# 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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