Count digit groupings of a number with given constraints

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 the number of groupings such that for every sub-group except the last one, the sum of digits in a sub-group is less than or equal to the 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.

Examples :

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
       the 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]

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Let “length” be the length of the 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 the naive recursive solution.

C++

// 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; 
      
    // sum of digits 
    int sum = 0;  
  
    // 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 Code 
int main() 
{ 
    char num[] = "1119"; 
    int len = strlen(num); 
    cout << countGroups(0, 0, len, num); 
    return 0; 
} 

Java

// Java 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. 
import java.io.*; 
  
class GFG  
{ 
  
// Function to find 
// the subgroups 
static int countGroups(int position,  
                       int previous_sum,  
                       int length,  
                       String num) 
{ 
    // Terminating Condition 
    if (position == length) 
        return 1; 
  
    int res = 0; 
      
    // sum of digits 
    int sum = 0;  
  
    // Traverse all digits from 
    // current position to rest 
    // of the length of string 
    for (int i = position; i < length; i++) 
    { 
        sum += (num.charAt(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 Code 
public static void main (String[] args) 
{ 
    String num = "1119"; 
    int len =num .length(); 
    System.out.println(countGroups(0, 0,  
                                   len, num)); 
} 
} 
  
// This code is contributed by anuj_67. 

Python3

# Python3 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. 
  
# Function to find the subgroups 
def countGroups(position, previous_sum, 
               length, num): 
  
    # Terminating Condition 
    if(position == length): 
        return 1
  
    res = 0
    # sum of digits 
    sum = 0
  
    # Traverse all digits from 
    # current position to rest 
    # of the length of string 
    for i in range(position, length): 
  
        sum = sum + int(num[i]) 
        # 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 = res + countGroups(i + 1, sum, length, num) 
  
    # Total number of subgroups 
    # till the current position 
    return res 
  
# Driver Code 
if __name__=='__main__': 
    num = "1119"
    len = len(num) 
    print(countGroups(0, 0, len, num)) 
  
# This code is contributed by 
# Sanjit_Prasad 

C#

// 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. 
using System; 
                      
class GFG 
{ 
  
// Function to find 
// the subgroups 
static int countGroups(int position,  
                       int previous_sum,  
                       int length,  
                       String num) 
{ 
    // Terminating Condition 
    if (position == length) 
        return 1; 
  
    int res = 0; 
  
    // sum of digits 
    int sum = 0;  
  
    // 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 Code 
public static void Main () 
{ 
    String num = "1119"; 
    int len = num.Length; 
    Console.Write(countGroups(0, 0, len, num)); 
} 
} 
  
// This code is contributed by 29AjayKumar 

PHP

<?php 
// PHP 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. 
  
// Function to find the subgroups 
function countGroups($position,  
                     $previous_sum,  
                     $length,$num) 
{ 
    // Terminating Condition 
    if ($position == $length) 
        return 1; 
  
    $res = 0; 
      
    // sum of digits 
    $sum = 0;  
  
    // Traverse all digits from  
    // current position to rest 
    // of the length of string 
    for ($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 Code 
$num = "1119"; 
$len = strlen($num); 
echo countGroups(0, 0, $len, $num); 
  
// This code is contributed by ajit 
?> 

Output :

If we take a closer look at the above recursive solution, we notice that there may be overlapping subproblems. For example, if the 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 number. 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 the implementation of the above algorithm.

C++

// 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 subproblems 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; 
      
    // sum of digits 
    int sum = 0;  
  
    // 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 Code 
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; 
} 

Java

      
// Java program to count the 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. 
class GFG  
{ 
  
// Maximum length of  
// input number string 
static int MAX = 40; 
  
// A memoization table to store  
// results of subproblems length 
// of string is 40 and maximum  
// sum will be 9 * 40 = 360. 
static int dp[][] = new int[MAX][9 * MAX + 1]; 
  
// Function to find the count 
// of splits with given condition 
static 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; 
      
    // sum of digits 
    int sum = 0;  
  
    // 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 Code 
public static void main(String[] args) 
{ 
    char num[] = "1119".toCharArray(); 
    int len = num.length; 
  
    // Initialize dp table 
    for(int i = 0; i < dp.length; i++) 
    { 
        for(int j = 0;j < 9 * MAX + 1; j++){ 
            dp[i][j] = -1; 
        } 
    } 
    System.out.println(countGroups(0, 0, len, num)); 
    } 
}  
  
// This code is contributed by PrinciRaj1992 

Python3

# Python3 program to count the 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. 
  
# Maximum length of 
# input number string 
MAX = 40
  
# A memoization table to store 
# results of subproblems length 
# of string is 40 and maximum 
# sum will be 9 * 40 = 360. 
dp = [[ -1 for i in range(9 * MAX + 1)]  
           for i in range(MAX)] 
  
# Function to find the count 
# of splits with given condition 
def countGroups(position, previous_sum, 
                           length, 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
  
    res = 0
  
    # sum of digits 
    sum = 0
  
    # Traverse all digits from 
    # current position to rest 
    # of the length of string 
    for i in range(position,length): 
        sum += (ord(num[i]) - ord('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 the current position 
    return res 
  
# Driver Code 
num = "1119"
len = len(num) 
  
print(countGroups(0, 0, len, num)) 
  
# This code is contributed by Mohit Kumar 

C#

// 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.  
using System;  
  
class GFG  
{  
    // Maximum length of  
    // input number string  
    static int MAX = 40;  
      
    // A memoization table to store  
    // results of subproblems length  
    // of string is 40 and maximum  
    // sum will be 9 * 40 = 360.  
    static int[,] dp = new int[MAX, 9 * MAX + 1];  
      
    // Function to find the count  
    // of splits with given condition  
    static 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;  
          
        // sum of digits  
        int sum = 0;  
      
        // 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 Code  
    static void Main()  
    {  
        char[] num = {'1', '1', '1', '9'};  
        int len = num.Length;  
      
        // Initialize dp table  
        for(int i = 0; i < MAX; i++) 
            for(int j = 0; j < 9 * MAX + 1; j++) 
                dp[i, j] = -1; 
      
        Console.Write(countGroups(0, 0, len, num));  
    } 
} 
  
// This code is contributed by DrRoot_ 

Output :

This article is contributed by Shashank Mishra ( Gullu ). This article is reviewed by team GeeksForGeeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

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