通过用它们的和替换相邻的一对数字来最小化形成的数字

给定字符串s表示一个数字。任务是找到将s的两个连续数字替换为它们的和后可以形成的最小数字。

例子：

Input: s = “1005”
Output: 105
Explanation: Select and replace any two consecutive digits with its sum 

Input: s = “56773″
Output: 11773
Explanation: Select and replace first two consecutive digits (5 and 6) with its sum (11) 

方法：这个问题可以通过使用贪心方法来解决。使用以下观察：

If there exists at least one pair having sum less than 10:

• The sum will always be greater than or equal to the elements of the adjacent pair.
• So replacing the last arrival of such pair will give the minimum number possible. Otherwise, the number will increase.
• For example, 2381 can be made 581 or 239 and 239 is the minimum between these two.
• This condition is given priority because it decreases the number of digits in the number and makes the value less than any number having same number of digits as the actual number.

If there is no such pair having sum less than 10:

• The sum will always be less than the number formed by the adjacent digits.
• So replacing the first occurrence of such a pair will give the minimum number as it makes the most significant part minimum.
• For example, 386 can be made 116 or 314. 116 is the lesser one.

请按照以下步骤使用上述观察解决给定的问题。

• 首先假设存在至少一对和小于 10的连续元素。
  • 从后面迭代数字字符串。
  • 找出总和小于10的第一对连续数字。
  • 用它们的总和替换它们并返回字符串。
• 现在，处理没有一对连续元素的总和小于 10的情况。
  • 从开始迭代数字字符串。
  • 和等于或大于10的第一对连续数字是最佳答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find minimum number after
// doing given operation
void findSmallestNumber(string str, int N)
{
    string ans = "";
    bool found = false;
 
    // Iterate through the string in
    // reverse order
    for (int i = N - 1; i >= 1; i--) {
        int a = (str[i] - '0');
        int b = (str[i - 1] - '0');
        if (a + b < 10) {
            found = true;
            ans = str.substr(0, i - 1);
            int sum = a + b;
            int last = sum % 10;
            char l = last + '0';
            sum /= 10;
 
            if (sum != 0) {
                char p = sum + '0';
                ans.push_back(p);
            }
            ans.push_back(l);
            ans += str.substr(i + 1,
                              N - i - 1);
            cout << ans;
            break;
        }
    }
 
    if (!found) {
        int sum = (str[0] - '0')
                  + (str[1] - '0');
        ans = (char)(sum / 10 + '0');
        sum %= 10;
        ans += (sum + '0');
        ans += str.substr(2, N - 2);
        cout << ans;
    }
}
 
// Driver code
int main()
{
    string s = "56773";
    int N = s.length();
 
    // Function Call
    findSmallestNumber(s, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG
{
 
  // Function to find minimum number after
  // doing given operation
  static void findSmallestNumber(String str, int N)
  {
    String ans = "";
    boolean found = false;
 
    // Iterate through the string in
    // reverse order
    for (int i = N - 1; i >= 1; i--) {
      int a = (str.charAt(i) - '0');
      int b = (str.charAt(i-1)- '0');
      if (a + b < 10) {
        found = true;
        ans = str.substring(0, i - 1);
        int sum = a + b;
        int last = sum % 10;
        char l = (char)(last + '0');
        sum /= 10;
 
        if (sum != 0) {
          char p = (char)(sum + '0');
          ans += p;
        }
        ans += l;
        ans += str.substring(i + 1,
                             N - i - 1);
        System.out.println(ans);
        break;
      }
    }
 
    if (!found) {
      int sum = (str.charAt(0) - '0')
        + (str.charAt(1) - '0');
      Integer num = sum / 10;
      ans = num.toString();
      Integer summ = sum % 10;
      ans += summ.toString();
      ans += str.substring(2, N );
      System.out.println(ans);
    }
  }
 
// Driver Code
public static void main(String args[])
{
    String s = "56773";
    int N = s.length();
 
    // Function Call
    findSmallestNumber(s, N);
}
}
 
// This code is contributed by sanjoy_62.

