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# Program to count number of perfect squares are added up to form a number in C++

Suppose we have a positive number n, we have to find the least number of perfect square numbers whose sum is same as n. So if the number is 10, then the output is 2, as the numbers are 10 = 9 + 1.

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

#include<bits/stdc++.h> using namespace std; #define INF 1e9 class Solution { public: int solve(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.solve(10); }

## Input

10

## Output

2

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