Sum of Digits of the Good Strings

A string is called good if it is made with only digits 0 to 9 and adjacent elements are different. The task is to find the sum of the digits of all possible good strings of length X that end with the given digit Y. The answer could be large so print the answer modulo 109 + 7.

Examples:

Input: X = 2, Y = 2
Output: 61
All possible strings of length 2 that end with 2 are:
02, 12, 32, 42, 52, 62, 72, 82, 92.
Now, ((0 + 2) + (1 + 2) + (3 + 2) + (4 + 2) + (5 + 2)
+ (6 + 2) + (7 + 2) + (8 + 2) + (9 + 2)) = 61



Input: X = 6, Y = 4
Output: 1567751

Approach: This problem can be solved by using dynamic programming. Lets define the following states:

  1. dp[i][j]: Sum of the digits of all possible good strings of length i that end with j.
  2. cnt[i][j]: Count of the good strings of length i that end with j.

The value of the previous state will have to be used to compute the value for the current state as the adjacent digits have to be compared whether they are equal or not. Now, the recurrence relation will be:

dp[i][j] = dp[i][j] + dp[i – 1][k] + cnt[i – 1][k] * j

Here, dp[i – 1][k] is the sum of the digits of good strings of length (i – 1) that end with k and k != j.
cnt[i -1][k] is the count of good strings of length (i – 1) that end with k and k != j.
So for position i, (cnt(i – 1)[k] * j) has to be added as j is being put at index i and the count of possible strings that have length (i – 1) is cnt[i – 1][k].

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define DIGITS 10
#define MAX 10000
#define MOD 1000000007
  
// To store the states of the dp
long dp[MAX][DIGITS], cnt[MAX][DIGITS];
  
// Function to fill the dp table
void precompute()
{
  
    // dp[i][j] : Sum of the digits of all
    // possible good strings of length
    // i that end with j
    // cnt[i][j] : Count of the good strings
    // of length i that end with j
  
    // Sum of digits of the string of length
    // 1 is i as i is only number in that string
    // and count of good strings of length 1
    // that end with i is also 1
    for (int i = 0; i < DIGITS; i++)
        dp[1][i] = i, cnt[1][i] = 1;
  
    for (int i = 2; i < MAX; i++) {
        for (int j = 0; j < DIGITS; j++) {
            for (int k = 0; k < DIGITS; k++) {
  
                // Adjacent digits are different
                if (j != k) {
                    dp[i][j] = dp[i][j]
                               + (dp[i - 1][k] + (cnt[i - 1][k] * j) % MOD)
                                     % MOD;
                    dp[i][j] %= MOD;
  
                    // Increment the count as digit at
                    // (i - 1)'th index is k and count
                    // of good strings is equal to this
                    // because at the end of the strings of
                    // length (i - 1) we are just
                    // putting digit j as the last digit
                    cnt[i][j] += cnt[i - 1][k];
                    cnt[i][j] %= MOD;
                }
            }
        }
    }
}
  
// Driver code
int main()
{
    long long int x = 6, y = 4;
  
    precompute();
  
    cout << dp[x][y];
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
    final static int DIGITS = 10;
    final static int MAX = 10000;
    final static int MOD = 1000000007;
      
    // To store the states of the dp 
    static int dp[][] = new int[MAX][DIGITS];
    static int cnt[][] = new int[MAX][DIGITS]; 
      
    // Function to fill the dp table 
    static void precompute() 
    
      
        // dp[i][j] : Sum of the digits of all 
        // possible good strings of length 
        // i that end with j 
        // cnt[i][j] : Count of the good strings 
        // of length i that end with j 
      
        // Sum of digits of the string of length 
        // 1 is i as i is only number in that string 
        // and count of good strings of length 1 
        // that end with i is also 1 
        for (int i = 0; i < DIGITS; i++)
        {
            dp[1][i] = i; 
            cnt[1][i] = 1
        }
      
        for (int i = 2; i < MAX; i++)
        
            for (int j = 0; j < DIGITS; j++)
            
                for (int k = 0; k < DIGITS; k++) 
                
      
                    // Adjacent digits are different 
                    if (j != k) 
                    
                        dp[i][j] = dp[i][j] + (dp[i - 1][k] +
                                             (cnt[i - 1][k] * j) % MOD) 
                                                                 % MOD; 
                        dp[i][j] %= MOD; 
      
