Minimum number of squares whose sum equals to given number n

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 X.

Examples :

Input:  n = 100
Output: 1
100 can be written as 10². Note that 100 can also be 
written as 5² + 5² + 5² + 5², but this
representation requires 4 squares.

Input:  n = 6
Output: 3

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

The idea is simple, we start from 1 and go till a number whose square is smaller than or equals to n. For every number x, we recur for n-x. Below is the recursive formula.

If n = 1 and x*x <= n

Below is a simple recursive solution based on above recursive formula.

C++

// A naive recursive C++ program to find minimum 
// number of squares whose sum is equal to a given number 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns count of minimum squares that sum to n 
int getMinSquares(unsigned int n) 
{ 
    // base cases 
    // if n is perfect square then Minimum squares required is 1 (144 = 12^2) 
    if (sqrt(n) - floor(sqrt(n)) == 0) 
        return 1; 
    if (n <= 3) 
        return n; 
  
    // getMinSquares rest of the table using recursive 
    // formula 
    int res = n; // Maximum squares required is n (1*1 + 1*1 + ..) 
  
    // Go through all smaller numbers 
    // to recursively find minimum 
    for (int x = 1; x <= n; x++) { 
        int temp = x * x; 
        if (temp > n) 
            break; 
        else
            res = min(res, 1 + getMinSquares(n - temp)); 
    } 
    return res; 
} 
  
// Driver program 
int main() 
{ 
    cout << getMinSquares(6); 
    return 0; 
} 

Java

// A naive recursive JAVA program to find minimum 
// number of squares whose sum is equal to a given number 
class squares { 
    // Returns count of minimum squares that sum to n 
    static int getMinSquares(int n) 
    { 
        // base cases 
        if (n <= 3) 
            return n; 
  
        // getMinSquares rest of the table using recursive 
        // formula 
        int res = n; // Maximum squares required is 
        // n (1*1 + 1*1 + ..) 
  
        // Go through all smaller numbers 
        // to recursively find minimum 
        for (int x = 1; x <= n; x++) { 
            int temp = x * x; 
            if (temp > n) 
                break; 
            else
                res = Math.min(res, 1 + getMinSquares(n - temp)); 
        } 
        return res; 
    } 
    public static void main(String args[]) 
    { 
        System.out.println(getMinSquares(6)); 
    } 
} 
/* This code is contributed by Rajat Mishra */

Python

# A naive recursive Python program to 
# find minimum number of squares whose 
# sum is equal to a given number 
  
# Returns count of minimum squares  
# that sum to n 
def getMinSquares(n): 
  
    # base cases 
    if n <= 3: 
        return n; 
  
    # getMinSquares rest of the table  
    # using recursive formula 
    res = n # Maximum squares required  
            # is n (1 * 1 + 1 * 1 + ..) 
  
    # Go through all smaller numbers 
    # to recursively find minimum 
    for x in range(1, n + 1): 
        temp = x * x; 
        if temp > n: 
            break
        else: 
            res = min(res, 1 + getMinSquares(n  
                                  - temp)) 
      
    return res; 
  
# Driver program  
print(getMinSquares(6)) 
  
# This code is contributed by nuclode 

C#

// A naive recursive C# program 
// to find minimum number of 
// squares whose sum is equal 
// to a given number 
using System; 
  
class GFG { 
  
    // Returns count of minimum 
    // squares that sum to n 
    static int getMinSquares(int n) 
    { 
  
        // base cases 
        if (n <= 3) 
            return n; 
  
        // getMinSquares rest of the 
        // table using recursive 
        // formula 
  
        // Maximum squares required is 
        // n (1*1 + 1*1 + ..) 
        int res = n; 
  
        // Go through all smaller numbers 
        // to recursively find minimum 
        for (int x = 1; x <= n; x++) { 
            int temp = x * x; 
            if (temp > n) 
                break; 
            else
                res = Math.Min(res, 1 + getMinSquares(n - temp)); 
        } 
        return res; 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        Console.Write(getMinSquares(6)); 
    } 
} 
  
