 1 or 3 only
- Count maximum elements of an array whose absolute difference does not exceed K
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