以“xAyB”形式编码字符串，其中 x 和 y 并基于位数

给定两个数字字符串N和M ，任务是以“ xAyB ”的形式对给定字符串进行编码，其中：

• x 是 N 和 M 中相同且出现在相同索引上的数字的计数
• y 是 N 和 M 中相同但出现在不同索引上的数字的计数

例子：

Input: N = 123, M = 321
Output: “1A2B”
Explanation: 
Digit 2 satisfies condition for x as count of digits that are same in N and M and are present on same indices
Digits 1 and 3 satisfy the condition for y as count of digits that are same in N and M but are present on different indices

Input: N = 123, M = 111
Output: “0A1B”

方法：可以使用散列和两指针方法解决问题。

• 将 N 和 M 转换为字符串以便于遍历
• 现在创建 2 个大小为 10 的散列来分别存储 N 和 M 中的数字频率
  • 现在从 0-9 遍历一个循环并且：
• 将 hashN[i] 和 hashM[i] 的最小值添加到变量计数中
• 现在使用两个指针遍历数字以查找相同的数字计数，并出现在 N 和 M 的相同索引上。将计数存储在变量 same_dig_cnt
• 因此 x = same_dig_cnt, y = count。
• 现在将最终字符串返回为“xAyB”

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
 
// Function to encode strings as "xAyB"
string encodeStrings(int N, int M)
{
 
    // Convert N and M to string
    // for ease of traversal
    string a = to_string(N), b = to_string(M);
 
    // Create 2 hash of size 10
    // to store frequency of digits
    // in N and M respectively
    vector hashN(10, 0);
    for (char c : a)
        hashN++;
    vector hashM(10, 0);
    for (char c : b)
        hashM++;
 
    int count = 0, same_dig_cnt = 0;
 
    // Count of common digits
    // irrespective of their positions
    for (int i = 0; i < 10; i++)
        count += min(hashN[i], hashM[i]);
 
    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.length() && b.length(); i++)
        if (a[i] == b[i])
            same_dig_cnt++;
 
    // Remove the count of digits that are
    // not at same indices in both numbers
    count -= same_dig_cnt;
 
    // Therefore x = same_dig_cnt, y = count.
    // Now return the final string as "xAyB"
    string ans = "" + to_string(same_dig_cnt) + "A"
                 + to_string(count) + "B";
    return ans;
}
 
// Driver code
int main()
{
 
    int N = 1807, M = 7810;
 
    cout << "\"" << encodeStrings(N, M) << "\"";
    return 0;
}

Java

// java implementation of the above approach
class GFG {
 
  // Function to encode Strings as "xAyB"
  static String encodeStrings(int N, int M)
  {
 
    // Convert N and M to String
    // for ease of traversal
    String a = Integer.toString(N), b = Integer.toString(M);
 
    // Create 2 hash of size 10
    // to store frequency of digits
    // in N and M respectively
    int[] hashN = new int[10];
    for (int i = 0; i < 10; i++) {
      hashN[i] = 0;
    }
 
    for(char c : a.toCharArray()) hashN++;
 
    int[] hashM = new int[10];
    for (int i = 0; i < 10; i++) {
      hashM[i] = 0;
    }
 
    for(char c : b.toCharArray()) hashM++;
 
    int count = 0, same_dig_cnt = 0;
 
    // Count of common digits
    // irrespective of their positions
    for (int i = 0; i < 10; i++)
      count += Math.min(hashN[i], hashM[i]);
 
    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.length() && i < b.length(); i++)
      if (a.charAt(i) == b.charAt(i))
        same_dig_cnt++;
 
    // Remove the count of digits that are
    // not at same indices in both numbers
    count -= same_dig_cnt;
 
    // Therefore x = same_dig_cnt, y = count.
    // Now return the final String as "xAyB"
    String ans = "" + Integer.toString(same_dig_cnt) + "A" + Integer.toString(count) + "B";
    return ans;
  }
 
  // Driver code
  public static void main(String args[])
  {
 
    int N = 1807, M = 7810;
 
