Find the Substring with maximum product

Given string str containing only lowercase English alphabets of size N, the task is to find the substring having the maximum product. 

Each English alphabet has a value such that val(‘a’) = 0, val(‘b’) = 1, val(‘c’) = 2, ……, val(‘z’) = 25.

Examples:

Input: str = “sdtfakdhdahdzz” 
Output: hdzz 
Here, the maximum product is for the substring “hdzz”. 
product = 7 * 3 * 25 * 25 = 13125
Input: str = “geeksforgeeks” 
Output: geeksforgeeks 

Approach:

• First, traverse through the given string while maintaining a maximum product value.
• Product will always keep increasing or will remain constant unless we encounter an ‘a’. Hence, start a new substring after each ‘a’ occurrence.
• Also, along with the maximum product value, we will also maintain the substring to which the maximum product corresponds.
• Once the entire string has been traversed, print the substring corresponding to the maximum product.

Below is the implementation of the above approach: 

C++

// C++ program to find the maximum product substring

#include <bits/stdc++.h>
using namespace std;

// Function to return the value of a character
int value(char x)
{
    return (int)(x - 'a');
}

// Function to find the maximum product substring
string maximumProduct(string str, int n)
{
    // To store substrings
    string answer = "", curr = "";
    long long maxProduct = 0, product = 1;

    for (int i = 0; i < n; i++) {
        product *= 1LL * value(str[i]);

        curr += str[i];

        // Check if current product is maximum
        // possible or not
        if (product >= maxProduct) {
            maxProduct = product;
            answer = curr;
        }

        // If product is 0
        if (product == 0) {
            product = 1;
            curr = "";
        }
    }

    // Return the substring with maximum product
    return answer;
}

// Driver code
int main()
{
    string str = "sdtfakdhdahdzz";

    int n = str.size();

    // Function call
    cout << maximumProduct(str, n) << endl;

    return 0;
}

Java

// Java program to find the maximum product subString

class GFG{

// Function to return the value of a character
static int value(char x)
{
    return (int)(x - 'a');
}

// Function to find the maximum product subString
static String maximumProduct(String str, int n)
{
    // To store subStrings
    String answer = "", curr = "";
    long maxProduct = 0, product = 1;

    for (int i = 0; i < n; i++) {
        product *= 1L * value(str.charAt(i));

        curr += str.charAt(i);

        // Check if current product is maximum
        // possible or not
        if (product >= maxProduct) {
            maxProduct = product;
            answer = curr;
        }

        // If product is 0
        if (product == 0) {
            product = 1;
            curr = "";
        }
    }

    // Return the subString with maximum product
    return answer;
}

// Driver code
public static void main(String[] args)
{
    String str = "sdtfakdhdahdzz";

    int n = str.length();

    // Function call
    System.out.print(maximumProduct(str, n) +"\n");

}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the maximum product sub

# Function to return the value of a character
def value(x):
    return (ord(x) - ord('a'))

# Function to find the maximum product sub
def maximumProduct(strr, n):

    # To store subs
    answer = ""
    curr = ""
    maxProduct = 0
    product = 1

    for i in range(n):
        product *=value(strr[i])

        curr += strr[i]

        # Check if current product is maximum
        # possible or not
        if (product >= maxProduct):
            maxProduct = product
            answer = curr

        # If product is 0
        if (product == 0):
            product = 1
            curr = ""

    # Return the sub with maximum product
    return answer

# Driver code
if __name__ == '__main__':
    strr = "sdtfakdhdahdzz"

    n = len(strr)

    # Function call
    print(maximumProduct(strr, n))

# This code is contributed by mohit kumar 29

C#

// C# program to find the maximum product subString

using System;

public class GFG{

// Function to return the value of a character
static int value(char x)
{
    return (int)(x - 'a');
}

// Function to find the maximum product subString
static String maximumProduct(String str, int n)
{
    // To store subStrings
    String answer = "", curr = "";
    long maxProduct = 0, product = 1;

    for (int i = 0; i < n; i++) {
        product *= 1L * value(str[i]);

        curr += str[i];

        // Check if current product is maximum
        // possible or not
        if (product >= maxProduct) {
            maxProduct = product;
            answer = curr;
        }

        // If product is 0
        if (product == 0) {
            product = 1;
            curr = "";
        }
    }

    // Return the subString with maximum product
    return answer;
}

// Driver code
public static void Main(String[] args)
{
    String str = "sdtfakdhdahdzz";

    int n = str.Length;

    // Function call
    Console.Write(maximumProduct(str, n) +"\n");

}
}
// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to find the
// maximum product subString

// Function to return the value of a character
function value(x)
{
    return (x.charCodeAt() - 'a'.charCodeAt());
}

// Function to find the maximum product subString
function maximumProduct(str, n)
{
    // To store subStrings
    let answer = "", curr = "";
    let maxProduct = 0, product = 1;

    for (let i = 0; i < n; i++) {
        product *= (1 * value(str[i]));

        curr += str[i];

        // Check if current product is maximum
        // possible or not
        if (product >= maxProduct) {
            maxProduct = product;
            answer = curr;
        }

        // If product is 0
        if (product == 0) {
            product = 1;
            curr = "";
        }
    }

    // Return the subString with maximum product
    return answer;
}


// Driver Code

     let str = "sdtfakdhdahdzz";

    let n = str.length;

    // Function call
    document.write(maximumProduct(str, n) +"<br/>");

</script>

Output:

hdzz

Time Complexity: O(N)

Auxiliary Space: O(1)