Added the solution of 0279 - Perfect Squares

cpp/0279-Perfect-Squares.cpp

@@ -0,0 +1,70 @@
+#include <iostream>
+#include <vector>
+using namespace std;
+
+/*
+ problem link: https://leetcode.com/problems/perfect-squares/description/
+ Given an integer n, return the least number of perfect square numbers that sum to n.
+
+ A perfect square is an integer that is the square of an integer; in other words, it is the
+ product of some integer with itself. For example, 1, 4, 9, and 16 are perfect squares while 3
+ and 11 are not.
+
+ Example 1:
+
+ Input: n = 12
+ Output: 3
+ Explanation: 12 = 4 + 4 + 4.
+
+ Example 2:
+
+ Input: n = 13
+ Output: 2
+ Explanation: 13 = 4 + 9.
+*/
+
+class Solution
+{
+public:
+ int numSquares(int n)
+ {
+ // Create a vector to store all perfect squares
+ vector<int> vectorOfPerfectSquare;
+
+ // Loop through the numbers from 1 to the square root of n
+ for (int i = 1; i * i <= n; i++)
+ {
+ // If the square of i is a perfect square, push it to the vector v
+ vectorOfPerfectSquare.push_back(i * i);
+ }
+
+ // If n is equal to 1, return 1
+ if (n == 1)
+ {
+ return 1;
+ }
+
+ // Define a variable Max equal to n + 1
+ int Max = n + 1;
+ // Create a vector dp with length equal to n + 1 and fill it with Max
+ vector<int> dp(n + 1, Max);
+ // Initialize the first element of dp as 0
+ dp[0] = 0;
+ // Loop through n from 1 to n
+ for (int i = 1; i <= n; i++)
+ {
+ // Loop through the vectorOfPerfectSquare
+ for (int j = 0; j < vectorOfPerfectSquare.size(); j++)
+ {
+ // If the value of i is greater than or equal to the current coin value
+ if (i - vectorOfPerfectSquare[j] >= 0)
+ {
+ // Update the value of dp[i] to the minimum of dp[i] and dp[i-vectorOfPerfectSquare[j]] + 1
+ dp[i] = min(dp[i], dp[i - vectorOfPerfectSquare[j]] + 1);
+ }
+ }
+ }
+ // Return the value of dp[n] if it is less than or equal to n, else return -1
+ return dp[n] > n ? -1 : dp[n];
+ }
+};