Minimum number of squares whose sum equals to given number nn

Any numbers can be represented by the sum of some perfect square numbers. In this problem, we need to find that how many minimum numbers of perfect square terms are needed to represent the given value.

let the value is 94,so 95 = 9² + 3² + 2² + 1². so the answer will be 4

The idea is to start from 1, we move further to get perfect squared numbers. When the value is 1 to 3, they must be formed with only 1s.

Input and Output

Input:
An integer number. Say 63.
Output:
Number of squared terms. Here the answer is 4.
63 =72 + 32 + 22 + 1

Algorithm

minSquareTerms(value)

Input: The given value.

Output: Minimum number of square terms to reach given value.

Begin
   define array sqList of size value + 1
   sqList[0] := 0, sqList[1] := 1, sqList[2] := 2, sqList[3] := 3

   for i in range 4 to n, do
      sqList[i] := i
      for x := 1 to i, do
         temp := x^2
         if temp > i, then
            break the loop
         else sqList[i] := minimum of sqList[i] and (1+sqList[i-temp])
      done
   done
   return sqList[n]
End

Example

#include<bits/stdc++.h>
using namespace std;

int min(int x, int y) {
   return (x < y)? x: y;
}

int minSquareTerms(int n) {
   int *squareList = new int[n+1];

   //for 0 to 3, there are all 1^2 needed to represent

   squareList[0] = 0;
   squareList[1] = 1;
   squareList[2] = 2;
   squareList[3] = 3;

   for (int i = 4; i <= n; i++) {
      squareList[i] = i; //initially store the maximum value as i

      for (int x = 1; x <= i; x++) {
         int temp = x*x;      //find a square term, lower than the number i
         if (temp > i)
            break;
         else squareList[i] = min(squareList[i], 1+squareList[itemp]);
      }
   }
   return squareList[n];
}

int main() {
   int n;
   cout << "Enter a number: "; cin >> n;
   cout << "Minimum Squared Term needed: " << minSquareTerms(n);
   return 0;
}

Output

Enter a number: 63
Minimum Squared Term needed: 4