                        // Increment the count as digit at 
                        // (i - 1)'th index is k and count 
                        // of good strings is equal to this 
                        // because at the end of the strings of 
                        // length (i - 1) we are just 
                        // putting digit j as the last digit 
                        cnt[i][j] += cnt[i - 1][k]; 
                        cnt[i][j] %= MOD; 
                    
                
            
        
    
      
    // Driver code 
    public static void main (String[] args)
    
        int x = 6, y = 4
      
        precompute(); 
      
        System.out.println(dp[x][y]); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach
DIGITS = 10;
MAX = 10000;
MOD = 1000000007;
  
# To store the states of the dp 
dp = [[0 for i in range(DIGITS)] 
         for i in range(MAX)];
cnt = [[0 for i in range(DIGITS)] 
          for i in range(MAX)];
  
# Function to fill the dp table 
def precompute():
  
    # dp[i][j] : Sum of the digits of all 
    # possible good strings of length 
    # i that end with j 
    # cnt[i][j] : Count of the good strings 
    # of length i that end with j 
  
    # Sum of digits of the string of length 
    # 1 is i as i is only number in that string 
    # and count of good strings of length 1 
    # that end with i is also 1 
    for i in range(DIGITS):
      
        dp[1][i] = i; 
        cnt[1][i] = 1
      
    for i in range(2, MAX):
        for j in range(DIGITS):
            for k in range(DIGITS):
                  
                # Adjacent digits are different 
                if (j != k):
                  
                    dp[i][j] = dp[i][j] + (dp[i - 1][k] +\
                                         (cnt[i - 1][k] * j) % MOD) % MOD; 
                    dp[i][j] %= MOD; 
  
                    # Increment the count as digit at 
                    # (i - 1)'th index is k and count 
                    # of good strings is equal to this 
                    # because at the end of the strings of 
                    # length (i - 1) we are just 
                    # putting digit j as the last digit 
                    cnt[i][j] += cnt[i - 1][k]; 
                    cnt[i][j] %= MOD; 
  
# Driver code 
x = 6; y = 4
  
precompute(); 
  
print(dp[x][y]); 
  
# This code is contributed by 29AjayKumar

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C#

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// C# implementation of the approach
using System;
      
class GFG 
{
    readonly static int DIGITS = 10;
    readonly static int MAX = 10000;
    readonly static int MOD = 1000000007;
      
    // To store the states of the dp 
    static int [,]dp = new int[MAX, DIGITS];
    static int [,]cnt = new int[MAX, DIGITS]; 
      
    // Function to fill the dp table 
    static void precompute() 
    
      
        // dp[i][j] : Sum of the digits of all 
        // possible good strings of length 
        // i that end with j 
        // cnt[i][j] : Count of the good strings 
        // of length i that end with j 
      
        // Sum of digits of the string of length 
        // 1 is i as i is only number in that string 
        // and count of good strings of length 1 
        // that end with i is also 1 
        for (int i = 0; i < DIGITS; i++)
        {
            dp[1, i] = i; 
            cnt[1, i] = 1; 
        }
      
        for (int i = 2; i < MAX; i++)
        
            for (int j = 0; j < DIGITS; j++)
            
                for (int k = 0; k < DIGITS; k++) 
                
      
                    // Adjacent digits are different 
                    if (j != k) 
                    
                        dp[i, j] = dp[i, j] + (dp[i - 1, k] +
                                             (cnt[i - 1, k] * j) % MOD) 
                                                                 % MOD; 
                        dp[i, j] %= MOD; 
      
                        // Increment the count as digit at 
                        // (i - 1)'th index is k and count 
                        // of good strings is equal to this 
                        // because at the end of the strings of 
                        // length (i - 1) we are just 
                        // putting digit j as the last digit 
                        cnt[i, j] += cnt[i - 1, k]; 
                        cnt[i, j] %= MOD; 
                    
                
            
        
    
      
    // Driver code 
    public static void Main (String[] args)
    
        int x = 6, y = 4; 
      
        precompute(); 
      
        Console.WriteLine(dp[x,y]); 
    
}
  
// This code is contributed by Rajput-Ji

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Output:

1567751

Time Complexity: O(N)



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