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

• Difficulty Level : Easy
• Last Updated : 24 Sep, 2021

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:
Since each sequence of consecutive 5 digits will contain a 0 so each time product will be zero so
the maximum product is zero.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Naive Approach:

1. Take all possible consecutive 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++

 `// Code is Improved by Surya Prakash Sharma` `// C++ implementation 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 = ``"3605356297"``;``    ``int` `m = 3;` `    ``maxProductSum(str, m);``}`

## Java

 `// Code is Improved by Surya Prakash Sharma` ` ``// Java implementation 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 = ``"3605356297"``;``    ``int` `m = ``3``;` `    ``maxProductSum(str, m);``    ``}``}``// This code is contributed by anuj_67..`

## Python 3

 `# Code is Improved by Surya Prakash Sharma` `# 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``+``1``) :``        ``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 ``=` `"3605356297"``    ``m ``=` `3``    ``maxProductSum(string, m)``        ` `# This code is contributed by ANKITRAI1`

## C#

 `// Code is Improved by Surya Prakash Sharma` `// C# implementation 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 = ``"3605356297"``;``    ``int` `m = 3;``    ` `    ``maxProductSum(str, m);``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

 ``

## Javascript

 ``
Output
```Maximum Product = 126
Maximum Sum = 18```

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++

 `// Code Improved By Surya Prakash Sharma` `// C++ implementation 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, ZeroesInWindow=0;` `    ``// find the sum and product of first K digits``    ``for` `(``int` `i = 0; i < m; i++) {``        ``sum += (str[i] - ``'0'``);``      ` `        ``if``(str[i]!=``'0'``)``        ``product *= (str[i] - ``'0'``);``        ``else``        ``ZeroesInWindow++;``    ``}` `    ``// Update maxProd and maxSum``    ``int` `maxProd = 0;``    ``if``(ZeroesInWindow==0)``      ``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``        ``if``(str[i]!=``'0'` `&& str[i-m]!=``'0'``)``        ``product = product * (str[i] - ``'0'``) / ((str[i - m]) - ``'0'``);``        ``else` `if``(str[i]!=``'0'` `&& str[i-m]==``'0'``)``        ``{ ``           ``product = product * (str[i] - ``'0'``);``           ``ZeroesInWindow--;``        ``}``        ``else` `if``(str[i]==``'0'` `&& str[i-m]!=``'0'``)``        ``{``           ``product = product / (str[i-m] - ``'0'``);``           ``ZeroesInWindow++;``        ``}``      ` `        ``// Update maxProd``        ``if``(ZeroesInWindow==0)``        ``maxProd = max(maxProd, product);` `        ``// Add the current digit and subtract``        ``// the first digit of previous window``        ``sum = sum + (str[i] - ``'0'``) - ((str[i - m]) - ``'0'``);` `        ``// Update maxSum``        ``maxSum = max(maxSum, sum);``    ``}` `    ``cout << ``"Maximum Product = "` `<< maxProd;``    ``cout << ``"\nMaximum Sum = "` `<< maxSum;``}` `// Driver code``int` `main()``{``    ``string str = ``"3601990545"``;``    ``int` `m = 3;` `    ``maxProductSum(str, m);``}`

## Java

 `// Code Improved By Surya Prakash Sharma` `// Java implementation 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``, ZeroesInWindow=``0``;` `    ``// find the sum and product of first K digits``    ``for` `(``int` `i = ``0``; i < m; i++)``    ``{``        ``sum += (str.charAt(i) - ``'0'``);``      ` `        ``if``(str.charAt(i)!=``'0'``)``        ``product *= (str.charAt(i) - ``'0'``);``        ``else``        ``ZeroesInWindow++;``        ` `    ``}` `    ``// Update maxProd and maxSum``    ``int` `maxProd = ``0``;``    ``if``(ZeroesInWindow==``0``)``      ``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``        ``if``(str.charAt(i)!=``'0'` `&& str.charAt(i-m)!=``'0'``)``        ``product = product * (str.charAt(i) - ``'0'``) / ((str.charAt(i-m)) - ``'0'``);``        ``else` `if``(str.charAt(i)!=``'0'` `&& str.charAt(i-m)==``'0'``)``        ``{ ``           ``product = product * (str.charAt(i) - ``'0'``);``           ``ZeroesInWindow--;``        ``}``        ``else` `if``(str.charAt(i)==``'0'` `&& str.charAt(i-m)!=``'0'``)``        ``{``           ``product = product / (str.charAt(i-m) - ``'0'``);``           ``ZeroesInWindow++;``        ``}``      ` `        ``// Update maxProd``        ``if``(ZeroesInWindow==``0``)``        ``maxProd = Math.max(maxProd, product);` `        ``// Add the current digit and subtract``        ``// the first digit of previous window``        ``sum = sum + (str.charAt(i) - ``'0'``) - ((str.charAt(i-m)) - ``'0'``);` `        ``// Update maxSum``        ``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 = ``"3601990545"``;``        ``int` `m = ``3``;``        ``maxProductSum(str, m);``    ``}``}` `// This code is contributed``// by ajit`

## Python 3

 `# Python 3 implementation 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# implementation 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

 ``

## Javascript

 ``
Output
```Maximum Product = 100
Maximum Sum = 19```

My Personal Notes arrow_drop_up