new soln

Author: shreymehul

633.JudgeSquareSum.cs

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+// 633. Sum of Square Numbers
+// Given a non-negative integer c, decide whether there're two integers a and b such that a2 + b2 = c.
+
+// Example 1:
+// Input: c = 5
+// Output: true
+// Explanation: 1 * 1 + 2 * 2 = 5
+// Example 2:
+// Input: c = 3
+// Output: false
+
+// Constraints:
+// 0 <= c <= 231 - 1
+
+
+//Approch 1 Fermat's theorem
+public class Solution {
+    public bool JudgeSquareSum(int c) {
+        if (c < 0) return false; // Negative numbers cannot be expressed as the sum of two squares
+        
+        int originalC = c; // Store the original value of c
+        
+        for(int divisor = 2; divisor * divisor <= c; divisor++) {
+            if (c % divisor == 0) {
+                int exponentCount = 0;
+                while (c % divisor == 0) {
+                    exponentCount++;
+                    c /= divisor;
+                }
+                if (divisor % 4 == 3 && exponentCount % 2 != 0) {
+                    return false;
+                }
+            }
+        }
+        
+        // If the remaining c is a prime number of the form 4k+3, it should appear an even number of times
+        // If after factorization, c is still greater than 1 and of the form 4k+3, it means it's a prime factor itself
+        return c % 4 != 3;
+    }
+}
+
+// The code provided is a solution to check if a given number c can be expressed as the sum of two squares. 
+// It implements Fermat's theorem on sums of two squares, which states that a number c can be expressed as the sum of two 
+// squares if every prime factor of the form 4k+3 in its prime factorization appears with an even exponent.
+
+//Approch 2 Two Sum
+public class Solution {
+    public bool JudgeSquareSum(int c) {
+        int left = 0;
+        int right = (int)Math.Sqrt(c);
+        
+        while (left <= right) {
+            long sum = (long)left * left + (long)right * right;
+            if (sum == c) {
+                return true;
+            } else if (sum < c) {
+                left++;
+            } else {
+                right--;
+            }
+        }
+        
+        return false;
+    }
+}