    System.out.println("\"" + encodeStrings(N, M) + "\"");
  }
}
 
// This code is contributed by Saurabh jaiswal

Python3

# Python code for the above approach
def encodeStrings(N, M):
 
    # Convert N and M to string
    # for ease of traversal
    a = str(N)
    b = str(M)
 
    # Create 2 hash of size 10
    # to store frequency of digits
    # in N and M respectively
    hashN = [0] * 10
    for c in range(len(a)):
        hashN[ord(a) - ord('0')] += 1
 
    hashM = [0] * 10
    for c in range(len(b)):
        hashM[ord(b) - ord('0')] += 1
 
    count = 0
    same_dig_cnt = 0
 
    # Count of common digits
    # irrespective of their positions
    for i in range(10):
        count += min(hashN[i], hashM[i])
 
    # Find the count of digits
    # that are same and occur on same indices
    # in both N and M.
    # Store the count in variable same_dig_cnt
    i = 0
    while(i < len(a) and len(b)):
        if (a[i] == b[i]):
            same_dig_cnt += 1
        i += 1
 
    # Remove the count of digits that are
    # not at same indices in both numbers
    count -= same_dig_cnt
 
    # Therefore x = same_dig_cnt, y = count.
    # Now return the final string as "xAyB"
    ans = str(same_dig_cnt) + "A" + str(count) + "B"
    return ans
 
# Driver code
N = 1807
M = 7810
 
print(f"\"{encodeStrings(N, M)}\"")
 
# This code is contributed by Saurabh jaiswal

C#

// C# implementation of the above approach
using System;
class GFG {
 
  // Function to encode strings as "xAyB"
  static string encodeStrings(int N, int M)
  {
 
    // Convert N and M to string
    // for ease of traversal
    string a = N.ToString(), b = M.ToString();
 
    // Create 2 hash of size 10
    // to store frequency of digits
    // in N and M respectively
    int[] hashN = new int[10];
    for (int i = 0; i < 10; i++) {
      hashN[i] = 0;
    }
 
    foreach(char c in a) hashN++;
 
    int[] hashM = new int[10];
    for (int i = 0; i < 10; i++) {
      hashM[i] = 0;
    }
 
    foreach(char c in b) hashM++;
 
    int count = 0, same_dig_cnt = 0;
 
    // Count of common digits
    // irrespective of their positions
    for (int i = 0; i < 10; i++)
      count += Math.Min(hashN[i], hashM[i]);
 
    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.Length && i < b.Length; i++)
      if (a[i] == b[i])
        same_dig_cnt++;
 
    // Remove the count of digits that are
    // not at same indices in both numbers
    count -= same_dig_cnt;
 
    // Therefore x = same_dig_cnt, y = count.
    // Now return the final string as "xAyB"
    string ans = "" + same_dig_cnt.ToString() + "A"
      + count.ToString() + "B";
    return ans;
  }
 
  // Driver code
  public static void Main()
  {
 
    int N = 1807, M = 7810;
 
    Console.Write("\"" + encodeStrings(N, M) + "\"");
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

        // JavaScript code for the above approach
        function encodeStrings(N, M) {
 
            // Convert N and M to string
            // for ease of traversal
            let a = (N).toString(), b = (M).toString();
 
            // Create 2 hash of size 10
            // to store frequency of digits
            // in N and M respectively
            let hashN = new Array(10).fill(0);
            for (let c = 0; c < a.length; c++)
                hashN[a.charCodeAt(0) - '0'.charCodeAt(0)]++;
            let hashM = new Array(10).fill(0);
            for (c = 0; c < b.length; c++)
                hashM[b.charCodeAt(0) - '0'.charCodeAt(0)]++;
 
            let count = 0, same_dig_cnt = 0;
 
            // Count of common digits
            // irrespective of their positions
            for (let i = 0; i < 10; i++)
                count += Math.min(hashN[i], hashM[i]);
 
            // Find the count of digits
            // that are same and occur on same indices
            // in both N and M.
            // Store the count in variable same_dig_cnt
            for (let i = 0; i < a.length && b.length; i++)
                if (a[i] == b[i])
                    same_dig_cnt++;
 
            // Remove the count of digits that are
            // not at same indices in both numbers
            count -= same_dig_cnt;
             
            // Therefore x = same_dig_cnt, y = count.
            // Now return the final string as "xAyB"
            let ans = (same_dig_cnt).toString() + "A"
                + (count).toString() + "B";
            return ans;
        }
 
        // Driver code
        let N = 1807, M = 7810;
 
        document.write(`"${encodeStrings(N, M)}"`);
         
       // This code is contributed by Potta Lokesh
    

输出

"1A3B"

时间复杂度： O(D)，其中 D 是 N 或 M 中的最大位数
辅助空间： O(D)，其中 D 是 N 或 M 中的最大位数