查找具有最大乘积的子字符串

给定一个仅包含大小为N的小写英文字母的字符串str ，任务是找到具有最大乘积的子字符串。

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

例子：

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

方法：

• 首先，遍历给定的字符串，同时保持最大乘积值。
• 除非我们遇到“ a”，否则乘积将始终保持增长或保持不变。因此，在每个“ a”出现之后开始一个新的子字符串。
• 此外，连同最大乘积值，我们还将维护最大乘积所对应的子字符串。
• 遍历整个字符串，请打印与最大乘积相对应的子字符串。

下面是上述方法的实现：

C++

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

输出：

hdzz