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Time Complexity: O(n*sum) Auxiliary Space: O(101*501) Another Method: We can easily count n digit numbers whose sum of digit equals to given sum by iterating all n digits and checking if current n digit number’s sum is equal to given sum, if it is then we will start increment number by 9 until it reaches to number whose sum of digit’s is greater than given sum, then again we will increment ...

There is a lottery with n coupons and n people take part in it. Each person picks exactly one coupon.Coupons are numbered consecutively from 1 to n, n being the maximum ticket number. The winner of the lottery is any person who owns a coupon where the sum of the digits on the coupon is equal to s.

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