# Sum of Digits of the Good Strings

• Last Updated : 25 May, 2021

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. Let’s 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 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 to 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++

 `// C++ implementation of the approach` `#include ` `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;` `}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``

Output:

`1567751`

Time Complexity: O(N)

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