# 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

```

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 ` `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` `    ``// 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 = 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` `        ``// Maximum squares required is` `        ``int` `res = n; ` `        ``// 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` `    ``# Maximum squares required ` `    ``# is n (1 * 1 + 1 * 1 + ..)` `    ``res ``=` `n `   `    ``# 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

 ` ``\$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 :

```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 ` `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;` `    ``dp = 1;` `    ``dp = 2;` `    ``dp = 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;` `    ``}` `  `  `    ``// Driver Code` `    ``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;` `        ``dp = 1;` `        ``dp = 2;` `        ``dp = 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

 ` ``\$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:

```3

```

Thanks to Gaurav Ahirwar for suggesting this solution.

Another Approach:
This problem can also be solved by using graph.Here is the basic idea how it can be done
We will use BFS(Breadth First Search) to find the minimum number of steps from given value of n to 0.
So for every node, we will push its next possible valid path which is not visited yet into a queue and,
and if it reaches the node 0, we will update our answer if it less than the answer.
Below is the implementation of the above approach:

## CPP

 `// CPP program for the above approach ` `#include ` `using` `namespace` `std; `   `// Function to count minimum ` `// squares that sum to n ` `int` `numSquares(``int` `n) ` `{` ` `  `  ``// Creating visited vector ` `  ``// of size n + 1` `  ``vector<``int``> visited(n + 1,0);` `  `  `  ``// Queue of pair to store node ` `  ``// and number of steps` `  ``queue< pair<``int``,``int``> >q;` `  `  `  ``// Initially ans variable is ` `  ``// initialized with inf ` `  ``int` `ans = INT_MAX;` `  `  `  ``// Push starting node with 0` `  ``// 0 indicate current number ` `  ``// of step to reach n` `  ``q.push({n,0});` `  `  `  ``// Mark starting node visited` `  ``visited[n] = 1;` `  ``while``(!q.empty())` `  ``{` `    ``pair<``int``,``int``> p;` `    ``p = q.front();` `    ``q.pop();`   `    ``// If node reaches its destination ` `    ``// 0 update it with answer` `    ``if``(p.first == 0)` `      ``ans=min(ans, p.second);`   `    ``// Loop for all possible path from ` `    ``// 1 to i*i <= current node(p.first)` `    ``for``(``int` `i = 1; i * i <= p.first; i++)` `    ``{` `      `  `      ``// If we are standing at some node ` `      ``// then next node it can jump to will ` `      ``// be current node- ` `      ``// (some square less than or equal n)` `      ``int` `path=(p.first - (i*i));`   `      ``// Check if it is valid and ` `      ``// not visited yet` `      ``if``(path >= 0 && ( !visited[path] ` `                             ``|| path == 0))` `      ``{` `        `  `        ``// Mark visited` `        ``visited[path]=1; ` `        `  `        ``// Push it it Queue` `        ``q.push({path,p.second + 1}); ` `      ``}` `    ``}` `  ``}` `  `  `  ``// Return ans to calling function` `  ``return` `ans; ` `}`   `// Driver program ` `int` `main() ` `{ ` `  ``cout << numSquares(12); ` `  ``return` `0; ` `} `

Output:

```3

```

The time complexity of the above problem is O(n)*sqrt(n) which is better than the Recursive approach
Also it is great to way to understand how BFS(Breadth First Search) work