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 10^{2}. Note that 100 can also be written as 5^{2}+ 5^{2}+ 5^{2}+ 5^{2}, 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.

## 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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