0012 problem and solution

L-I/0012 Robot Bounded In Circle/README.md

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+# Problem
+
+- On an infinite plane, a robot initially stands at (0, 0) and faces north. Note that:
+
+The north direction is the positive direction of the y-axis.
+The south direction is the negative direction of the y-axis.
+The east direction is the positive direction of the x-axis.
+The west direction is the negative direction of the x-axis.
+The robot can receive one of three instructions:
+
+"G": go straight 1 unit.
+"L": turn 90 degrees to the left (i.e., anti-clockwise direction).
+"R": turn 90 degrees to the right (i.e., clockwise direction).
+The robot performs the instructions given in order, and repeats them forever.
+
+- Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.
+
+# Solution
+
+```javascript
+var isRobotBounded = function (ins) {
+  let x = 0,
+    y = 0,
+    i = 0,
+    d = [
+      [0, 1],
+      [1, 0],
+      [0, -1],
+      [-1, 0],
+    ];
+
+  for (let j = 0; j < ins.length; ++j) {
+    if (ins.charAt(j) === "R") i = (i + 1) % 4;
+    else if (ins.charAt(j) === "L")
+      i = (i + 3) % 4; // one left turn = 3 right turn
+    else {
+      x += d[i][0];
+      y += d[i][1];
+    }
+  }
+
+  return (x === 0 && y === 0) || i > 0;
+};
+```
+
+## how it works
+
+1. creating the function using variable expression
+2. Function get the instructions as parameter
+3. Initialise the variables x, y are the 2D position, i is the direction and d is the possible solution to move.
+4. Then run the loop using for loop through entire characters to check the cycle move.
+5. if charcter is as R or L only the direction change. Other wise same direction.
+6. finally after the loop finishes, there are conditions to check to give the result.
+7. if last and current position (0,0) , or i > 0
+
+## references
+
+[Leetcode][https://leetcode.com/problems/robot-bounded-in-circle/description/?envType=study-plan-v2&envId=programming-skills]

L-I/0012 Robot Bounded In Circle/Robot Bounded In Circle.js

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+var isRobotBounded = function(ins) {
+    let x = 0,
+        y = 0,
+        i = 0,
+        d = [
+          [0, 1],
+          [1, 0],
+          [0, -1],
+          [-1, 0]
+        ];
+
+      for (let j = 0; j < ins.length; ++j) {
+        if (ins.charAt(j) === 'R') i = (i + 1) % 4;
+        else if (ins.charAt(j) === 'L') i = (i + 3) % 4;
+        else {
+          x += d[i][0];
+          y += d[i][1];
+        }
+      }
+
+      return x === 0 && y === 0 || i > 0;
+};