square root

leetcode/Sqrt(x).cpp

@@ -0,0 +1,57 @@
+
+#include <iostream>
+
+using namespace std;
+int floorSqrt(int x)
+{
+    // Base cases
+    if (x == 0 || x == 1)
+        return x;
+
+    // Do Binary Search for floor(sqrt(x))
+    int start = 1, end = x / 2, ans;
+    while (start <= end)
+    {
+        int mid = (start + end) / 2;
+
+        // If x is a perfect square
+        int sqr = mid * mid;
+        if (sqr == x)
+            return mid;
+
+        // Since we need floor, we update answer when
+        // mid*mid is smaller than x, and move closer to
+        // sqrt(x)
+
+        /*
+           if(mid*mid<=x)
+                   {
+                           start = mid+1;
+                           ans = mid;
+                   }
+            Here basically if we multiply mid with itself so
+           there will be integer overflow which will throw
+           tle for larger input so to overcome this
+           situation we can use long or we can just divide
+            the number by mid which is same as checking
+           mid*mid < x
+
+           */
+        if (sqr <= x)
+        {
+            start = mid + 1;
+            ans = mid;
+        }
+        else // If mid*mid is greater than x
+            end = mid - 1;
+    }
+    return ans;
+}
+
+// Driver program
+int main()
+{
+    int x = 11;
+    cout << floorSqrt(x) << endl;
+    return 0;
+}