Minimum Pair Removal to Sort Array I

• 1. Initialize operations = 0.
• 2. Create a List<Long> from the input nums array to handle potential overflows and allow dynamic resizing.
• 3. Loop indefinitely:
  • a. Check if the list is non-decreasing. If it is, break the loop.
  • b. Initialize minSum to a very large value and minIndex to -1.
  • c. Iterate through the list from i = 0 to size - 2 to find the index minIndex of the adjacent pair with the minimum sum.
  • d. Get the values at minIndex and minIndex + 1.
  • e. Remove the two elements at minIndex + 1 and minIndex (in this order to avoid index shifting issues).
  • f. Insert their sum at minIndex.
  • g. Increment the operations counter.
• 4. Return the final operations count.

The algorithm is a straightforward simulation. We start by converting the input array into a List to make element removal and insertion easier. We then enter a loop that continues until the list is sorted in non-decreasing order.

Inside the loop, we first have a helper function, isNonDecreasing, which iterates through the list to check if list.get(i) <= list.get(i + 1) for all valid i. If this condition holds for the entire list, it means the list is sorted, and we can terminate the simulation.

If the list is not sorted, we proceed to find the adjacent pair with the minimum sum. We iterate through all adjacent pairs, keeping track of the minimum sum found so far and the index of the first element of that pair. The problem specifies that in case of a tie in sums, the leftmost pair should be chosen, which our linear scan naturally handles.

Once the minimum sum pair is identified at minIndex, we perform the merge operation. We calculate the sum of the two elements, remove them from the list, and insert the sum back at minIndex. Each merge operation counts as one, so we increment an operations counter.

The loop then repeats with the modified, smaller list. The process is guaranteed to terminate because the list size decreases by one in each iteration.

1import java.util.ArrayList;2import java.util.List;3 4class Solution {5    public int minimumOperations(int[] nums) {6        if (nums.length <= 1) {7            return 0;8        }9 10        List<Long> currentNums = new ArrayList<>();11        for (int num : nums) {12            currentNums.add((long) num);13        }14 15        int operations = 0;16        while (true) {17            if (isNonDecreasing(currentNums)) {18                break;19            }20 21            int minIndex = -1;22            long minSum = Long.MAX_VALUE;23 24            for (int i = 0; i < currentNums.size() - 1; i++) {25                long currentSum = currentNums.get(i) + currentNums.get(i + 1);26                if (currentSum < minSum) {27                    minSum = currentSum;28                    minIndex = i;29                }30            }31 32            long sum = currentNums.get(minIndex) + currentNums.get(minIndex + 1);33            34            currentNums.remove(minIndex + 1);35            currentNums.remove(minIndex);36            37            currentNums.add(minIndex, sum);38            operations++;39        }40 41        return operations;42    }43 44    private boolean isNonDecreasing(List<Long> list) {45        for (int i = 0; i < list.size() - 1; i++) {46            if (list.get(i) > list.get(i + 1)) {47                return false;48            }49        }50        return true;51    }52}

• 1. Create a doubly linked list from the nums array. Each node stores its value and pointers to its neighbors.
• 2. Create a min-priority queue to store Pair objects, where each Pair represents two adjacent nodes and their sum.
• 3. Populate the priority queue with pairs from the initial linked list.
• 4. Initialize operations = 0.
• 5. Loop until the list is sorted:
  • a. Extract the Pair with the minimum sum from the priority queue.
  • b. Check if the pair is stale (i.e., its nodes have been modified or are no longer adjacent). If so, discard it and continue to the next pair.
  • c. If the pair is valid, perform the merge: update the left node's value with the sum, and adjust pointers in the linked list to remove the right node.
  • d. Add new pairs formed by the merged node and its new neighbors to the priority queue.
  • e. Increment operations.
• 6. Return operations.

To improve the time complexity, we need to optimize the step of finding the minimum sum pair. A min-priority queue is the ideal data structure for this, as it can provide the minimum element in logarithmic time.

However, using a priority queue introduces a new challenge: when we merge a pair, the adjacencies in the array change. For example, if we merge (a, b) into c, the old pairs involving a and b (like (x, a) and (b, y)) become invalid, and new pairs (x, c) and (c, y) are formed. The priority queue must reflect these changes.

We can solve this by using a doubly linked list to represent the array. Each node in the list holds a number. The priority queue will store objects representing adjacent pairs of nodes, ordered by their sum.

When we extract the minimum pair from the queue, we must first verify it's not 'stale'. A pair is stale if one of its nodes has already been merged into another. We can check this by adding a flag to each node or by verifying that the two nodes are still adjacent in the linked list (node1.next == node2).

If the pair is valid, we merge them. This involves updating the value of the first node, and then updating the pointers of the surrounding nodes to bypass the second node, effectively removing it. We then add new pairs (formed by the merged node and its new neighbors) to the priority queue. We don't need to explicitly remove stale pairs from the queue; the staleness check upon extraction is sufficient.

This process continues until the array becomes sorted. The overall approach is much faster asymptotically but requires careful implementation.

A conceptual sketch of the data structures:

1// Node for the doubly linked list2class Node {3    long val;4    Node prev, next;5    boolean isMerged = false;6    Node(long val) { this.val = val; }7}8 9// Object to store in the Priority Queue10class Pair implements Comparable<Pair> {11    long sum;12    Node left, right;13 14    Pair(Node l, Node r) {15        this.left = l;16        this.right = r;17        this.sum = l.val + r.val;18    }19 20    @Override21    public int compareTo(Pair other) {22        if (this.sum != other.sum) {23            return Long.compare(this.sum, other.sum);24        }25        // Tie-breaking rule is not explicitly handled here but would be based on original index.26        // For this problem, the PQ's internal tie-breaking is usually sufficient.27        return 0; 28    }29}30 31// Main logic sketch32public int minimumOperationsWithPQ(int[] nums) {33    // 1. Build doubly linked list and initial priority queue.34    // 2. Loop until sorted, performing operations:35    //    a. Poll from PQ.36    //    b. Check for staleness (if (left.isMerged || right.isMerged || left.next != right) continue;)37    //    c. Merge nodes in the linked list.38    //    d. Mark right node as merged.39    //    e. Add new pairs involving the merged node to the PQ.40    //    f. Increment operations count.41    // 3. Return operations.42    return 0; // Placeholder for full implementation43}