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Problem solution · Java

Minimum Number of Increasing Subsequence to Be Removed

Minimum Number of Increasing Subsequence to Be Removed: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
43 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Minimum Number of Increasing Subsequence to Be Removed, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 43 lines of Java from the credited upstream file 3231.java.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 3 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Number of Increasing Subsequence to Be Removed · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minOperations(int[] nums) {    // The length of the longest non-increasing subsequence is equal to the    // number of strictly increasing subsequences needed to cover the entire    // array. This is because any number in the non-increasing subsequence must    // use one number from each of the strictly increasing subsequences. e.g.,    // [4, 3, 1, 2] has 3 strictly increasing subsequences: [4], [3], and [1,    // 2]. The longest non-increasing subsequences are [4, 3, 1] or [4, 3, 2].    int[] reversedNums = new int[nums.length];    for (int i = 0; i < nums.length; ++i)      reversedNums[i] = nums[nums.length - 1 - i];    return lengthOfLIS(reversedNums);  }   // Similar to 300. Longest Increasing Subsequence  private int lengthOfLIS(int[] nums) {    // tails[i] := the minimum tail of all the increasing subsequences having    // length i + 1    List<Integer> tails = new ArrayList<>();     for (final int num : nums)      if (tails.isEmpty() || num >= tails.get(tails.size() - 1))        tails.add(num);      else        tails.set(firstGreater(tails, num), num);     return tails.size();  }   private int firstGreater(List<Integer> arr, int target) {    int l = 0;    int r = arr.size();    while (l < r) {      final int m = (l + r) / 2;      if (arr.get(m) > target)        r = m;      else        l = m + 1;    }    return l;  }}

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