Create Minimum Number of Squares

Author: akshayavb99

Course 2 - Data Structures in JAVA/Lecture 18 - Dynamic Programming I/Minimum Number of Squares

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+/*
+Given an integer N, find and return the count of minimum numbers required to represent N as a sum of squares.
+That is, if N is 4, then we can represent it as : {1^2 + 1^2 + 1^2 + 1^2} and {2^2}. 
+The output will be 1, as 1 is the minimum count of numbers required to represent N as sum of squares.
+
+Input format :
+The first and the only line of input contains an integer value, 'N'.
+
+Output format :
+Print the minimum count of numbers required.
+
+Constraints :
+0 <= n <= 10 ^ 4
+Time Limit: 1 sec
+
+Sample Input 1 :
+12
+Sample Output 1 :
+3
+Explanation of Sample Output 1 :
+12 can be represented as : 
+A) (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1)
+
+B) (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1) + (1^1)  + (2 ^ 2)
+
+C) (1^1) + (1^1) + (1^1) + (1^1) + (2 ^ 2) + (2 ^ 2)
+
+D) (2 ^ 2) + (2 ^ 2) + (2 ^ 2)
+
+As we can see, the output should be 3.
+
+Sample Input 2 :
+9
+Sample Output 2 :
+1
+*/
+public class Solution {
+
+	public static int minCount(int n) {
+		//Your code goes here
+        int dp[] = new int[n+1];
+        for (int i=0;i<n+1;i++)
+            dp[i]=-1;
+        
+        return minCountHelper(n,dp);
+	}
+    
+    private static int minCountHelper(int n, int[] dp)
+    {
+        if (n==0)
+            return 0;
+        
+        int minVal = Integer.MAX_VALUE;
+        for (int i=1;i*i<=n;i++)
+        {
+            if (dp[n-(i*i)]==-1)
+                dp[n-(i*i)]=minCountHelper(n-(i*i),dp);
+            int currVal = dp[n-(i*i)];
+            if (currVal<minVal)
+                minVal=currVal;
+        }
+        
+        return minVal+1;
+    }
+
+}