Count total number of N digit numbers such that the difference between sum of even and odd digits is 1
Given a number n, we need to count the total number of n digit numbers such that the sum of even digits is 1 more than the sum of odd digits. Here even and odd means positions of digits are like array indexes, for example, the leftmost (or leading) digit is considered as even digit, next to leftmost is considered as odd and so on.
Input: n = 2 Output: Required Count of 2 digit numbers is 9 Explanation : 10, 21, 32, 43, 54, 65, 76, 87, 98. Input: n = 3 Output: Required Count of 3 digit numbers is 54 Explanation: 100, 111, 122, ......, 980
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This problem is mainly an extension of Count of n digit numbers whose sum of digits equals to given sum. Here the solution of subproblems depends on four variables: digits, esum (current even sum), osum (current odd sum), isEven(A flag to indicate whether the current digit is even or odd).
Below is Memoization based solution for the same.
Count of 3 digit numbers is 54
Thanks to Gaurav Ahirwar for providing above solution.
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