Python3

# Python program for the above approach
 
# Function to find minimum number after
# doing given operation
def findSmallestNumber(str, N):
    ans = ""
    found = False
 
    # Iterate through the string in
    # reverse order
    for i in range(N - 1, 1, -1):
        a = (ord(str[i]) - ord('0'))
        b = (ord(str[i - 1]) - ord('0'))
        if (a + b < 10):
            found = True
            ans = str[0: i - 1]
            sum = a + b
            last = sum % 10
            l = (str)(last)
            sum = sum // 10
 
            if (sum != 0):
                p = (str)(sum)
                ans += p
 
            ans += (l)
            ans += str[i + 1: i + 1 + N - i - 1]
            print(ans)
            break
 
    if (not found):
        sum = (ord(str[0]) - ord('0')) + (ord(str[1]) - ord('0'))
        ans = f'{sum // 10}'
 
        sum %= 10
        ans += f'{sum}'
        ans += str[2: 2 + N - 2]
        print(ans)
 
# Driver code
s = "56773"
N = len(s)
 
# Function Call
findSmallestNumber(s, N)
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
using System.Collections;
class GFG
{
  // Function to find minimum number after
  // doing given operation
  static void findSmallestNumber(string str, int N)
  {
    string ans = "";
    bool found = false;
 
    // Iterate through the string in
    // reverse order
    for (int i = N - 1; i >= 1; i--) {
      int a = (str[i] - '0');
      int b = (str[i - 1] - '0');
      if (a + b < 10) {
        found = true;
        ans = str.Substring(0, i - 1);
        int sum = a + b;
        int last = sum % 10;
        char l = (char)(last + '0');
        sum /= 10;
 
        if (sum != 0) {
          char p = (char)(sum + '0');
          ans += p;
        }
        ans += l;
        ans += str.Substring(i + 1,
                             N - i - 1);
        Console.Write(ans);
        break;
      }
    }
 
    if (!found) {
      int sum = (str[0] - '0')
        + (str[1] - '0');
      int num = sum / 10;
      ans = num.ToString();
      sum %= 10;
      ans += sum.ToString();
      ans += str.Substring(2, N - 2);
      Console.Write(ans);
    }
  }
 
  // Driver code
  public static void Main()
  {
    string s = "56773";
    int N = s.Length;
 
    // Function Call
    findSmallestNumber(s, N);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for the above approach
 
    // Function to find minimum number after
    // doing given operation
    const findSmallestNumber = (str, N) => {
        let ans = "";
        let found = false;
 
        // Iterate through the string in
        // reverse order
        for (let i = N - 1; i >= 1; i--) {
            let a = (str.charCodeAt(i) - '0'.charCodeAt(0));
            let b = (str.charCodeAt(i - 1) - '0'.charCodeAt(0));
            if (a + b < 10) {
                found = true;
                ans = str.substring(0, i - 1);
                let sum = a + b;
                let last = sum % 10;
                let l = last.toString();
                sum = parseInt(sum / 10);
 
                if (sum != 0) {
                    let p = sum.toString();
                    ans += p;
                }
                ans += (l);
                ans += str.substring(i + 1, i + 1 + N - i - 1);
                document.write(ans);
                break;
            }
        }
 
        if (!found) {
            let sum = (str.charCodeAt(0) - '0'.charCodeAt(0))
                + (str.charCodeAt(1) - '0'.charCodeAt(0));
            ans = (parseInt(sum / 10)).toString();
 
            sum %= 10;
            ans += sum.toString();
            ans += str.substring(2, 2 + N - 2);
            document.write(ans);
        }
    }
 
    // Driver code
    let s = "56773";
    let N = s.length;
 
    // Function Call
    findSmallestNumber(s, N);
 
 // This code is contributed by rakeshsahni
 

输出

11773

时间复杂度： O(N) 其中 N 是字符串的大小
辅助空间： O(N)