# 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 <= 3, then return n
Else
minSquares(n) = min {1 + minSquares(n - x*x)}
where x >= 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 <= 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
```

Output:
`3`

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 <= 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)
{
// 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 <= 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

# 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, 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
```

Output:
`3`

Thanks to Gaurav Ahirwar for suggesting this solution in a comment here.

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