# Minimum number of squares whose sum equals to given number N | set 2

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, the task is to represent N as the sum of minimum square numbers.

Examples:

Input : 10
Output : 1 + 9
These are all possible ways
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1 + 1 + 4
1 + 1 + 4 + 4
1 + 9
Choose one with minimum numbers

Input : 25
Output : 25

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : This is a typical application of dynamic programming. 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 the same subproblems are called again, this problem has Overlapping Subproblems 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 the same subproblems can be avoided by constructing a temporary array table[][] in a bottom-up manner.

Below is the implementation of the above approach:

## C++

 `// C++ program to represent N as the ` `// sum of minimum square numbers. ` `#include ` `using` `namespace` `std; ` ` `  `// Function for finding ` `// minimum square numbers ` `vector<``int``> minSqrNum(``int` `n) ` `{ ` `    ``// A[i] of array arr store ` `    ``// minimum count of ` `    ``// square number to get i ` `    ``int` `arr[n + 1], k; ` ` `  `    ``// sqrNum[i] store last ` `    ``// square number to get i ` `    ``int` `sqrNum[n + 1]; ` `    ``vector<``int``> v; ` ` `  `    ``// Initialize ` `    ``arr = 0; ` `    ``sqrNum = 0; ` ` `  `    ``// Find minimun count of ` `    ``// square number for ` `    ``// all value 1 to n ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``// In worst case it will ` `        ``// be arr[i-1]+1 we use all ` `        ``// combination of a[i-1] and add 1 ` `        ``arr[i] = arr[i - 1] + 1; ` `        ``sqrNum[i] = 1; ` ` `  `        ``k = 1; ` `        ``// Check for all square ` `        ``// number less or equal to i ` `        ``while` `(k * k <= i) { ` `            ``// if it gives less ` `            ``// count then update it ` `            ``if` `(arr[i] > arr[i - k * k] + 1) { ` `                ``arr[i] = arr[i - k * k] + 1; ` `                ``sqrNum[i] = k * k; ` `            ``} ` ` `  `            ``k++; ` `        ``} ` `    ``} ` ` `  `    ``// Vector v stores optimum ` `    ``// square number whose sum give N ` `    ``while` `(n > 0) { ` `        ``v.push_back(sqrNum[n]); ` `        ``n -= sqrNum[n]; ` `    ``} ` `    ``return` `v; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 10; ` ` `  `    ``vector<``int``> v; ` ` `  `    ``// Calling funcion ` `    ``v = minSqrNum(n); ` ` `  `    ``// Printing vector ` `    ``for` `(``auto` `i = v.begin(); ` `         ``i != v.end(); i++) { ` `        ``cout << *i; ` `        ``if` `(i + 1 != v.end()) ` `            ``cout << ``" + "``; ` `    ``} ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program to represent N as the ` `# sum of minimum square numbers. ` ` `  `# Function for finding ` `# minimum square numbers ` `def` `minSqrNum(n): ` ` `  `    ``# arr[i] of array arr store ` `    ``# minimum count of ` `    ``# square number to get i ` `    ``arr ``=` `[``0``] ``*` `(n ``+` `1``) ` `     `  `    ``# sqrNum[i] store last ` `    ``# square number to get i ` `    ``sqrNum ``=` `[``0``] ``*` `(n ``+` `1``) ` `    ``v ``=` `[] ` ` `  `    ``# Find minimun count of ` `    ``# square number for ` `    ``# all value 1 to n ` `    ``for` `i ``in` `range``(n ``+` `1``): ` `         `  `        ``# In worst case it will ` `        ``# be arr[i-1]+1 we use all ` `        ``# combination of a[i-1] and add 1 ` `        ``arr[i] ``=` `arr[i ``-` `1``] ``+` `1` `        ``sqrNum[i] ``=` `1` ` `  `        ``k ``=` `1``; ` `         `  `        ``# Check for all square ` `        ``# number less or equal to i ` `        ``while` `(k ``*` `k <``=` `i): ` `             `  `            ``# If it gives less ` `            ``# count then update it ` `            ``if` `(arr[i] > arr[i ``-` `k ``*` `k] ``+` `1``): ` `                ``arr[i] ``=` `arr[i ``-` `k ``*` `k] ``+` `1` `                ``sqrNum[i] ``=` `k ``*` `k ` ` `  `            ``k ``+``=` `1` ` `  `    ``# v stores optimum ` `    ``# square number whose sum give N ` `    ``while` `(n > ``0``): ` `        ``v.append(sqrNum[n]) ` `        ``n ``-``=` `sqrNum[n]; ` `         `  `    ``return` `v ` ` `  `# Driver code ` `n ``=` `10` ` `  `# Calling funcion ` `v ``=` `minSqrNum(n) ` ` `  `# Printing vector ` `for` `i ``in` `range``(``len``(v)): ` `    ``print``(v[i], end ``=` `"") ` `     `  `    ``if` `(i < ``len``(v) ``-` `1``): ` `        ``print``(``" + "``, end ``=` `"") ` `         `  `# This article is contributed by Apurvaraj `

Output:

```1 + 9
```

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