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Minimum digits to remove to make a number Perfect Square

  • Difficulty Level : Hard
  • Last Updated : 01 Jul, 2021

Given a integer n, we need to find how many digits remove from the number to make it a perfect square.

Examples : 

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

Input : 57 
Output : -1



The idea is to generate all possible subsequences and return optimal string using set bits. Let’s suppose we have a string 8314. And using set bits we form all possible subsequences i.e.,
8, 3, 83, 1, 81, 31, 831, 4, 84, 34, 834, 14, 814, 314, 8314.
After forming all possible subsequences, we check which one is the perfect square. And we return a perfect square number which has the minimum length.

In above example, three perfect squares are 1 4 and 81, so answer would be 81 because 81 has the max length 2.  

C++




// C++ program to find required minimum digits
// need to remove to make a number perfect square
#include <bits/stdc++.h>
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 square
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 square
 
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 square
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




<?php
// PHP program to find required
// minimum digits need to remove
// to make a number perfect square
 
// function to check minimum
// number of digits should be
// removed to make this number
// a perfect square
function perfectSquare($s)
{
    // size of the string
    $n = strlen($s);
 
    // our final answer
    $ans = -1;
 
    // to store string which
    // is perfect square.
    $num = "";
 
    // We make all possible
    // subsequences
    for ($i = 1; $i < (1 << $n); $i++)
    {
        $str = "";
        for ($j = 0; $j < $n; $j++)
        {
             
            // 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 = 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)
?>

Javascript




<script>
 
// Javascript program to find required
// minimum digits need to remove
// to make a number perfect square
 
// Function to check minimum
// number of digits should be
// removed to make this number
// a perfect square
function perfectSquare(s)
{
     
    // Size of the string
    let n = s.length;
 
    // Our final answer
    let ans = -1;
 
    // To store string which
    // is perfect square.
    let num = "";
 
    // We make all possible
    // subsequences
    for(let i = 1; i < (1 << n); i++)
    {
        let str = "";
        for(j = 0; j < n; j++)
        {
 
            // 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
            let temp = 0;
            for(let j = 0; j < str.length; j++)
                temp = temp * 10 + (str[j].charCodeAt(0) -
                                       '0'.charCodeAt(0));
 
            k = Math.floor(Math.sqrt(temp));
 
            // Checking temp is perfect
            // square or not.
            if ((k * k) == temp)
            {
                 
                // Taking maximum sized string
                if (ans < str.length)
                {
                    ans = str.length;
                    num = str;
                }
            }
        }
    }
 
    if (ans == -1)
        return ans;
    else
    {
         
        // Print PerfectSquare
        document.write(num + " ");
        return (n - ans);
    }
}
 
// Driver code
document.write(perfectSquare("8314") + "<br>");
document.write(perfectSquare("753") + "<br>");
 
// This code is contributed by gfgking
 
</script>
Output : 
81 2
-1

 




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