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(10N)
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 -1int 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 initialized 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 Codeint 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;} |
Java
// Java code to implement the approachimport java.io.*;import java.util.*;class GFG {static int MOD = 1000000000 + 7;// DP table initialized with -1static int dp[][] = new int[1000001][3];// Recursive Function to count ways to create// number of N digits such that its digit sum// is even.static 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 initialized 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.static int countWaysTo(int N){ // Initializing dp array with - 1 for(int i=0;i<1000001;i++) { for(int j=0;j<3;j++) { dp[i][j]=-1; } } return recur(N, 0);} // Driver codepublic static void main(String[] args){ // Input 1 int N = 1; // Function Call System.out.println(countWaysTo(N)); // Input 2 int N1 = 2; // Function Call System.out.println(countWaysTo(N1));}} |
Python3
MOD = int(1e9 + 7)# DP table initialized with -1dp = [[-1 for _ in range(3)] for _ in range(1000001)]# Recursive Function to count ways to create# number of N digits such that its digit sum# is even.def recur(i: int, j: int) -> int: # 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 initialized with zero ans = 0 # start state of automata machine if j == 0: # If odd digit is picked then it # will move to automata state 1 for k in range(1, 10, 2): ans += recur(i-1, 1) ans %= MOD # If even digit is picked then it # will move to automata state 2 for k in range(2, 9, 2): ans += recur(i-1, 2) ans %= MOD # Odd state of automata machine elif j == 1: # If odd digit picked then it # will move to automata state 2 for k in range(1, 10, 2): ans += recur(i-1, 2) ans %= MOD # If even digit picked then it # will remain on same state 1 for k in range(2, 9, 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 k in range(1, 10, 2): ans += recur(i-1, 1) ans %= MOD # If even digit picked then it # will remain on current state 2. for k in range(0, 9, 2): ans += recur(i-1, 2) ans %= MOD # Save and return dp value dp[i][j] = ans return ans# Function to count ways to create# number of N digits such that its digit# sum is even.def countWaysTo(N: int) -> int: return recur(N, 0)# Driver Codeif __name__ == "__main__": # Input 1 N = 1 # Function Call print(countWaysTo(N)) # Input 2 N1 = 2 # Function Call print(countWaysTo(N1)) |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class GFG { static int MOD = 1000000007; // DP table initialized with -1 static int[,] dp=new int[1000001,3]; // Recursive Function to count ways to create // number of N digits such that its digit sum // is even. static 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 initialized 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. static int countWaysTo(int N) { // Initializing dp array with - 1 for(int i=0; i<1000001; i++) { for(int j=0;j<3; j++) dp[i,j]=-1; } return recur(N, 0); } // Driver Code public static void Main() { // Input 1 int N = 1; // Function Call Console.Write(countWaysTo(N)+"\n"); // Input 2 int N1 = 2; // Function Call Console.Write(countWaysTo(N1)+"\n"); }} |
Javascript
// Javascript code to implement the approachconst MOD = 1e9 + 7;// DP table initialized with -1let dp=new Array(1000001)for(let i=0; i<1000001; i++) dp[i]=new Array(3).fill(-1);// Recursive Function to count ways to create// number of N digits such that its digit sum// is even.function recur( i, 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 initialized with zero let ans = 0; // start state of automata machine if (j == 0) { // If odd digit is picked then it // will move to automata state 1 for (let 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 (let 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 (let 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 (let 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 (let 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 (let 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.function countWaysTo( N){ return recur(N, 0);}// Driver Code// Input 1let N = 1;// Function Callconsole.log(countWaysTo(N));// Input 2let N1 = 2;// Function Callconsole.log(countWaysTo(N1)); |
Output
4 45
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Articles:
- Introduction to Dynamic Programming – Data Structures and Algorithms Tutorials
- Introduction of Finite Automata
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