Create Water_Jug_Problem.js

add the water jug problem solving function

Author: swappy-2003

Diff for: Recursive/Water_Jug_Problem.js

@@ -0,0 +1,34 @@
+function canMeasureWater(jug1Capacity, jug2Capacity, targetAmount) {
+    // Base case: If the target amount is greater than the sum of both jugs, it's not possible.
+    if (targetAmount > jug1Capacity + jug2Capacity) return false;
+
+    // Use BFS to explore possible states.
+    let visited = new Set(); // To keep track of visited states.
+    let queue = [[0, 0]]; // Starting with both jugs empty.
+
+    while (queue.length > 0) {
+        let [jug1, jug2] = queue.shift();
+
+        // If we've reached the target amount in either jug.
+        if (jug1 === targetAmount || jug2 === targetAmount || jug1 + jug2 === targetAmount) {
+            return true;
+        }
+
+        // If this state has been visited, skip it.
+        let state = `${jug1},${jug2}`;
+        if (visited.has(state)) continue;
+        visited.add(state);
+
+        // Add all possible next states to the queue:
+        queue.push([jug1Capacity, jug2]); // Fill Jug 1
+        queue.push([jug1, jug2Capacity]); // Fill Jug 2
+        queue.push([0, jug2]); // Empty Jug 1
+        queue.push([jug1, 0]); // Empty Jug 2
+        queue.push([Math.max(0, jug1 - (jug2Capacity - jug2)), Math.min(jug2Capacity, jug1 + jug2)]); // Pour Jug 1 into Jug 2
+        queue.push([Math.min(jug1Capacity, jug1 + jug2), Math.max(0, jug2 - (jug1Capacity - jug1))]); // Pour Jug 2 into Jug 1
+    }
+
+    // If no solution is found
+    return false;
+}
+