Number of integers of size N having even sum of digits

By Ann Roberts On Dec 28, 2022

Given the number N, the task is to count a number of ways to create a number with digit size N with the sum of digits even. print the answer modulo 10⁹ + 7. (1 <= N <= 10⁵)

Examples:

Input: N = 1
Output: 4
Explanation: 2, 4, 6 and 8 are the numbers.

Input: N = 2
Output: 45
Explanation: 11, 13, 15, 17, 19, 20, 22, 24, 26, 28, 31, 33, 35, 37, 39, 40, 42, 44, 46, 48, 51, 53, 55, 57, 59, 60, 62, 64, 66, 68, 71, 73, 75, 77, 79, 80, 82, 84, 86, 88, 91, 93, 95, 97, and 99 are the possible numbers.

Naive approach: The basic way to solve the problem is as follows:

The basic way to solve this problem is to generate all possible combinations by using a recursive approach.

Time Complexity: O(10^N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following idea:

Dynamic programming can be used to solve this problem. For solving this problem its very simple to make finite automata machine for states with following transitions:

dp[i][j] represents number of ways of creating numbers with i digits and j represents current state of state machine.

It can be observed that the recursive function is called exponential times. That means that some states are called repeatedly.

So the idea is to store the value of each state. This can be done by using the stored value of a state and whenever the function is called, return the stored value without computing again.

Follow the steps below to solve the problem:

Create a recursive function that takes two parameters representing i’th position to be filled with digit and j representing the current state of Finite State Machine.
Call the recursive function for choosing all digits from 0 to 9.
Base case if digit of size N formed return 1 if the current state even else returns 0.
Create a 2d array of dp[N][3] initially filled with -1.
If the answer for a particular state is computed then save it in dp[i][j].
If the answer for a particular state is already computed then just return dp[i][j].

Below is the implementation of the above approach:

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

const int MOD = 1e9 + 7;

// DP table initialized with -1
int dp[1000001][3];

// Recursive Function to count ways to create
// number of N digits such that its digit sum
// is even.
int recur(int i, int j)
{
    // Base case
    if (i == 0) {

        // If current sum is even
        if (j == 2)
            return 1;

        // Else if sum is odd
        else
            return 0;
    }

    // If answer for current state is already
    // calculated then just return dp[i][j]
    if (dp[i][j] != -1)
        return dp[i][j];

    // Answer initiallized with zero
    int ans = 0;

    // start state of automata machine
    if (j == 0) {
        // If odd digit is picked then it
        // will move to automata state 1
        for (int k = 1; k <= 9; k += 2) {
            ans += recur(i - 1, 1);
            ans %= MOD;
        }

        // If even digit is picked then it
        // will move to automata state 2
        for (int k = 2; k <= 8; k += 2) {
            ans += recur(i - 1, 2);
            ans %= MOD;
        }
    }

    // Odd state of automata machine
    else if (j == 1) {

        // If odd digit picked then it
        // will move to automata state 2
        for (int k = 1; k <= 9; k += 2) {
            ans += recur(i - 1, 2);
            ans %= MOD;
        }

        // If even digit picked then it
        // will remain on same state 1
        for (int k = 2; k <= 8; k += 2) {
            ans += recur(i - 1, 1);
            ans %= MOD;
        }
    }

    // Even state of automata machine
    else {

        // If odd digit picked then it
        // will move to state 1
        for (int k = 1; k <= 9; k += 2) {
            ans += recur(i - 1, 1);
            ans %= MOD;
        }

        // If even digit picked then it
        // will remain on current state 2.
        for (int k = 0; k <= 8; k += 2) {

            ans += recur(i - 1, 2);
            ans %= MOD;
        }
    }

    // Save and return dp value
    return dp[i][j] = ans;
}

// Function to count ways to create
// number of N digits such that its digit
// sum is even.
int countWaysTo(int N)
{

    // Initializing dp array with - 1
    memset(dp, -1, sizeof(dp));

    return recur(N, 0);
}

// Driver Code
int main()
{
    // Input 1
    int N = 1;

    // Function Call
    cout << countWaysTo(N) << endl;

    // Input 2
    int N1 = 2;

    // Function Call
    cout << countWaysTo(N1) << endl;
    return 0;
}

Time Complexity: O(N)
Auxiliary Space: O(N)

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