Water Jug Problem

C++/Leetcode/waterjug.cpp

@@ -0,0 +1,80 @@
+#include <iostream>
+#include <algorithm>
+#include <cstdlib>
+using namespace std;
+
+// Utility function to print an array
+void printArray(int arr[], int n, string msg)
+{
+    cout << msg << " : ";
+    for (int i = 0; i < n; i++) {
+        cout << arr[i] << " ";
+    }
+    cout << endl;
+}
+
+// This function is similar to the partition routine of the Quicksort algorithm.
+// It partitions the specified array `A[low, high]` around `pivot`
+// and returns the partition index.
+int partition(int A[], int low, int high, int pivot)
+{
+    int pIndex = low;
+    for (int j = low; j < high; j++)
+    {
+        if (A[j] < pivot)
+        {
+            swap(A[pIndex], A[j]);
+            pIndex++;
+        }
+        else if (A[j] == pivot)
+        {
+            swap(A[j], A[high]);
+            j--;
+        }
+    }
+
+    swap(A[pIndex], A[high]);
+    return pIndex;
+}
+
+// Group the jugs into pairs of red and blue jugs that hold the same amount of water.
+// This function is similar to the Quicksort-like routine.
+void findMatchingPairs(int red[], int blue[], int low, int high)
+{
+    // base case: if there is only one red jug and one blue jug,
+    // they must be of the same size
+    if (low >= high) {
+        return;
+    }
+
+    // Randomly select a jug from either red or blue jugs
+    int r = rand() % (high - low + 1) + low;
+    int chosenJug = red[r];     // or use `blue[r]`
+
+    // Partition the red jugs around the chosen jug
+    int pivot = partition(red, low, high, chosenJug);
+
+    // Now partition the blue jugs around the chosen jug
+    partition(blue, low, high, chosenJug);
+
+    // `pivot` now points to an index of the chosen jug in red/blue jugs.
+
+    // Recur, once the red and blue jugs are divided into two groups
+    // around the chosen jug
+    findMatchingPairs(red, blue, low, pivot - 1);
+    findMatchingPairs(red, blue, pivot + 1, high);
+}
+
+int main()
+{
+    int red[] = { 6, 3, 2, 8, 7 };
+    int blue[] = { 8, 6, 7, 2, 3 };
+
+    int n = sizeof(red) / sizeof(red[0]);
+    findMatchingPairs(red, blue, 0, n - 1);
+
+    printArray(red, n, "Red jugs ");
+    printArray(blue, n, "Blue jugs");
+
+    return 0;
+}