# Maximum sum and product of the M consecutive digits in a number

Given a number in the form of a string. The task is to find the maximum sum and product of m consecutive digits that are taken from the number string.

Examples:

Input: N = 3675356291, m = 5
Output: 3150
There are 6 sequences of 5 digits 36753, 67535, 75356, 53562, 35629, 56291
6 x 7 x 5 x 3 x 5 gives the maximum product.

Input: N = 2709360626, m = 5
Output: 0
Since each sequence of consecutive 5 digits will contain a 0 so each time product will be zero so
the maximum product is zero.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:

1. Take all possible consequtive sequences of m characters from the given string.
2. Add them and Multiply them by changing the characters into integers.
3. Compare the product and sum of each sequence and find the maximum product and sum.

Below is the implementation of the above approach:

## C++

 // C++ implemenattion of the above approach #include using namespace std;    // Function to find the maximum product void maxProductSum(string str, int m) {     int n = str.length();     int maxProd = INT_MIN, maxSum = INT_MIN;     for (int i = 0; i < n - m; i++) {         int product = 1, sum = 0;            for (int j = i; j < m + i; j++) {             product = product * (str[j] - '0');             sum = sum + (str[j] - '0');         }            maxProd = max(maxProd, product);         maxSum = max(maxSum, sum);     }     cout << "Maximum Product = " << maxProd;     cout << "\nMaximum Sum = " << maxSum; }    // Driver code int main() {     string str = "3675356291";     int m = 5;        maxProductSum(str, m); }

## Java

 // Java implemenattion of the above approach    import java.io.*;    class GFG {        // Function to find the maximum product  static void maxProductSum(String str, int m) {     int n = str.length();     int maxProd = Integer.MIN_VALUE, maxSum = Integer.MIN_VALUE;     for (int i = 0; i < n - m; i++) {         int product = 1, sum = 0;            for (int j = i; j < m + i; j++) {             product = product * (str.charAt(j) - '0');             sum = sum + (str.charAt(j) - '0');         }            maxProd = Math.max(maxProd, product);         maxSum = Math.max(maxSum, sum);     }     System.out.println("Maximum Product = " + maxProd);     System.out.print( "\nMaximum Sum = " + maxSum); }    // Driver code        public static void main (String[] args) {         String str = "3675356291";     int m = 5;        maxProductSum(str, m);     } } // This code is contributed by anuj_67..

## Python 3

 # Python implementation of # above approach import sys    # Function to find the maximum product def maxProductSum(string, m) :        n = len(string)     maxProd , maxSum = (-(sys.maxsize) - 1,                          -(sys.maxsize) - 1)        for i in range(n - m) :         product, sum = 1, 0            for j in range(i, m + i) :             product = product * (ord(string[j]) -                                  ord('0'))             sum = sum + (ord(string[j]) -                           ord('0'))            maxProd = max(maxProd, product)         maxSum = max(maxSum, sum)        print("Maximum Product =", maxProd)     print("Maximum sum =", maxSum)        # Driver code if __name__ == "__main__" :        string = "3675356291"     m = 5     maxProductSum(string, m)            # This code is contributed by ANKITRAI1

## C#

 // C# implemenattion of the above approach using System; class GFG  {        // Function to find the maximum product static void maxProductSum(string str, int m) {     int n = str.Length;     int maxProd = int.MinValue,         maxSum = int.MinValue;     for (int i = 0; i < n - m; i++)      {         int product = 1, sum = 0;            for (int j = i; j < m + i; j++)          {             product = product * (str[j] - '0');             sum = sum + (str[j] - '0');         }            maxProd = Math.Max(maxProd, product);         maxSum = Math.Max(maxSum, sum);     }     Console.WriteLine("Maximum Product = " + maxProd);     Console.Write( "\nMaximum Sum = " + maxSum); }    // Driver code public static void Main ()  {     string str = "3675356291";     int m = 5;            maxProductSum(str, m); } }    // This code is contributed  // by Akanksha Rai

## PHP



Output:

Maximum Product = 3150
Maximum Sum = 26

Efficient Approach: The idea is to use the Sliding Window concept . First find the sum and product of M consecutive digits and update the maxProd and maxSum.
Then start traversing from Mth index and add current digit to the sum and subtract str[i-M] from the sum i.e. considering only M elements/digits. Similarly for the product. And keep updating the maxSum and maxProd.

