Closest perfect square and its distance

Given a positive integer $N$ . The task is to find the perfect square number closest to N and steps required to reach this number from N.

Note: The closest perfect square to N can be either less than, equal to or greater than N and steps is referred to the difference between N and the closest perfect square.

Examples:

Input: N = 1500
Output: Perfect square = 1521, Steps = 21
For N = 1500
Closest perfect square greater than N is 1521.
So steps required is 21.
Closest perfect square less than N is 1444.
So steps required is 56.
The minimum of these two is 1521 with steps 21.

Input: N = 2
Output: Perfect Square = 1, Steps = 1
For N = 2
Closest perfect square greater than N is 4.
So steps required is 2.
Closest perfect square less than N is 1.
So steps required is 1.
The minimum of these two is 1.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach:

If N is a perefct square then print N and steps as 0.
Else, find the first perfect square number > N and note its difference with N.
Then, find the first perfect square number < N and note its difference with N.
And print the perfect square resulting in the minimum of these two differences obtained and also the difference as the minimum steps.

Below is the implementation of the above approach:

C++

// CPP program to find the closest perfect square 
// taking minimum steps to reach from a number 
  
#include<bits/stdc++.h> 
using namespace std; 
  
  
    // Function to check if a number is 
    // perfect square or not 
    bool isPerfect(int N) 
    { 
        if ((sqrt(N) - floor(sqrt(N))) != 0) 
            return false; 
        return true; 
    } 
  
    // Function to find the closest perfect square 
    // taking minimum steps to reach from a number 
    void getClosestPerfectSquare(int N) 
    { 
        if (isPerfect(N)) 
        { 
            cout<<N<<" "<<"0"<<endl; 
            return; 
        } 
  
        // Variables to store first perfect 
        // square number 
        // above and below N 
        int aboveN = -1, belowN = -1; 
        int n1; 
  
        // Finding first perfect square 
        // number greater than N 
        n1 = N + 1; 
        while (true) { 
            if (isPerfect(n1)) { 
                aboveN = n1; 
                break; 
            } 
            else
                n1++; 
        } 
  
        // Finding first perfect square 
        // number less than N 
        n1 = N - 1; 
        while (true) { 
            if (isPerfect(n1)) { 
                belowN = n1; 
                break; 
            } 
            else
                n1--; 
        } 
  
        // Variables to store the differences 
        int diff1 = aboveN - N; 
        int diff2 = N - belowN; 
  
        if (diff1 > diff2) 
            cout<<belowN<<" "<<diff2; 
        else
            cout<<aboveN<<" "<<diff1; 
    } 
  
    // Driver code 
    int main() 
    { 
        int N = 1500; 
  
        getClosestPerfectSquare(N); 
    } 
//This code is contributed by 
// Surendra_Gangwar 

Java

// Java program to find the closest perfect square 
// taking minimum steps to reach from a number 
  
class GFG { 
  
    // Function to check if a number is 
    // perfect square or not 
    static boolean isPerfect(int N) 
    { 
        if ((Math.sqrt(N) - Math.floor(Math.sqrt(N))) != 0) 
            return false; 
        return true; 
    } 
  
    // Function to find the closest perfect square 
    // taking minimum steps to reach from a number 
    static void getClosestPerfectSquare(int N) 
    { 
        if (isPerfect(N)) { 
            System.out.println(N + " "
                               + "0"); 
            return; 
        } 
  
        // Variables to store first perfect 
        // square number 
        // above and below N 
        int aboveN = -1, belowN = -1; 
        int n1; 
  
        // Finding first perfect square 
        // number greater than N 
        n1 = N + 1; 
        while (true) { 
            if (isPerfect(n1)) { 
                aboveN = n1; 
                break; 
            } 
            else
                n1++; 
        } 
  
        // Finding first perfect square 
        // number less than N 
        n1 = N - 1; 
        while (true) { 
            if (isPerfect(n1)) { 
                belowN = n1; 
                break; 
            } 
            else
                n1--; 
        } 
  
        // Variables to store the differences 
        int diff1 = aboveN - N; 
        int diff2 = N - belowN; 
  
        if (diff1 > diff2) 
            System.out.println(belowN + " " + diff2); 
        else
            System.out.println(aboveN + " " + diff1); 
    } 
  
    // Driver code 
    public static void main(String args[]) 
    { 
        int N = 1500; 
  
        getClosestPerfectSquare(N); 
    } 
} 

Python3

# Python3 program to find the closest  
# perfect square taking minimum steps 
# to reach from a number  
  
# Function to check if a number is  
# perfect square or not 
from math import sqrt, floor  
def isPerfect(N): 
    if (sqrt(N) - floor(sqrt(N)) != 0): 
        return False
    return True
  
