Given a number N and a sum S, find the count of numbers upto N that have digit sum equal to S.
Examples: Input : N = 100, S = 4 Output : 5 Upto 100 only 5 numbers(4, 13, 22, 31, 40) can produce 4 as their sum of digits. Input : N = 1000, S = 1 Output : 4 Upto 1000 only 4 numbers(1, 10, 100 and 1000) can produce 1 as their sum of digits.
We can do a digit DP having state (current index, whether the currently constructed number of i digits is equal or less than the number formed by first i digits of N, the sum of digits of currently constructed number).
Let dp[i][tight][sum_so_far] denotes the count of numbers whose first i digits have been considered and tight denotes whether the currently constructed number is equal or less than number formed by first i digits of N. If tight is true, then it means that currently constructed number is equal to the number constructed by first i digits of N. If it is false then it means that currently constructed number is less than the number constructed by first i digits of N. sum_so_far denotes the sum of digits of currently constructed number.
If i = number of digits in N, sum_so_far = Sum, then answer = 1 else answer is 0.
For filling (i+1)th digit we can consider following –
If tight is true, then it means that our constructed number is still equal to number formed by first i digits of N. We can try our current possible digit value from 0 to (i+1)th digit of N. If we try the digit more than (i+1)th digit, then the constructed number will become greater than number formed by first i digits of N, which will violate the property that our constructed number should be <= N.
If tight is false, then it means that number constructed from first i – 1 digits has become less than number constructed from the first i – 1 digit of N, So it means that our number can never exceed N, so we can chose the any digit from 0 to 9.
nsum_so_far can be obtained by adding sum_so_far and current digit(currdigit).
Finally we will return answer which is count of numbers upto N that have digit sum equal to S.
Time Complexity: 2(tight) * Sum * log 10(N) * 10(transitions) = 20*log 10(N)*Sum.
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- Count numbers formed by given two digit with sum having given digits
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- Count numbers with difference between number and its digit sum greater than specific value
- Count total number of N digit numbers such that the difference between sum of even and odd digits is 1
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