Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

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 <bits/stdc++.h>
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




<?php
// Code is Improved by Surya Prakash Sharma
   
// PHP implementation of the above approach
 
// Function to find the maximum product
function maxProductSum($str, $m)
{
    $n = strlen($str);
    $maxProd = PHP_INT_MIN;
    $maxSum = PHP_INT_MIN;
    for ($i = 0; $i <= ($n - $m); $i++)
    {
        $product = 1;
        $sum = 0;
 
        for ($j = $i; $j < ($m + $i); $j++)
        {
            $product = $product *
                      ($str[$j] - '0');
            $sum = $sum + ($str[$j] - '0');
        }
 
        $maxProd = max($maxProd, $product);
        $maxSum = max($maxSum, $sum);
    }
    echo "Maximum Product = " ,$maxProd;
    echo "\nMaximum Sum = " , $maxSum;
}
 
// Driver code
$str = "3605356297";
$m = 3;
 
maxProductSum($str, $m);
 
// This code is contributed by @Tushil.
?>

Javascript




<script>
    // Code is Improved by Surya Prakash Sharma
     
    // Javascript implementation of the above approach
     
    // Function to find the maximum product
    function maxProductSum(str, m)
    {
        let n = str.length;
        let maxProd = Number.MIN_VALUE,
            maxSum = Number.MIN_VALUE;
        for (let i = 0; i <= n - m; i++)
        {
            let product = 1, sum = 0;
 
            for (let 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);
        }
        document.write("Maximum Product = " + maxProd + "</br>");
        document.write( "Maximum Sum = " + maxSum);
    }
     
    let str = "3605356297";
    let m = 3;
       
    maxProductSum(str, m);
     
</script>
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 <bits/stdc++.h>
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




<?php
// PHP implementation of the above approach
 
// Function to find the maximum
// product and sum
function maxProductSum($str, $m)
{
    $n = strlen($str);
    $product = 1;
    $sum = 0;
     
    // find the sum and product of
    // first K digits
    for ($i = 0; $i < $m; $i++)
    {
        $sum += ($str[$i] - '0');
        $product *= ($str[$i] - '0');
    }
 
    // Update maxProd and maxSum
    $maxProd = $product;
    $maxSum = $sum;
 
    // Start traversing the next element
    for ($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);
    }
 
    echo "Maximum Product = " , $maxProd;
    echo "\nMaximum Sum = " , $maxSum;
}
 
// Driver code
$str = "3675356291";
$m = 5;
maxProductSum($str, $m);
 
// This code is contributed by ajit
?>

Javascript




<script>   
    // Javascript implementation of the above approach
     
    // Function to find the maximum product and sum
    function maxProductSum(str, m)
    {
        let n = str.length;
        let product = 1, sum = 0;
 
        // find the sum and product of first K digits
        for (let i = 0; i < m; i++)
        {
            sum += (str[i].charCodeAt() - '0'.charCodeAt());
            product *= (str[i].charCodeAt() - '0'.charCodeAt());
        }
 
        // Update maxProd and maxSum
        let maxProd = product;
        let maxSum = sum;
 
        // Start traversing the next element
        for (let i = m; i < n; i++)
        {
 
            // Multiply with the current digit and divide by
            // the first digit of previous window
            product = product * (str[i].charCodeAt() - '0'.charCodeAt()) / ((str[i - m].charCodeAt()) - '0'.charCodeAt());
 
            // Add the current digit and subtract
            // the first digit of previous window
            sum = sum + (str[i].charCodeAt() - '0'.charCodeAt()) - ((str[i - m].charCodeAt()) - '0'.charCodeAt());
 
            // Update maxProd and maxSum
            maxProd = Math.max(maxProd, product);
            maxSum = Math.max(maxSum, sum);
        }
 
        document.write("Maximum Product = " + maxProd + "</br>");
        document.write("Maximum Sum = " + maxSum);
    }
     
    let str = "3675356291";
    let m = 5;
  
    maxProductSum(str, m);
 
</script>
Output
Maximum Product = 100
Maximum Sum = 19



My Personal Notes arrow_drop_up
Recommended Articles
Page :