Hello there! In this unit, we're offering an engaging coding lesson that highlights the performance efficiencies offered by the utilization of HashMap in Java. We'll address a list-based problem that requires us to make an optimal choice to minimize the size of our list. Excited? So am I! Let's get started.

In this unit's task, we'll manipulate a list of integers. You are required to construct a Java function titled minimalMaxBlock(). This function should accept a list as an input and compute an intriguing property related to contiguous blocks within the list.

More specifically, you must select a particular integer, k, from the list. Once you've selected k, the function should remove all occurrences of k from the list, thereby splitting it into several contiguous blocks or remaining sub-lists. A unique feature of k is that it is chosen such that the maximum length among these blocks is minimized.

For instance, consider the list {1, 2, 2, 3, 1, 4, 4, 4, 1, 2, 5}. If we eliminate all instances of 2 (our k), the remaining blocks would be {1}, {3, 1, 4, 4, 4, 1}, {5}, with the longest containing 6 elements. Now, if we instead remove all instances of 1, the new remaining blocks would be {2, 2, 3}, {4, 4, 4}, {2, 5}, the longest of which contains 3 elements. As such, the function should return 1 in this case, as it leads to a minimally maximal block length.

An initial way to approach this problem can be through a brute force method. Each possible value in the list could be tested in turn by removing it from the list and then checking the resulting sub-list sizes. This approach entails iteratively stepping through the list for each possible value in the list.

Java
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1import java.util.*;
2
3public class MinimalMaxBlock {
4
5    public static int minimalMaxBlockBruteforce(List<Integer> list) {
6        int minMaxBlockSize = Integer.MAX_VALUE;
7        int minNum = -1;
8
9        Set<Integer> uniqueElements = new HashSet<>(list);
10
11        for (int num : uniqueElements) {  // Avoid duplicates.
12            List<Integer> indices = new ArrayList<>();
13            for (int i = 0; i < list.size(); ++i) {
14                if (list.get(i) == num) {
15                    indices.add(i);
16                }
17            }
18            indices.add(0, -1);
19            indices.add(list.size());
20
21            int maxBlockSize = 0;
22            for (int i = 1; i < indices.size(); ++i) {
23                maxBlockSize = Math.max(maxBlockSize, indices.get(i) - indices.get(i - 1) - 1);
24            }
25
26            if (maxBlockSize < minMaxBlockSize) {
27                minMaxBlockSize = maxBlockSize;
28                minNum = num;
29            }
30        }
31
32        return minNum;
33    }
34}

This method has a time complexity of $O(n^2)$, as it involves two nested loops: the outer loop cycling through each potential k value and the inner loop sweeping through the list for each of these k values.

However, this approach becomes increasingly inefficient as the size n of the list grows, due to its quadratic time complexity. For larger lists or multiple invocations, the computation time can noticeably increase, demonstrating the need for a more efficient solution.

Before we delve into the steps of the optimized HashMap solution, let's understand why the proposed method is superior in terms of time and space complexity when compared to the brute force approach.

1. Time Complexity: The HashMap solution only requires a single traversal of the list. This results in a linear time complexity, $O(n)$, where n is the number of elements in the list. This is significantly more efficient than the $O(n^2)$ complexity of the brute force approach, which needs to traverse the list for every distinct number.

2. Space Complexity: The brute-force approach maintains a set of unique elements and an indices list for each element. Although the indices list can have up to n elements across some particular element, the total space required for all indices combined is still n, leading to an overall space complexity of $O(n)$. Similarly, the HashMap solution maintains two maps for storing the last occurrence and maximum block size for each number. In a worst-case scenario, every element in the list is unique, leading to an $O(n)$ space complexity, where n is the number of elements in the list.

Now let's jump into the step-by-step implementation of the solution.

To find the number that, when removed from the list, would split the list into several contiguous blocks such that the length of the longest block is minimized, we need to track every position of the same elements, the block size when removing each of the encountered elements, and store the maximum of these blocks.

But, in this case, we don't need to store all positions. We only need the last occurrence position since the blocks we are interested in are between two adjacent same elements, the beginning of the list and the first occurrence of an element, and between the last occurrence of an element and the end of the list. For the first two cases, we will keep the maximum block size for each element and update it whenever we get a bigger one, and for the last case, we will process it separately after the list traversal.

To do this, we need to:

• Initialize two HashMaps. The lastOccurrence map will store the last occurrence index of each number, while maxBlockSizes will map each number to the maximum size of the blocks formed when the number is removed.

• Traverse the list from left to right, and for each number:

  • If it's the first time we encounter the number, we regard the block from the start of the list up to its current position as a block formed when this number is removed, and store the size of this block in maxBlockSizes for this number.
  • If the number has appeared before, we calculate the size of the block it forms (excluding the number itself) by subtracting the last occurrence index from the current index and subtracting 1. If the size is larger than the current maximum stored in maxBlockSizes for this number, we update it.
  • Store the current index as the last occurrence of the number.

• After finishing the list traversal, we need to calculate the size of the "tail block" (i.e., the block between the last occurrence of a number and the end of the list) for each number, and update its maximum block size in maxBlockSizes if necessary.

• Find the number that gives the smallest maximum block size and return it as the result.

First, we initialize two HashMaps. The lastOccurrence map stores the last occurrence index of each number, while maxBlockSizes maps each number to the maximum size of the blocks formed when the number is removed.

Java
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1import java.util.*;
2
3public class MinimalMaxBlock {
4
5    public static int minimalMaxBlock(List<Integer> list) {
6        HashMap<Integer, Integer> lastOccurrence = new HashMap<>();
7        HashMap<Integer, Integer> maxBlockSizes = new HashMap<>();

Next, we iterate over the list. For each number:

• If it's the first time the number is encountered, we regard the block from the start of the list up to its current position as a block formed when this number is removed, and store the size of this block in maxBlockSizes for this number.
• If it has appeared before, we calculate the size of the block it forms by subtracting the last occurrence index from the current index, and subtracting 1 (since block length doesn't include the number itself). We update maxBlockSizes for this number if necessary.
• Store the current index as the last occurrence of this number.

