Perfect Squares in C++


Suppose we have a positive integer n, find the least number of perfect square numbers whose sum is n. So if the number is 13, then the output is 2, as the numbers are 13 = 9 + 4

To solve this, we will follow these steps −

  • create one table for dynamic programming, of length n + 1, and fill it with infinity
  • dp[0] := 0
  • for i := 1, when i*i <= n
    • x = i * i
    • for j := x to n
      • dp[j] := minimum of dp[j] and 1 + dp[j – x]
  • return dp[n]

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
#define INF 1e9
class Solution {
   public:
   int numSquares(int n) {
      vector < int > dp(n+1,INF);
      dp[0] = 0;
      for(int i =1;i*i<=n;i++){
         int x = i*i;
         for(int j = x;j<=n;j++){
            dp[j] = min(dp[j],1+dp[j-x]);
         }
      }
      return dp[n];
   }
};
main(){
   Solution ob;
   cout << (ob.numSquares(147));
}

Input

147

Output

3

Updated on: 04-May-2020

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