Minimum number of squares whose sum equals to a given number N | Set-3

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A number can always be represented as a sum of squares of other numbers. Note that 1 is a square, and we can always break a number as (1*1 + 1*1 + 1*1 + …). Given a number n, find the minimum number of squares that sum to N.

Examples:

Input: N = 13
Output: 2
Explanation:
13 can be expressed as, 13 = 3² + 2². Hence, the output is 2.

Input: N = 100
Output: 1
Explanation:
100 can be expressed as 100 = 10². Hence, the output is 1.

Naive Approach: For [O(N*sqrt(N))] approach, please refer to Set 2 of this article.

Efficient Approach: To optimize the naive approach the idea is to use Lagrange’s Four-Square Theorem and Legendre’s Three-Square Theorem. The two theorems are discussed below:

Lagrange’s four-square theorem, also known as Bachet’s conjecture, states that every natural number can be represented as the sum of four integer squares, where each integer is non-negative.

Legendre’s three-square theorem states that a natural number can be represented as the sum of three squares of integers if and only if n is not of the form : n = 4^a (8b+7), for non-negative integers a and b.

Therefore, it is proved that the minimum number of squares to represent any number N can only be within the set {1, 2, 3, 4}. Thus, only checking for these 4 possible values, the minimum number of squares to represent any number N can be found. Follow the steps below:

If N is a perfect square, then the result is 1.
If N can be expressed as the sum of two squares, then the result is 2.
If N cannot be expressed in the form of N = 4^a (8b+7), where a and b are non-negative integers, then the result is 3 by Legendre’s three-square theorem.
If all the above conditions are not satisfied, then by Lagrange’s four-square theorem, the result is 4.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if N
// is a perfect square
bool isPerfectSquare(int N)
{
    int floorSqrt = sqrt(N);
 
    return (N == floorSqrt * floorSqrt);
}
 
// Function that returns true check if
// N is sum of three squares
bool legendreFunction(int N)
{
    // Factor out the powers of 4
    while (N % 4 == 0)
        N /= 4;
 
    // N is NOT of the
    // form 4^a * (8b + 7)
    if (N % 8 != 7)
        return true;
    else
        return false;
}
 
// Function that finds the minimum
// number of square whose sum is N
int minSquares(int N)
{
 
    // If N is perfect square
    if (isPerfectSquare(N))
        return 1;
 
    // If N is sum of 2 perfect squares
    for (int i = 1; i * i < N; i++) {
        if (isPerfectSquare(N - i * i))
            return 2;
    }
 
    // If N is sum of 3 perfect squares
    if (legendreFunction(N))
        return 3;
 
    // Otherwise, N is the
    // sum of 4 perfect squares
    return 4;
}
 
// Driver code
int main()
{
    // Given number
    int N = 123;
 
    // Function call
    cout << minSquares(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function that returns true if N
// is a perfect square
static boolean isPerfectSquare(int N)
{
    int floorSqrt = (int)Math.sqrt(N);
 
    return (N == floorSqrt * floorSqrt);
}
 
// Function that returns true check if
// N is sum of three squares
static boolean legendreFunction(int N)
{
     
    // Factor out the powers of 4
    while (N % 4 == 0)
        N /= 4;
 
    // N is NOT of the
    // form 4^a * (8b + 7)
    if (N % 8 != 7)
        return true;
    else
        return false;
}
 
// Function that finds the minimum
// number of square whose sum is N
static int minSquares(int N)
{
 
    // If N is perfect square
    if (isPerfectSquare(N))
        return 1;
 
    // If N is sum of 2 perfect squares
    for(int i = 1; i * i < N; i++) 
    {
        if (isPerfectSquare(N - i * i))
            return 2;
    }
 
    // If N is sum of 3 perfect squares
    if (legendreFunction(N))
        return 3;
 
    // Otherwise, N is the
    // sum of 4 perfect squares
    return 4;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given number
    int N = 123;
 
    // Function call
    System.out.print(minSquares(N));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
from math import sqrt, floor, ceil
 
# Function that returns True if N
# is a perfect square
def isPerfectSquare(N):
   
    floorSqrt = floor(sqrt(N))
    return (N == floorSqrt * floorSqrt)
   
# Function that returns True check if
# N is sum of three squares
def legendreFunction(N):
   
    # Factor out the powers of 4
    while (N % 4 == 0):
        N //= 4
 
    # N is NOT of the
    # form 4^a * (8b + 7)
    if (N % 8 != 7):
        return True
    else:
        return False
 
# Function that finds the minimum
# number of square whose sum is N
def minSquares(N):
   
    # If N is perfect square
    if (isPerfectSquare(N)):
        return 1
 
    # If N is sum of 2 perfect squares
    for i in range(N):
        if i * i < N:
            break
        if (isPerfectSquare(N - i * i)):
            return 2
           
    # If N is sum of 3 perfect squares
    if (legendreFunction(N)):
        return 3
       
    # Otherwise, N is the
    # sum of 4 perfect squares
    return 4
   
# Driver code
if __name__ == '__main__':
 
    # Given number
    N = 123
 
    # Function call
    print(minSquares(N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function that returns true if N
// is a perfect square
static bool isPerfectSquare(int N)
{
    int floorSqrt = (int)Math.Sqrt(N);
 
    return (N == floorSqrt * floorSqrt);
}
 
// Function that returns true check 
// if N is sum of three squares
static bool legendreFunction(int N)
{
     
    // Factor out the powers of 4
    while (N % 4 == 0)
        N /= 4;
 
    // N is NOT of the
    // form 4^a * (8b + 7)
    if (N % 8 != 7)
        return true;
    else
        return false;
}
 
// Function that finds the minimum
// number of square whose sum is N
static int minSquares(int N)
{
 
    // If N is perfect square
    if (isPerfectSquare(N))
        return 1;
 
    // If N is sum of 2 perfect squares
    for(int i = 1; i * i < N; i++) 
    {
        if (isPerfectSquare(N - i * i))
            return 2;
    }
 
    // If N is sum of 3 perfect squares
    if (legendreFunction(N))
        return 3;
 
    // Otherwise, N is the
    // sum of 4 perfect squares
    return 4;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given number
    int N = 123;
 
    // Function call
    Console.Write(minSquares(N));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// Javascript program for the above approach
 
// Function that returns true if N
// is a perfect square
function isPerfectSquare(N)
{
    let floorSqrt = Math.floor(Math.sqrt(N));
   
    return (N == floorSqrt * floorSqrt);
}
   
// Function that returns true check if
// N is sum of three squares
function legendreFunction(N)
{
       
    // Factor out the powers of 4
    while (N % 4 == 0)
        N = Math.floor(N / 4);
   
    // N is NOT of the
    // form 4^a * (8b + 7)
    if (N % 8 != 7)
        return true;
    else
        return false;
}
   
// Function that finds the minimum
// number of square whose sum is N
function minSquares(N)
{
   
    // If N is perfect square
    if (isPerfectSquare(N))
        return 1;
   
    // If N is sum of 2 perfect squares
    for(let i = 1; i * i < N; i++) 
    {
        if (isPerfectSquare(N - i * i))
            return 2;
    }
   
    // If N is sum of 3 perfect squares
    if (legendreFunction(N))
        return 3;
   
    // Otherwise, N is the
    // sum of 4 perfect squares
    return 4;
}
 
// Driver Code
 
    // Given number
    let N = 123;
   
    // Function call
    document.write(minSquares(N));
 
</script>

Output:

Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)

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