lcb-0155

1.000

2/2 tests· algorithms

Proposed solution

I'll solve this step by step.

## Problem Understanding

1. We have an array of numbers
2. We can repeatedly select the leftmost adjacent pair with minimum sum and replace it with their sum
3. Goal: Find minimum operations to make array non-decreasing
4. An array is non-decreasing if each element ≥ previous element

## Key Insights

Let me trace through Example 1:
- Initial: [5,2,3,1]
- Find pairs: (5,2)=7, (2,3)=5, (3,1)=4
- Minimum sum is 4, so pair (3,1) → [5,2,4]
- Find pairs: (5,2)=7, (2,4)=6
- Minimum sum is 6, so pair (2,4) → [5,6]
- Result: [5,6] - non-decreasing ✓

## Approach

This looks like a greedy approach where we always pick the leftmost minimum sum pair. However, I need to think about when to stop.

Let me reconsider the problem:
- We keep doing operations until the array becomes non-decreasing
- Each operation replaces two adjacent elements with their sum
- We want to minimize total operations

## Better Approach

I'll simulate the process:
1. While the array is not non-decreasing:
   - Find all adjacent pairs and their sums
   - Pick the leftmost pair with minimum sum
   - Replace it with the sum
2. Count operations

Let me trace Example 1 again:
- [5,2,3,1] → pairs: (5,2)=7, (2,3)=5, (3,1)=4 → min=4, leftmost is (3,1) → [5,2,4]
- [5,2,4] → pairs: (5,2)=7, (2,4)=6 → min=6, leftmost is (2,4) → [5,6]  
- [5,6] → already non-decreasing

## Implementation Strategy

1. Create a function to check if array is non-decreasing
2. While not non-decreasing:
   - Find minimum sum adjacent pair from left
   - Replace it and increment counter
3. Return counter

## Code Implementation

```python
class Solution:
    def minimumPairRemoval(self, nums: List[int]) -> int:
        def is_non_decreasing(arr):
            for i in range(1, len(arr)):
                if arr[i] < arr[i-1]:
                    return False
            return True
        
        operations = 0
        current_nums = nums[:]
        
        while not is_non_decreasing(current_nums):
            min_sum = float('inf')
            min_index = -1
            
            # Find the leftmost adjacent pair with minimum sum
            for i in range(len(current_nums) - 1):
                pair_sum = current_nums[i] + current_nums[i+1]
                if pair_sum < min_sum:
                    min_sum = pair_sum
                    min_index = i
            
            # Replace the pair with their sum
            new_nums = []
            for i in range(len(current_nums)):
                if i == min_index:
                    new_nums.append(current_nums[i] + current_nums[i+1])
                    # Skip next element as it's already included
                    i += 1
                elif i == min_index + 1:
                    # Skip this element as it was merged
                    continue
                else:
                    new_nums.append(current_nums[i])
            
            current_nums = new_nums
            operations += 1
        
        return operations
```

Wait, let me reconsider the logic for building the new array. Let me fix that:

```python
class Solution:
    def minimumPairRemoval(self, nums: List[int]) -> int:
        def is_non_decreasing(arr):
            for i in range(1, len(arr)):
                if arr[i] < arr[i-1]:
                    return False
            return True
        
        operations = 0
        current_nums = nums[:]
        
        while not is_non_decreasing(current_nums):
            min_sum = float('inf')
            min_index = -1
            
            # Find the leftmost adjacent pair with minimum sum
            for i in range(len(current_nums) - 1):
                pair_sum = current_nums[i] + current_nums[i+1]
                if pair_sum < min_sum:
                    min_sum = pair_sum
                    min_index = i
            
            # Replace the pair with their sum
            new_nums = []
            i = 0
            while i < len(current_nums):
                if i == min_index:
                    new_nums.append(current_nums[i] + current_nums[i+1])
                    i += 2  # Skip next element as it's merged
                else:
                    new_nums.append(current_nums[i])
                    i += 1
            
            current_nums = new_nums
            operations += 1
        
        return operations
```

Test output (stdout)

..                                                                       [100%]
2 passed in 0.01s

System prompt

You are an expert programmer. Solve the task exactly as specified. Output your solution as fenced code blocks using the required file name(s) and the exact function/type signatures requested. Prefer correctness; do not include prose outside code unless asked.