# 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 <= 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.

`// 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;`
`}`

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 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

`// 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;`
`}`

Output:

`3`

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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