batCoder95 · batCoder95 · Jul 16, 2016 · Jul 15, 2016 · Jul 15, 2016 · Jul 15, 2016
diff --git a/BalancedParantheses.java b/BalancedParantheses.java
@@ -0,0 +1,37 @@
+/**
+ * Created by Vatsal Gosaliya on 15-Jul-16.
+ *
+ * PROBLEM: Check if the given string contains a balanced pattern of parantheses.
+ */
+
+import java.util.Stack;
+class BalancedParantheses {
+
+    public boolean solution(String a){
+        int size = a.length();
+        int check = 0;
+        Stack s = new Stack();
+
+        for(int i=0;i<size;i++){
+            if(a.charAt(i)!='(' && a.charAt(i)!=')') check = -1;
+            if(a.charAt(i)=='(') s.push(a.charAt(i));
+            else{
+                if(!s.empty()) s.pop();
+                else return false;
+            }
+        }
+        if(check==-1) return false;
+        else{
+            if(s.empty()) return true;
+            else return false;
+        }
+
+    }
+    public static void main(String[] args) {
+        String a = "((()))";
+        boolean result = new BalancedParantheses().solution(a);
+        System.out.println(result);
+
+    }
+
+}
diff --git a/FrogJumps.java b/FrogJumps.java
@@ -0,0 +1,32 @@
+/**
+ * Created by Vatsal on 15-Jul-16.
+ * PROBLEM: Given the current position(x) of a frog and destination(y),
+ *          compute the number of jumps required to reach from x to y,
+ *          if the covers D blocks in a single jump.
+ */
+
+class FrogJumps {
+
+    public int solution(int x, int y, int D){
+        int jumps;
+
+        if(x==y) return 0;
+
+        else{
+            int distance = Math.abs((y-x));
+            if(distance%D==0) jumps = distance/D;
+            else jumps = (distance/D)+1;
+
+            return jumps;
+        }
+    }
+
+    public static void main(String[] args) {
+        FrogJumps t = new FrogJumps();
+        int jumps = t.solution(0, 10, 3);
+        System.out.println("No. of jumps = "+jumps);
+
+
+    }
+
+}
diff --git a/OverlappingRectangles.java b/OverlappingRectangles.java
@@ -0,0 +1,51 @@
+/**
+ * Created by Vatsal Gosaliya on 15-Jul-16.
+ *
+ * PROBLEM: Given are two rectangles defined over the 2-D cartesian coordinate system.
+ *          Each rectangle is expressed using the top-left corner, width and height.
+ *          The aim is to detect whether they overlap, and compute the overlapping
+ *          area, if they do.
+ */
+
+class OverlappingRectangles {
+	private double x,y,width,height;
+
+	OverlappingRectangles(double x, double y, double width, double height){
+		this.x = x;
+		this.y = y;
+		this.width = width;
+		this.height = height;
+	}
+
+	OverlappingRectangles intersection(OverlappingRectangles rect2) {
+        double newX = Math.max(this.x, rect2.x);
+        double newY = Math.min(this.y, rect2.y);
+        //System.out.println("newX = "+newX);
+        //System.out.println("newY = "+newY);
+
+        double newWidth = Math.min(this.x + this.width, rect2.x + rect2.width) - newX;
+        double newHeight = newY - Math.max(this.y - this.height, rect2.y - rect2.height);
+        //System.out.println("newWidth = "+newWidth);
+        //System.out.println("newHeight = "+newHeight);
+
+        if (newWidth <= 0d || newHeight <= 0d){
+            System.out.print("No intersection.");
+            return null;
+        }
+
+        return new OverlappingRectangles(newX, newY, newWidth, newHeight);
+    }
+
+
+	public double getArea(){
+		return this.width*this.height;
+	}
+
+	public static void main(String args[]){
+		OverlappingRectangles r1 = new OverlappingRectangles(2,8,3,4);
+		OverlappingRectangles r2 = new OverlappingRectangles(1,7,6,2);
+		OverlappingRectangles r = r1.intersection(r2);
+		double areaOfIntersection = r == null ? 0 : r.getArea();
+		System.out.print("Area of intersection : "+areaOfIntersection);
+	}
+}