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Shortest Subarray to be Removed to Make Array Sorted

Problem Overview

The goal is to find the minimum number of elements to remove from a given array in order to make the entire array sorted in non-decreasing order. The solution needs to identify and utilize already sorted segments of the array to minimize the number of elements that must be removed.

Approach

General Steps

  1. Identify the Longest Sorted Prefix:

    • Start from the beginning of the array and move forward until you find a break in the non-decreasing order. This part of the array is the longest "prefix" that is already sorted.

  2. Identify the Longest Sorted Suffix:

    • Starting from the end of the array, move backward until you find a break in the non-decreasing order. This part of the array is the longest "suffix" that is already sorted.

  3. Check if Entire Array is Sorted:

    • If the longest sorted prefix already covers the entire array, no removal is needed, so the result is 0.

  4. Calculate Initial Minimum Removals:

    • If the entire array is not sorted, calculate the initial minimum number of elements to remove by either:
      • Removing the entire suffix (from the end of the prefix to the array’s end).
      • Removing the entire prefix (from the start to the beginning of the suffix).

  5. Optimize Removal with Two Pointers:

    • Use a two-pointer technique to attempt merging parts of the prefix and suffix.
    • Adjust pointers to find if a smaller section in between can be removed to achieve the sorted order, updating the minimum elements to remove if a smaller solution is found.

Solutions in Multiple Languages

C++ Code - Step-by-Step Explanation

  1. Find the Longest Sorted Prefix: Initialize a pointer from the start and increment it while the next element is larger than or equal to the current. Stop when the order breaks.

  2. Find the Longest Sorted Suffix: Initialize a pointer from the end and decrement it while the previous element is smaller than or equal to the current. Stop when the order breaks.

  3. Check Full Array Condition: If the prefix pointer reaches the end, the array is already sorted. Return 0.

  4. Calculate Initial Minimum Removals: Compute the minimum removals by comparing two options:

    • Removing all elements from the end of the prefix to the array's end.
    • Removing all elements from the start to the beginning of the suffix.

  5. Merge with Two-Pointer Technique: Set two pointers on the prefix and suffix. Adjust them while maintaining sorted order, minimizing the number of elements to remove for a sorted result.

Java Code - Step-by-Step Explanation

  1. Identify Longest Sorted Prefix: Use a loop to traverse from the beginning, stopping when the non-decreasing order is broken.

  2. Identify Longest Sorted Suffix: Traverse from the end towards the beginning, stopping when the order breaks.

  3. Full Array Check: If the prefix pointer reaches the last index, the array is already sorted, so return 0.

  4. Calculate Minimum Removals:

    • Calculate the number of elements to remove if only keeping the prefix.
    • Calculate the number of elements to remove if only keeping the suffix.
    • Set the minimum of these values as the initial result.

  5. Optimize with Two Pointers: Using pointers on both the prefix and suffix, try merging sections and update the minimum removals if a smaller solution is possible.

JavaScript Code - Step-by-Step Explanation

  1. Longest Sorted Prefix: Start from index 0, increment until reaching a break in the order.

  2. Longest Sorted Suffix: Start from the last index, decrement until reaching a break in the order.

  3. Check if Sorted: If the entire array is already sorted (i.e., the prefix reaches the end), return 0.

  4. Calculate Initial Minimum: Evaluate the removals required by only keeping the prefix or only keeping the suffix and set the minimum.

  5. Two-Pointer Optimization: Use two pointers on the prefix and suffix to find a merged solution, minimizing the required removals by adjusting the pointers to maintain order.

Python Code - Step-by-Step Explanation

  1. Find the Longest Non-Decreasing Prefix: Initialize a pointer and move it forward while each element is less than or equal to the next.

  2. Find the Longest Non-Decreasing Suffix: Initialize a pointer at the end and move it backward while each element is greater than or equal to the previous.

  3. Full Array Check: If the prefix reaches the last element, the array is fully sorted, and the answer is 0.

  4. Initial Minimum Removals: Calculate removals by keeping either the prefix or the suffix, whichever results in fewer removals.

  5. Two-Pointer Technique for Optimal Merging: Attempt to merge the prefix and suffix while keeping them in order. Adjust pointers to find the smallest number of elements that need to be removed.

Go Code - Step-by-Step Explanation

  1. Identify Longest Sorted Prefix: Set a pointer to start and increment as long as each element is in non-decreasing order with the next.

  2. Identify Longest Sorted Suffix: Set a pointer to the end and decrement while each element is in non-decreasing order with the previous.

  3. Entire Array Check: If the prefix covers the whole array, it’s already sorted, so return 0.

  4. Compute Initial Minimum Removals: Calculate the required removals if keeping only the prefix or the suffix. Set the result to the minimum of these two values.

  5. Two-Pointer Optimization for Minimal Removals: Using two pointers on the prefix and suffix, move them towards each other while keeping the array sorted to find the minimal number of elements to remove.

Complexity Analysis

  • Time Complexity: (O(n)) - Each solution involves linear scans to identify the sorted prefix and suffix and a two-pointer pass, making the solution efficient.
  • Space Complexity: (O(1)) - Only a few variables are used for indices and result calculation, leading to constant extra space usage.

Each solution leverages the initial sorted segments of the array (prefix and suffix) to minimize the removal count, making it a highly efficient approach to solve this problem.