最多改变一位数可以被3整除的不同数字的计数

给定一个表示数字的字符串str 有N个数字，任务是通过改变最多一位数来计算使给定数能被 3 整除的方法数。

例子：

Input: str[] = “23”
Output: 7
Explanation: Below are the numbers that can be made from the string which are divisible by 3 – 03, 21, 24, 27, 33, 63, 93
1.Change 2 to 0 (0+3)=3 divisible by 3
2.Change 3 to 1 (2+1)=3 divisible by 3
3 change 3 to 4 (2+4)=6 divisible by 3
4 change 2 to 3 sum is 6 divisible by 3
Similarly there are total 7 number of ways to make the given number divisible by 3

Input: str[] = “235”
Output: 9

方法：解决这个问题的思路很简单。计算给定数字的数字总和，然后对于每个索引，删除该数字并尝试从0到9的所有可能数字，看看总和是否可以被3整除。请按照以下步骤解决问题：

• 将变量sum初始化为0以存储数字的位数之和。
• 使用变量i迭代范围[0, N]并执行以下步骤：
  • 在变量sum的第 i 个索引处添加 digit 的值。
• 将变量count初始化为0以存储答案。
• 如果数字本身可以被 3 整除，则将 count 加一。
• 使用变量i迭代范围[0, N]并执行以下步骤：
  • 将变量remaining_sum初始化为sum-(number.charAt(i)-48)。
  • 使用变量j迭代范围[0, 9]并执行以下步骤：
    • 将j的值加到变量remaining_sum中，如果剩余和可被3整除且j不等于第 i 个索引处的数字，则将count的值加1。
• 执行上述步骤后，打印count的值作为答案。

下面是上述方法的实现..

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to count the number of
// possible numbers divisible by 3
void findCount(string number)
{
  
    // Calculate the sum
    int sum = 0;
    for (int i = 0; i < number.length(); ++i) {
        sum += number[i] - 48;
    }
  
    // Store the answer
    int count = 0;
  
    // Consider the edge case when
    // the number itself is divisible by 3
    // The count will be added by 1
    if (sum % 3 == 0)
        count++;
  
    // Iterate over the range
    for (int i = 0; i < number.length(); ++i) {
  
        // Decreasing the sum
        int remaining_sum = sum - (number[i] - 48);
  
        // Iterate over the range
        for (int j = 0; j <= 9; ++j) {
  
            // Checking if the new sum
            // is divisible by 3 or not
            if ((remaining_sum + j) % 3 == 0
                && j != number[i] - 48) {
  
                // If yes increment
                // the value of the count
                ++count;
            }
        }
    }
    cout << count;
}
  
// Driver Code
int main()
{
    // Given number
    string number = "235";
  
    findCount(number);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
  
class GFG {
  
    // Function to count the number of
    // possible numbers divisible by 3
    public static void findCount(String number)
    {
  
        // Calculate the sum
        int sum = 0;
        for (int i = 0; i < number.length(); ++i) {
            sum += number.charAt(i) - 48;
        }
  
        // Store the answer
        int count = 0;
        if (sum % 3 == 0)
            count++;
  
        // Iterate over the range
        for (int i = 0; i < number.length(); ++i) {
  
            // Decreasing the sum
            int remaining_sum
                = sum - (number.charAt(i) - 48);
  
            // Iterate over the range
            for (int j = 0; j <= 9; ++j) {
  
                // Checking if the new sum
                // is divisible by 3 or not
                if ((remaining_sum + j) % 3 == 0
                    && j != number.charAt(i) - 48) {
  
                    // If yes increment
                    // the value of the count
                    ++count;
                }
            }
        }
        System.out.println(count);
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        // Given number
        String number = "235";
  
        findCount(number);
    }
}

Python3

# Python program for the above approach
  
# Function to count the number of
# possible numbers divisible by 3
  
  
def findCount(number):
  
    # Calculate the sum
    sum = 0
    for i in range(len(number)):
        sum += int(number[i]) - 48
  
    # Store the answer
    count = 0
    if(sum % 3 == 0):
      count += 1
  
    # Iterate over the range
    for i in range(len(number)):
  
        # Decreasing the sum
        remaining_sum = sum - (int(number[i]) - 48)
  
        # Iterate over the range
        for j in range(10):
  
            # Checking if the new sum
            # is divisible by 3 or not
            if ((remaining_sum + j) % 3 == 0 and j != int(number[i]) - 48):
  
                # If yes increment
                # the value of the count
                count += 1
  
    print(count)
  
# Driver Code
  
  
# Given number
number = "235"
findCount(number)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
  
class GFG {
  
    // Function to count the number of
    // possible numbers divisible by 3
    static void findCount(string number)
    {
  
        // Calculate the sum
        int sum = 0;
        for (int i = 0; i < number.Length; ++i) {
            sum += (int)number[i] - 48;
        }
  
        // Store the answer
        int count = 0;
        if (sum % 3 == 0)
            count++;
  
        // Iterate over the range
        for (int i = 0; i < number.Length; ++i) {
  
            // Decreasing the sum
            int remaining_sum = sum - ((int)number[i] - 48);
  
            // Iterate over the range
            for (int j = 0; j <= 9; ++j) {
  
                // Checking if the new sum
                // is divisible by 3 or not
                if ((remaining_sum + j) % 3 == 0
                    && j != number[i] - 48) {
  
                    // If yes increment
                    // the value of the count
                    ++count;
                }
            }
        }
        Console.Write(count);
    }
  
    // Driver Code
    public static void Main()
    {
  
        // Given number
        string number = "235";
  
        findCount(number);
    }
}

Javascript

// Javascript program for the above approach
  
// Function to count the number of
// possible numbers divisible by 3
function findCount(number) {
  // Calculate the sum
  let sum = 0;
  for (let i = 0; i < number.length; ++i) {
    sum += number[i] - 48;
  }
  
  // Store the answer
  let count = 0;
  if(sum % 3 == 0) count++;
  // Iterate over the range
  for (let i = 0; i < number.length; ++i) {
    // Decreasing the sum
    let remaining_sum = sum - (number[i] - 48);
  
    // Iterate over the range
    for (let j = 0; j <= 9; ++j) {
      // Checking if the new sum
      // is divisible by 3 or not
      if ((remaining_sum + j) % 3 == 0 && j != number[i] - 48) {
        // If yes increment
        // the value of the count
        ++count;
      }
    }
  }
  document.write(count);
}
  
// Driver Code
  
// Given number
let number = "235";
  
findCount(number);
  
// This code is contributed by gfgking
  

输出

9

时间复杂度： O(N)
辅助空间： O(1)