1713. Minimum Operations to Make a Subsequence LeetCode Solution

Last updated: February 24, 2025

In this guide, you will get 1713. Minimum Operations to Make a Subsequence LeetCode Solution with the best time and space complexity. The solution to Minimum Operations to Make a Subsequence problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Minimum Operations to Make a Subsequence solution in C++
Minimum Operations to Make a Subsequence solution in Java
Minimum Operations to Make a Subsequence solution in Python
Additional Resources

1713. Minimum Operations to Make a Subsequence LeetCode Solution image

Problem Statement of Minimum Operations to Make a Subsequence

You are given an array target that consists of distinct integers and another integer array arr that can have duplicates.
In one operation, you can insert any integer at any position in arr. For example, if arr = [1,4,1,2], you can add 3 in the middle and make it [1,4,3,1,2]. Note that you can insert the integer at the very beginning or end of the array.
Return the minimum number of operations needed to make target a subsequence of arr.
A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements’ relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.

Example 1:

Input: target = [5,1,3], arr = [9,4,2,3,4]
Output: 2
Explanation: You can add 5 and 1 in such a way that makes arr = [5,9,4,1,2,3,4], then target will be a subsequence of arr.

Example 2:

Input: target = [6,4,8,1,3,2], arr = [4,7,6,2,3,8,6,1]
Output: 3

Constraints:

1 <= target.length, arr.length <= 105
1 <= target[i], arr[i] <= 109
target contains no duplicates.

Complexity Analysis

Time Complexity: O(n\log n)
Space Complexity: O(n)

1713. Minimum Operations to Make a Subsequence LeetCode Solution in C++

class Solution {
 public:
  int minOperations(vector<int>& target, vector<int>& arr) {
    vector<int> indices;
    unordered_map<int, int> numToIndex;

    for (int i = 0; i < target.size(); ++i)
      numToIndex[target[i]] = i;

    for (const int a : arr)
      if (const auto it = numToIndex.find(a); it != numToIndex.end())
        indices.push_back(it->second);

    return target.size() - lengthOfLIS(indices);
  }

 private:
  // Same as 300. Longest Increasing Subsequence
  int lengthOfLIS(vector<int>& nums) {
    // tails[i] := the minimum tail of all the increasing subsequences having
    // length i + 1
    vector<int> tails;
    for (const int num : nums)
      if (tails.empty() || num > tails.back())
        tails.push_back(num);
      else
        tails[firstGreaterEqual(tails, num)] = num;
    return tails.size();
  }

 private:
  int firstGreaterEqual(const vector<int>& arr, int target) {
    return ranges::lower_bound(arr, target) - arr.begin();
  }
};
/* code provided by PROGIEZ */

1713. Minimum Operations to Make a Subsequence LeetCode Solution in Java

class Solution {
  public int minOperations(int[] target, int[] arr) {
    List<Integer> indices = new ArrayList<>();
    Map<Integer, Integer> numToIndex = new HashMap<>();

    for (int i = 0; i < target.length; ++i)
      numToIndex.put(target[i], i);

    for (final int a : arr)
      if (numToIndex.containsKey(a))
        indices.add(numToIndex.get(a));

    return target.length - lengthOfLIS(indices);
  }

  // Same as 300. Longest Increasing Subsequence
  private int lengthOfLIS(List<Integer> nums) {
    // tails[i] := the minimum tail of all the increasing subsequences with
    // 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(firstGreaterEqual(tails, num), num);
    return tails.size();
  }

  private int firstGreaterEqual(List<Integer> arr, int target) {
    final int i = Collections.binarySearch(arr, target);
    return i < 0 ? -i - 1 : i;
  }
}
// code provided by PROGIEZ

1713. Minimum Operations to Make a Subsequence LeetCode Solution in Python

class Solution:
  def minOperations(self, target: list[int], arr: list[int]) -> int:
    indices = []
    numToIndex = {num: i for i, num in enumerate(target)}

    for a in arr:
      if a in numToIndex:
        indices.append(numToIndex[a])

    return len(target) - self._lengthOfLIS(indices)

  # Same as 300. Longest Increasing Subsequence
  def _lengthOfLIS(self, nums: list[int]) -> int:
    # tails[i] := the minimum tail of all the increasing subsequences having
    # length i + 1
    tails = []
    for num in nums:
      if not tails or num > tails[-1]:
        tails.append(num)
      else:
        tails[bisect.bisect_left(tails, num)] = num
    return len(tails)
 # code by PROGIEZ

Additional Resources

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