删除的最小位数以形成一个数字完美的平方

给定一个整数n，我们需要找出从数字中删除多少个数字，使其成为一个完美的平方。

例子 ：

Input : 8314
Output: 81 2
Explanation: If we remove 3 and 4 number becomes 81 which is a perfect square.

Input : 57
Output : -1

想法是生成所有可能的子序列，并使用设置的位返回最佳字符串。假设我们有一个字符串8314。并且使用置位，我们形成了所有可能的子序列，即

8、3、83、1、81、31、831、4、84、34、834、14、814、314、8314。

形成所有可能的子序列后，我们检查哪一个是理想的正方形。然后我们返回一个最小长度的理想平方数。

在上面的示例中，三个完美的平方分别是1 4和81，因此答案将是81，因为81的最大长度为2。

C++

// C++ program to find required minimum digits
// need to remove to make a number perfect squre
#include 
using namespace std;
  
// function to check minimum number of digits
// should be removed to make this number
// a perfect square
int perfectSquare(string s)
{
    // size of the string
    int n = s.size();
  
    // our final answer
    int ans = -1;
  
    // to store string which is perfect square.
    string num;
  
    // We make all possible subsequences
    for (int i = 1; i < (1 << n); i++) {
        string str = "";
          
        for (int j = 0; j < n; j++) {
              
            // to check jth bit is set or not.
            if ((i >> j) & 1) {
                str += s[j];
            }
        }
  
        // we do not consider a number with leading zeros
        if (str[0] != '0') {
              
            // convert our temporary string into integer
            int temp = 0;
            for (int j = 0; j < str.size(); j++)
                temp = temp * 10 + (int)(str[j] - '0');
  
            int k = sqrt(temp);
  
            // checking temp is perfect square or not.
            if (k * k == temp) {
                  
                // taking maximum sized string
                if (ans < (int)str.size()) {
                    ans = (int)str.size();
                    num = str;
                }
            }
        }
    }
  
    if (ans == -1)
        return ans;
    else {
          
        // print PerfectSquare
        cout << num << " ";
        return n - ans;
    }
}
  
// Driver code
int main()
{
    cout << perfectSquare("8314") << endl;
    cout << perfectSquare("753") << endl;  
    return 0;
}

Java

// Java program to find required minimum digits
// need to remove to make a number perfect squre
import java.io.*;
import java.lang.*;
  
public class GFG {
      
    // function to check minimum 
    // number of digits should 
    // be removed to make this 
    // number a perfect square
    static int perfectSquare(String s)
    {
        // size of the string
        int n = s.length();
      
        // our final answer
        int ans = -1;
      
        // to store string which
        // is perfect square.
        String num = "";
      
        // We make all possible subsequences
        for (int i = 1; i < (1 << n); i++) {
            String str = "";
              
            for (int j = 0; j < n; j++) {
                  
                // to check jth bit is set or not.
                if (((i >> j) & 1) == 1) {
                    str += s.charAt(j);
                }
            }
      
            // we do not consider a number 
            // with leading zeros
            if (str.charAt(0) != '0') {
                  
                // convert our temporary
                // string into integer
                int temp = 0;
                for (int j = 0; j < 
                              str.length(); j++)
                    temp = temp * 10 + 
                      (int)(str.charAt(j) - '0');
      
                int k = (int)Math.sqrt(temp);
      
                // checking temp is perfect
                // square or not.
                if (k * k == temp) {
                      
                    // taking maximum sized string
                    if (ans < (int)str.length()) {
                        ans = (int)str.length();
                        num = str;
                    }
                }
            }
        }
      
        if (ans == -1)
            return ans;
        else {
              
            // print PerfectSquare
            System.out.print(num + " ");
            return n - ans;
        }
    }
      
    // Driver code
    public static void main(String args[])
    {
        System.out.println(perfectSquare("8314"));
        System.out.println(perfectSquare("753"));
    }
}
  
// This code is contributed by 
// Manish Shaw (manishshaw1)