# Function to find the closest perfect square  
# taking minimum steps to reach from a number  
def getClosestPerfectSquare(N): 
    if (isPerfect(N)):  
        print(N, "0")  
        return
  
    # Variables to store first perfect  
    # square number above and below N  
    aboveN = -1
    belowN = -1
    n1 = 0
  
    # Finding first perfect square  
    # number greater than N  
    n1 = N + 1
    while (True): 
        if (isPerfect(n1)): 
            aboveN = n1  
            break
        else: 
            n1 += 1
  
    # Finding first perfect square  
    # number less than N  
    n1 = N - 1
    while (True):  
        if (isPerfect(n1)):  
            belowN = n1  
            break
        else: 
            n1 -= 1
              
    # Variables to store the differences  
    diff1 = aboveN - N  
    diff2 = N - belowN  
  
    if (diff1 > diff2): 
        print(belowN, diff2)  
    else: 
        print(aboveN, diff1) 
  
# Driver code  
N = 1500
getClosestPerfectSquare(N) 
  
# This code is contributed  
# by sahishelangia 

C#

// C# program to find the closest perfect square 
// taking minimum steps to reach from a number 
using System; 
  
class GFG { 
  
    // Function to check if a number is 
    // perfect square or not 
    static bool isPerfect(int N) 
    { 
        if ((Math.Sqrt(N) - Math.Floor(Math.Sqrt(N))) != 0) 
            return false; 
        return true; 
    } 
  
    // Function to find the closest perfect square 
    // taking minimum steps to reach from a number 
    static void getClosestPerfectSquare(int N) 
    { 
        if (isPerfect(N)) { 
            Console.WriteLine(N + " "
                            + "0"); 
            return; 
        } 
  
        // Variables to store first perfect 
        // square number 
        // above and below N 
        int aboveN = -1, belowN = -1; 
        int n1; 
  
        // Finding first perfect square 
        // number greater than N 
        n1 = N + 1; 
        while (true) { 
            if (isPerfect(n1)) { 
                aboveN = n1; 
                break; 
            } 
            else
                n1++; 
        } 
  
        // Finding first perfect square 
        // number less than N 
        n1 = N - 1; 
        while (true) { 
            if (isPerfect(n1)) { 
                belowN = n1; 
                break; 
            } 
            else
                n1--; 
        } 
  
        // Variables to store the differences 
        int diff1 = aboveN - N; 
        int diff2 = N - belowN; 
  
        if (diff1 > diff2) 
            Console.WriteLine(belowN + " " + diff2); 
        else
            Console.WriteLine(aboveN + " " + diff1); 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int N = 1500; 
  
        getClosestPerfectSquare(N); 
    } 
} 
// This code is contributed by anuj_67.. 

PHP

<?php 
// PHP program to find the closest perfect  
// square taking minimum steps to reach  
// from a number 
  
// Function to check if a number is 
// perfect square or not 
function isPerfect($N) 
{ 
    if ((sqrt($N) - floor(sqrt($N))) != 0) 
        return false; 
    return true; 
} 
  
// Function to find the closest perfect square 
// taking minimum steps to reach from a number 
function getClosestPerfectSquare($N) 
{ 
    if (isPerfect($N)) 
    { 
        echo $N, " ", "0", "\n"; 
        return; 
    } 
  
    // Variables to store first perfect 
    // square number 
    // above and below N 
    $aboveN = -1; 
    $belowN = -1; 
    $n1; 
  
    // Finding first perfect square 
    // number greater than N 
    $n1 = $N + 1; 
    while (true) 
    { 
        if (isPerfect($n1)) 
        { 
            $aboveN = $n1; 
            break; 
        } 
        else
            $n1++; 
    } 
  
    // Finding first perfect square 
    // number less than N 
    $n1 = $N - 1; 
    while (true) 
    { 
        if (isPerfect($n1)) 
        { 
            $belowN = $n1; 
            break; 
        } 
        else
            $n1--; 
    } 
  
    // Variables to store the differences 
    $diff1 = $aboveN - $N; 
    $diff2 = $N - $belowN; 
  
    if ($diff1 > $diff2) 
        echo $belowN, " " , $diff2; 
    else
        echo $aboveN, " ", $diff1; 
} 
  
// Driver code 
$N = 1500; 
getClosestPerfectSquare($N); 
  
// This code is contributed by ajit. 
?> 

Output:

1521 21

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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