# 1. Introduction

This blog post explores a unique problem in array processing: truncating an array such that 2×min is greater than max, with minimal removals. The elements can be removed from either the start or end of the array. This problem tests one's understanding of array manipulation and optimization in C++.

## Problem

Given an array of positive integers, the task is to truncate it in a way that 2×min becomes more than max, minimizing the total number of removals. Here, min and max refer to the minimum and maximum elements of the array, respectively.

Examples:

- Input: [4, 6, 1, 7, 5, 9, 2]

Output: 4

- Input: [4, 2, 6, 4, 9]

Output: 3

# 2. Solution Steps

1. Use two pointers: one at the start and the other at the end of the array.

2. Find the minimum and maximum in the current window.

3. Check if 2×min is more than max. If not, increment the removal count and move one of the pointers inwards.

4. Repeat the process until the condition is met or pointers overlap.

5. Return the total number of removals.

# 3. Code Program

``````#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int minRemovalsForCondition(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
int removals = nums.size();

while (left <= right) {
int minNum = *min_element(nums.begin() + left, nums.begin() + right + 1);
int maxNum = *max_element(nums.begin() + left, nums.begin() + right + 1);

if (2 * minNum > maxNum) {
removals = min(removals, left + (nums.size() - 1 - right));
break;
}

if (nums[left] < nums[right]) left++;
else right--;
}

return removals;
}

int main() {
vector<int> nums = {4, 6, 1, 7, 5, 9, 2};
cout << "Minimum removals: " << minRemovalsForCondition(nums) << endl;
return 0;
}

``````

### Output:

```Minimum removals: 4
```

### Explanation:

The minRemovalsForCondition function uses two pointers to determine the minimal number of elements to remove from the array to meet the condition 2×min > max. It iterates through the array, adjusting the pointers based on the current minimum and maximum values in the window. If the condition is met, it calculates the number of removals needed. This approach ensures the minimum number of removals required to satisfy the condition, as demonstrated in the output for the given example.