// This code is contributed by nitin mittal 

PHP

<?php 
// A naive recursive PHP program 
// to find minimum number of  
// squares whose sum is equal  
// to a given number 
  
// Returns count of minimum  
// squares that sum to n 
function getMinSquares($n) 
{ 
    // base cases 
    if ($n <= 3) 
        return $n; 
  
    // getMinSquares rest of the  
    // table using recursive formula 
      
    // Maximum squares required 
    // is n (1*1 + 1*1 + ..) 
    $res = $n;  
  
    // Go through all smaller numbers 
    // to recursively find minimum 
    for ($x = 1; $x <= $n; $x++) 
    { 
        $temp = $x * $x; 
        if ($temp > $n) 
            break; 
        else
            $res = min($res, 1 +  
                       getMinSquares($n -  
                                     $temp)); 
    } 
    return $res; 
} 
  
// Driver Code 
echo getMinSquares(6); 
  
// This code is contributed 
// by nitin mittal. 
?> 

Output :

The time complexity of above solution is exponential. If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. For example, when we start from n = 6, we can reach 4 by subtracting one 2 times and by subtracting 2 one times. So the subproblem for 4 is called twice.
Since same suproblems are called again, this problem has Overlapping Subprolems property. So min square sum problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array table[][] in bottom up manner. Below is Dynamic Programming based solution

C++

// A dynamic programming based C++ program to find minimum 
// number of squares whose sum is equal to a given number 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns count of minimum squares that sum to n 
int getMinSquares(int n) 
{ 
    // Create a dynamic programming table 
    // to store sq 
    int* dp = new int[n + 1]; 
  
    // getMinSquares table for base case entries 
    dp[0] = 0; 
    dp[1] = 1; 
    dp[2] = 2; 
    dp[3] = 3; 
  
    // getMinSquares rest of the table using recursive 
    // formula 
    for (int i = 4; i <= n; i++) { 
        // max value is i as i can always be represented 
        // as 1*1 + 1*1 + ... 
        dp[i] = i; 
  
        // Go through all smaller numbers to 
        // to recursively find minimum 
        for (int x = 1; x <= ceil(sqrt(i)); x++) { 
            int temp = x * x; 
            if (temp > i) 
                break; 
            else
                dp[i] = min(dp[i], 1 + dp[i - temp]); 
        } 
    } 
  
    // Store result and free dp[] 
    int res = dp[n]; 
    delete[] dp; 
  
    return res; 
} 
  
// Driver program 
int main() 
{ 
    cout << getMinSquares(6); 
    return 0; 
} 

Java

// A dynamic programming based JAVA program to find minimum 
// number of squares whose sum is equal to a given number 
class squares { 
    // Returns count of minimum squares that sum to n 
    static int getMinSquares(int n) 
    { 
  
        // We need to add a check here for n. If user enters 0 or 1 or 2 
        // the below array creation will go OutOfBounds. 
        if (n <= 3) 
            return n; 
  
        // Create a dynamic programming table 
        // to store sq 
        int dp[] = new int[n + 1]; 
  
        // getMinSquares table for base case entries 
        dp[0] = 0; 
        dp[1] = 1; 
        dp[2] = 2; 
        dp[3] = 3; 
  
        // getMinSquares rest of the table using recursive 
        // formula 
        for (int i = 4; i <= n; i++) { 
            // max value is i as i can always be represented 
            // as 1*1 + 1*1 + ... 
            dp[i] = i; 
  
            // Go through all smaller numbers to 
            // to recursively find minimum 
            for (int x = 1; x <= Math.ceil(Math.sqrt(i)); x++) { 
                int temp = x * x; 
                if (temp > i) 
                    break; 
                else
                    dp[i] = Math.min(dp[i], 1 + dp[i - temp]); 
            } 
        } 
  
        // Store result and free dp[] 
        int res = dp[n]; 
  
        return res; 
    } 
    public static void main(String args[]) 
    { 
        System.out.println(getMinSquares(6)); 
    } 
} /* This code is contributed by Rajat Mishra */

