How do you solve Minimum Operations to Make a Subsequence on LeetCode?

Minimum Operations to Make a Subsequence (LeetCode #1713) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Mapping elements to indices allows us to focus on the order rather than the values.

What is the time complexity of Minimum Operations to Make a Subsequence?

The optimal solution for Minimum Operations to Make a Subsequence runs in O(n log n) time and uses O(n) space. The complexity is O(n log n) due to the binary search used to find the position in the LIS, and O(n) for transforming the array.

What is the brute force solution for Minimum Operations to Make a Subsequence?

The brute force approach for Minimum Operations to Make a Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subsequence of `arr` to see if it matches `target`. This is inefficient but helps us understand the problem's nature. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Operations to Make a Subsequence?

Minimum Operations to Make a Subsequence uses the following concepts: Array, Hash Table, Binary Search, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Make a Subsequence?

Always consider if a problem can be transformed into a well-known problem like LIS. Explain your thought process clearly; interviewers appreciate understanding your reasoning. Practice coding the LIS algorithm as it frequently appears in various forms.

What are common mistakes when solving Minimum Operations to Make a Subsequence?

Failing to recognize that the problem can be reduced to finding the LIS. Not handling duplicates in `arr` correctly when transforming to indices.

#1713

Minimum Operations to Make a Subsequence

Hard

ArrayHash TableBinary SearchGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

Instead of generating subsequences, we can map elements of `arr` to their indices in `target` to find the longest increasing subsequence (LIS). The number of operations needed will be the difference between the length of `target` and the length of the LIS.

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Algorithm

4 steps

1Step 1: Create a mapping of each element in `target` to its index.
2Step 2: Transform `arr` into a new array of indices based on the mapping from `target` (ignore elements not in `target`).
3Step 3: Find the length of the longest increasing subsequence (LIS) in this new array.
4Step 4: The minimum operations needed is the difference between the length of `target` and the length of the LIS.

solution.py15 lines

1# Full working Python code
2from bisect import bisect_left
3
4def min_operations_optimal(target, arr):
5    index_map = {num: i for i, num in enumerate(target)}
6    transformed = [index_map[num] for num in arr if num in index_map]
7    lis = []
8    for num in transformed:
9        pos = bisect_left(lis, num)
10        if pos == len(lis):
11            lis.append(num)
12        else:
13            lis[pos] = num
14    return len(target) - len(lis)
15

ℹ

Complexity note: The complexity is O(n log n) due to the binary search used to find the position in the LIS, and O(n) for transforming the array.

1Mapping elements to indices allows us to focus on the order rather than the values.
2Finding the longest increasing subsequence helps minimize the number of insertions needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.