feat: Add leetcode problem 441

leetcode/src/441.c

@@ -0,0 +1,35 @@
+// Given n coins, find out how many rows can be filled with them if each row has
+// one more coin than the last (starting from 1 coin. Last row may or may not be
+// full)
+int arrangeCoins(int n)
+{
+    // We need to find the point where n(n + 1) / 2 is just bigger than input -
+    // that minus 1 is our answer, since the sum will be that number. So we will
+    // use binarySearch for it
+    int low = 1, high = 10e5;  // 10e5 square is bigger than constraints
+    while (low <= high)
+    {
+        int mid = (low + high) / 2;
+        long long int guess = mid * (mid + 1L) / 2;
+        long long int prevGuess = (mid - 1L) * mid / 2;
+        if (guess == n)
+        {  // Edge case - if it's equal then the last row is full and we have
+           // reached our answer
+            return mid;
+        }
+        else if (guess > n && prevGuess < n)
+        {
+            return mid - 1;
+        }
+        else if (guess > n)
+        {
+            high = mid - 1;
+        }
+        else
+        {
+            low = mid + 1;
+        }
+    }
+    return -1;  // Return -1 to satisfy the compiler although early exit via
+                // `return` is guaranteed by problem constraints
+}