Added the solution of 0279 - Perfect Squares

Author: Abhinav200207

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];
+    }
+};