Python

# A dynamic programming based Python 
# program to find minimum number of 
# squares whose sum is equal to a  
# given number 
from math import ceil, sqrt 
  
# Returns count of minimum squares  
# that sum to n 
def getMinSquares(n): 
  
    # Create a dynamic programming table 
    # to store sq and getMinSquares table 
    # for base case entries 
    dp = [0, 1, 2, 3] 
  
    # getMinSquares rest of the table  
    # using recursive formula 
    for i in range(4, n + 1): 
          
        # max value is i as i can always  
        # be represented as 1 * 1 + 1 * 1 + ... 
        dp.append(i) 
  
        # Go through all smaller numbers  
        # to recursively find minimum 
        for x in range(1, int(ceil(sqrt(i))) + 1): 
            temp = x * x; 
            if temp > i: 
                break
            else: 
                dp[i] = min(dp[i], 1 + dp[i-temp]) 
  
    # Store result 
    return dp[n] 
  
# Driver program 
print(getMinSquares(6)) 
  
# This code is contributed by nuclode 

C#

// A dynamic programming based 
// C# program to find minimum 
// number of squares whose sum 
// is equal to a given number 
using System; 
  
class squares { 
    // Returns count of minimum 
    // squares that sum to n 
    static int getMinSquares(int n) 
    { 
  
        // We need to add a check here for n. If user enters 0 or 1 or 2 
        // the below array creation will go OutOfBounds. 
  
        if (n <= 3) 
            return n; 
  
        // Create a dynamic programming 
        // table to store sq 
        int[] dp = new int[n + 1]; 
  
        // getMinSquares table for base 
        // case entries 
        dp[0] = 0; 
        dp[1] = 1; 
        dp[2] = 2; 
        dp[3] = 3; 
  
        // getMinSquares for rest of the 
        // table using recursive formula 
        for (int i = 4; i <= n; i++) { 
            // max value is i as i can 
            // always be represented 
            // as 1 * 1 + 1 * 1 + ... 
            dp[i] = i; 
  
            // Go through all smaller numbers to 
            // to recursively find minimum 
            for (int x = 1; x <= Math.Ceiling(Math.Sqrt(i)); x++) { 
                int temp = x * x; 
                if (temp > i) 
                    break; 
                else
                    dp[i] = Math.Min(dp[i], 1 + dp[i - temp]); 
            } 
        } 
  
        // Store result and free dp[] 
        int res = dp[n]; 
  
        return res; 
    } 
  
    // Driver Code 
    public static void Main(String[] args) 
    { 
        Console.Write(getMinSquares(6)); 
    } 
} 
  
// This code is contributed by Nitin Mittal. 

PHP

<?php 
// A dynamic programming based  
// PHP program to find minimum 
// number of squares whose sum 
// is equal to a given number 
  
// Returns count of minimum 
// squares that sum to n 
function getMinSquares($n) 
{ 
    // Create a dynamic programming 
    // table to store sq 
    $dp; 
  
    // getMinSquares table for 
    // base case entries 
    $dp[0] = 0; 
    $dp[1] = 1; 
    $dp[2] = 2; 
    $dp[3] = 3; 
  
    // getMinSquares rest of the  
    // table using recursive formula 
    for ($i = 4; $i <= $n; $i++) 
    { 
        // max value is i as i can  
        // always be represented 
        // as 1*1 + 1*1 + ... 
        $dp[$i] = $i; 
  
        // Go through all smaller  
        // numbers to recursively  
        // find minimum 
for ($x = 1; $x <= ceil(sqrt($i)); $x++)  
        { 
            $temp = $x * $x; 
            if ($temp > $i) 
                break; 
            else $dp[$i] = min($dp[$i],  
                              (1 + $dp[$i - $temp])); 
        } 
    } 
  
    // Store result  
    // and free dp[] 
    $res = $dp[$n]; 
  
// delete $dp; 
  
    return $res; 
} 
  
// Driver Code 
echo getMinSquares(6); 
      
// This code is contributed 
// by shiv_bhakt.  
?> 

Output:

Thanks to Gaurav Ahirwar for suggesting this solution.

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

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