## C++

 // C++ implemenattion of the above approach #include using namespace std;    // Function to find the maximum product and sum void maxProductSum(string str, int m) {     int n = str.length();     int product = 1, sum = 0;        // find the sum and product of first K digits     for (int i = 0; i < m; i++) {         sum += (str[i] - '0');         product *= (str[i] - '0');     }        // Update maxProd and maxSum     int maxProd = product;     int maxSum = sum;        // Start traversing the next element     for (int i = m; i < n; i++) {            // Multiply with the current digit and divide by         // the first digit of previous window         product = product * (str[i] - '0') / ((str[i - m]) - '0');            // Add the current digit and subtract         // the first digit of previous window         sum = sum + (str[i] - '0') - ((str[i - m]) - '0');            // Update maxProd and maxSum         maxProd = max(maxProd, product);         maxSum = max(maxSum, sum);     }        cout << "Maximum Product = " << maxProd;     cout << "\nMaximum Sum = " << maxSum; }    // Driver code int main() {     string str = "3675356291";     int m = 5;        maxProductSum(str, m); }

## Java

 // Java implemenattion of the above approach import java.util.Arrays;  import java.io.*;    class GFG {        // Function to find the maximum product and sum static void maxProductSum(String str, int m) {     int n = str.length();     int product = 1, sum = 0;        // find the sum and product of first K digits     for (int i = 0; i < m; i++)      {         sum += (str.charAt(i) - '0');         product *= (str.charAt(i) - '0');     }        // Update maxProd and maxSum     int maxProd = product;     int maxSum = sum;        // Start traversing the next element     for (int i = m; i < n; i++)      {            // Multiply with the current digit and divide by         // the first digit of previous window         product = product * (str.charAt(i) - '0') / ((str.charAt(i-m)) - '0');            // Add the current digit and subtract         // the first digit of previous window         sum = sum + (str.charAt(i) - '0') - ((str.charAt(i-m)) - '0');            // Update maxProd and maxSum         maxProd = Math.max(maxProd, product);         maxSum = Math.max(maxSum, sum);     }        System.out.println("Maximum Product = " + maxProd);     System.out.println("\nMaximum Sum = " + maxSum); }    // Driver code     public static void main (String[] args) {         String str = "3675356291";         int m = 5;         maxProductSum(str, m);     } }    // This code is contributed  // by ajit

## Python 3

# Python 3 implemenattion of the above approach

# Function to find the maximum product and sum
def maxProductSum(str, m):

n = len(str)
product = 1
sum = 0

# find the sum and product of first K digits
for i in range(m):
sum += (ord(str[i]) – ord(‘0’))
product *= (ord(str[i]) – ord(‘0’))

# Update maxProd and maxSum
maxProd = product
maxSum = sum

# Start traversing the next element
for i in range(m, n) :

# Multiply with the current digit and divide
# by the first digit of previous window
product = (product * (ord(str[i]) – ord(‘0’)) //
((ord(str[i – m])) – ord(‘0’)))

# Add the current digit and subtract
# the first digit of previous window
sum = (sum + (ord(str[i]) – ord(‘0’)) –
((ord(str[i – m])) – ord(‘0’)))

# Update maxProd and maxSum
maxProd = max(maxProd, product)
maxSum = max(maxSum, sum)

print(“Maximum Product =”, maxProd)
print(“Maximum Sum =”, maxSum)

# Driver code
if __name__ == “__main__”:

str = “3675356291”
m = 5

maxProductSum(str, m)

# This code is contributed by ita_c

## C#

 // C# implemenattion of the above approach using System;    class GFG { // Function to find the maximum product and sum static void maxProductSum(string str, int m) {     int n = str.Length;     int product = 1, sum = 0;        // find the sum and product of first K digits     for (int i = 0; i < m; i++)      {         sum += (str[i] - '0');         product *= (str[i] - '0');     }        // Update maxProd and maxSum     int maxProd = product;     int maxSum = sum;        // Start traversing the next element     for (int i = m; i < n; i++)      {            // Multiply with the current digit and divide by         // the first digit of previous window         product = product * (str[i] - '0') / ((str[i - m]) - '0');            // Add the current digit and subtract         // the first digit of previous window         sum = sum + (str[i] - '0') - ((str[i - m]) - '0');            // Update maxProd and maxSum         maxProd = Math.Max(maxProd, product);         maxSum = Math.Max(maxSum, sum);     }        Console.Write("Maximum Product = " + maxProd);     Console.Write("\nMaximum Sum = " + maxSum); }    // Driver code public static void Main() {     string str = "3675356291";     int m = 5;        maxProductSum(str, m); } }    // This code is contributed  // by Akanksha Rai

## PHP



Output:

Maximum Product = 3150
Maximum Sum = 26

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