Python3

# C++ program to find required minimum 
# digits need to remove to make a 
# number perfect squre
  
import math
# function to check minimum number of
# digits should be removed to make
# this number a perfect square
def perfectSquare(s) :
      
    # size of the string
    n = len(s)
  
    # our final answer
    ans = -1
  
    # to store string which is
    # perfect square.
    num = ""
  
    # We make all possible subsequences
    for i in range(1, (1 << n)) :
        str = ""
          
        for j in range(0, n) :
              
            # to check jth bit is
            # set or not.
            if ((i >> j) & 1) :
                str = str + s[j]
  
        # we do not consider a number 
        # with leading zeros
        if (str[0] != '0') :
              
            # convert our temporary 
            # string into integer
            temp = 0;
            for j in range(0, len(str)) :
                temp = (temp * 10 + 
                 (ord(str[j]) - ord('0')))
  
            k = int(math.sqrt(temp))
  
            # checking temp is perfect 
            # square or not.
            if (k * k == temp) :
                  
                # taking maximum sized
                # string
                if (ans < len(str)) :
                    ans = len(str)
                    num = str
  
    if (ans == -1) :
        return ans
    else :         
          
        # print PerfectSquare
        print ("{} ".format(num), end="")
        return n - ans
      
# Driver code
print (perfectSquare("8314"))
print (perfectSquare("753"));
  
# This code is contributed by 
# manishshaw1.

C#

// C# program to find required minimum digits
// need to remove to make a number perfect squre
using System;
class GFG {
      
    // function to check minimum 
    // number of digits should 
    // be removed to make this 
    // number a perfect square
    static int perfectSquare(string s)
    {
        // size of the string
        int n = s.Length;
      
        // our final answer
        int ans = -1;
      
        // to store string which
        // is perfect square.
        string num = "";
      
        // We make all possible subsequences
        for (int i = 1; i < (1 << n); i++) {
            string str = "";
              
            for (int j = 0; j < n; j++) {
                  
                // to check jth bit is set or not.
                if (((i >> j) & 1) == 1) {
                    str += s[j];
                }
            }
      
            // we do not consider a number 
            // with leading zeros
            if (str[0] != '0') {
                  
                // convert our temporary
                // string into integer
                int temp = 0;
                for (int j = 0; j < str.Length; j++)
                    temp = temp * 10 + (int)(str[j] - '0');
      
                int k = (int)Math.Sqrt(temp);
      
                // checking temp is perfect
                // square or not.
                if (k * k == temp) {
                      
                    // taking maximum sized string
                    if (ans < (int)str.Length) {
                        ans = (int)str.Length;
                        num = str;
                    }
                }
            }
        }
      
        if (ans == -1)
            return ans;
        else {
              
            // print PerfectSquare
            Console.Write(num + " ");
            return n - ans;
        }
    }
      
    // Driver code
    public static void Main()
    {
        Console.WriteLine(perfectSquare("8314"));
        Console.WriteLine(perfectSquare("753"));
    }
}
  
// This code is contributed by 
// Manish Shaw (manishshaw1)

PHP

> $j) & 1) 
            {
                $str = $str.$s[$j];
            }
        }
  
        // we do not consider a 
        // number with leading zeros
        if ($str[0] != '0') 
        {
            // convert our temporary
            // string into integer
            $temp = 0;
            for ($j = 0; $j < strlen($str); $j++)
                $temp = $temp * 10 + 
                        (ord($str[$j]) - ord('0'));
  
            $k = (int)(sqrt($temp));
  
            // checking temp is perfect
            // square or not.
            if (($k * $k) == $temp)
            {
                  
                // taking maximum sized string
                if ($ans < strlen($str)) 
                {
                    $ans = strlen($str);
                    $num = $str;
                }
            }
        }
    }
  
    if ($ans == -1) 
        return $ans;
    else 
    {         
        // print PerfectSquare
        echo ($num." ");
        return ($n - $ans);
    }
} 
  
// Driver code
echo (perfectSquare("8314"). "\n");
echo (perfectSquare("753"). "\n");
  
// This code is contributed by 
// Manish Shaw (manishshaw1)
?>

输出 ：

81 2
-1

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