Java
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1        for (int i = 0; i < list.size(); ++i) {
2            int num = list.get(i);
3            if (!lastOccurrence.containsKey(num)) {
4                maxBlockSizes.put(num, i);
5            } else {
6                int blockSize = i - lastOccurrence.get(num) - 1;
7                maxBlockSizes.put(num, Math.max(maxBlockSizes.get(num), blockSize));
8            }
9            lastOccurrence.put(num, i);
10        }

Tail blocks are defined as blocks formed from the last occurrence of a number to the end of the list. For each number, we calculate the size of its tail block and update maxBlockSizes if necessary.

Java
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1        for (Map.Entry<Integer, Integer> entry : lastOccurrence.entrySet()) {
2            int num = entry.getKey();
3            int pos = entry.getValue();
4            int blockSize = list.size() - pos - 1;
5            maxBlockSizes.put(num, Math.max(maxBlockSizes.get(num), blockSize));
6        }

Finally, we find the number associated with the smallest maximum block size in maxBlockSizes, and return it.

Java
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1        int minNum = -1;
2        int minBlockSize = Integer.MAX_VALUE;
3        for (Map.Entry<Integer, Integer> entry : maxBlockSizes.entrySet()) {
4            if (entry.getValue() < minBlockSize) {
5                minBlockSize = entry.getValue();
6                minNum = entry.getKey();
7            }
8        }
9
10        return minNum;
11    }
12}

Putting it all together, here's the full code:

Java
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1import java.util.*;
2
3public class MinimalMaxBlock {
4
5    public static int minimalMaxBlockBruteforce(List<Integer> list) {
6        int minMaxBlockSize = Integer.MAX_VALUE;
7        int minNum = -1;
8
9        Set<Integer> uniqueElements = new HashSet<>(list);
10
11        for (int num : uniqueElements) {
12            List<Integer> indices = new ArrayList<>();
13            for (int i = 0; i < list.size(); ++i) {
14                if (list.get(i) == num) {
15                    indices.add(i);
16                }
17            }
18            indices.add(0, -1);
19            indices.add(list.size());
20
21            int maxBlockSize = 0;
22            for (int i = 1; i < indices.size(); ++i) {
23                maxBlockSize = Math.max(maxBlockSize, indices.get(i) - indices.get(i - 1) - 1);
24            }
25
26            if (maxBlockSize < minMaxBlockSize) {
27                minMaxBlockSize = maxBlockSize;
28                minNum = num;
29            }
30        }
31
32        return minNum;
33    }
34
35    public static int minimalMaxBlock(List<Integer> list) {
36        HashMap<Integer, Integer> lastOccurrence = new HashMap<>();
37        HashMap<Integer, Integer> maxBlockSizes = new HashMap<>();
38
39        for (int i = 0; i < list.size(); ++i) {
40            int num = list.get(i);
41            if (!lastOccurrence.containsKey(num)) {
42                maxBlockSizes.put(num, i);
43            } else {
44                int blockSize = i - lastOccurrence.get(num) - 1;
45                maxBlockSizes.put(num, Math.max(maxBlockSizes.get(num), blockSize));
46            }
47            lastOccurrence.put(num, i);
48        }
49
50        for (Map.Entry<Integer, Integer> entry : lastOccurrence.entrySet()) {
51            int num = entry.getKey();
52            int pos = entry.getValue();
53            int blockSize = list.size() - pos - 1;
54            maxBlockSizes.put(num, Math.max(maxBlockSizes.get(num), blockSize));
55        }
56
57        int minNum = -1;
58        int minBlockSize = Integer.MAX_VALUE;
59        for (Map.Entry<Integer, Integer> entry : maxBlockSizes.entrySet()) {
60            if (entry.getValue() < minBlockSize) {
61                minBlockSize = entry.getValue();
62                minNum = entry.getKey();
63            }
64        }
65
66        return minNum;
67    }
68
69    public static void main(String[] args) {
70        List<Integer> list = Arrays.asList(1, 2, 2, 3, 1, 4, 4, 4, 1, 2, 5);
71
72        // Timing brute force method
73        long startTimeBruteForce = System.nanoTime();
74        int bruteForceResult = minimalMaxBlockBruteforce(list);
75        long endTimeBruteForce = System.nanoTime();
76        long durationBruteForce = endTimeBruteForce - startTimeBruteForce;
77
78        // Timing optimized method
79        long startTimeOptimized = System.nanoTime();
80        int optimizedResult = minimalMaxBlock(list);
81        long endTimeOptimized = System.nanoTime();
82        long durationOptimized = endTimeOptimized - startTimeOptimized;
83
84        // Output results
85        System.out.println("Brute Force Result: " + bruteForceResult);
86        System.out.println("Brute Force Time: " + durationBruteForce + " nanoseconds");
87
88        System.out.println("Optimized Result: " + optimizedResult);
89        System.out.println("Optimized Time: " + durationOptimized + " nanoseconds");
90    }
91}

Excellent work! The lesson of this unit was quite comprehensive — we revisited the concept of HashMap, learned how they enhance the performance of code, and even constructed a function to locate a specific number in a list that minimizes the maximum chunk size upon removal.

Now that we've mastered the basics, the logical next step is to apply your newfound knowledge. In the upcoming practice session, a variety of intriguing challenges await you that delve further into HashMap, list manipulation, and innovative optimizations. So brace yourself, and let's dive in!