We are given a string consisting of digits, we may group these digits into sub-groups (but maintaining their original order). The task is to count number of groupings such that for every sub-group except the last one, sum of digits in a sub-group is less than or equal to sum of the digits in the sub-group immediately on its right.

For example, a valid grouping of digits of number 1119 is (1-11-9). Sum of digits in first subgroup is 1, next subgroup is 2, and last subgroup is 9. Sum of every subgroup is less than or equal to its immediate right.

Input : "1119" Output: 7 Sub-groups: [1-119], [1-1-19], [1-11-9], [1-1-1-9], [11-19] and [111-9]. Note : Here we have included [1119] in the group and sum of digits is 12 and this group has no immediate right. Input : "1234" Output: 6 Sub-groups : [1234], [1-234], [12-34], [1-2-3-4], [12-3-4] and [1-2-34]

Let “length” be the length of input number. A recursive solution is to consider every position from 0 length-1. For every position, recursively count all possible subgroups after it. Below is C++ implementation of naive recursive solution.

// C++ program to count number of ways to group digits of // a number such that sum of digits in every subgroup is // less than or equal to its immediate right subgroup. #include<bits/stdc++.h> using namespace std; // Function to find the subgroups int countGroups(int position, int previous_sum, int length, char *num) { /* Terminating Condition */ if (position == length) return 1; int res = 0; int sum = 0; // sum of digits // Traverse all digits from current position to rest // of the length of string for (int i = position; i < length; i++) { sum += (num[i] - '0'); // If forward_sum is greater than the previous sum, // then call the method again if (sum >= previous_sum) // Note : We pass current sum as previous sum res += countGroups(i + 1, sum, length, num); } // Total number of subgroups till current position return res; } // Driver program to test the case int main() { char num[] = "1119"; int len = strlen(num); cout << countGroups(0, 0, len, num); return 0; }

Output :

7

If we take a closer look at above recursive solution, we notice that there may be overlapping subproblems. For example, if input number is 12345, then for position = 3 and previous_sum = 3, we recur two times. Similarly for position 4 and previous_sum = 7, we recur two times. Therefore the above solution can be optimized using **Dynamic Programming**. Below is a Dynamic Programming based solution for this problem.

- The maximum sum of digits can be 9*length where ‘length’ is length of input num.
- Create a 2D array int dp[MAX][9*MAX] where MAX is maximum possible length of input numebr. A value dp[position][previous] is going to store result for ‘position’ and ‘previous_sum’.
- If current subproblem has been evaluated i.e; dp[position][previous_sum] != -1, then use this result, else recursively compute its value.
- If by including the current position digit in sum i.e; sum = sum + num[position]-‘0’, sum becomes greater than equal to previous sum, then increment the result and call the problem for next position in the num.
- If position == length, then we have been traversed current subgroup successfully and we return 1;

Below is C++ implementation of above algorithm.

// C++ program to count number of ways to group digits of // a number such that sum of digits in every subgroup is // less than or equal to its immediate right subgroup. #include<bits/stdc++.h> using namespace std; // Maximum length of input number string const int MAX = 40; // A memoization table to store results of subprobllems // length of string is 40 and maximum sum will // be 9*40 = 360. int dp[MAX][9*MAX + 1]; // Function to find the count of splits with given condition int countGroups(int position, int previous_sum, int length, char *num) { /* Terminating Condition */ if (position == length) return 1; /* If already evaluated for a given sub problem then return the value */ if (dp[position][previous_sum] != -1) return dp[position][previous_sum]; // countGroups for current sub-group is 0 dp[position][previous_sum] = 0; int res = 0; int sum = 0; // sum of digits // Traverse all digits from current position to rest // of the length of string for (int i = position; i < length; i++) { sum += (num[i] - '0'); // If forward_sum is greater than the previous sum, // then call the method again if (sum >= previous_sum) // Note : We pass current sum as previous sum res += countGroups(i + 1, sum, length, num); } dp[position][previous_sum] = res; // total number of subgroups till current position return res; } // Driver program to test the case int main() { char num[] = "1119"; int len = strlen(num); // Initialize dp table memset(dp, -1, sizeof(dp)); cout << countGroups(0, 0, len, num); return 0; }

Output:

7

This article is contributed by **Shashank Mishra ( Gullu )**. This article is reviewed by team GeeksForGeeks.

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