LeetCode 1713: Minimum Operations to Make a Subsequence Solution
Master LeetCode problem 1713 (Minimum Operations to Make a Subsequence), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1713. Minimum Operations to Make a Subsequence
Problem Explanation
Explanation
To solve this problem, we can first create a mapping of elements in the target array to their indices. Then, we iterate through the arr array and for each element that is present in the target array, we update the longest increasing subsequence (LIS) ending with that element. Finally, the answer to the problem will be the length of the target array minus the length of the LIS we found.
Algorithm:
- Create a mapping of elements in
targetto their indices. - Initialize an empty array to store the LIS ending with each element in
arr. - Iterate through each element in
arr.- If the element is present in the
targetarray, update the LIS ending with that element.
- If the element is present in the
- Return the answer as the length of
targetminus the length of the longest LIS found.
Time Complexity: O(nlogn) where n is the length of the arr array.
Space Complexity: O(n) where n is the length of the arr array.
Solution Code
class Solution {
public int minOperations(int[] target, int[] arr) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < target.length; i++) {
map.put(target[i], i);
}
List<Integer> lis = new ArrayList<>();
for (int val : arr) {
if (map.containsKey(val)) {
int idx = map.get(val);
int pos = Collections.binarySearch(lis, idx);
if (pos < 0) {
pos = -(pos + 1);
}
if (pos == lis.size()) {
lis.add(idx);
} else {
lis.set(pos, idx);
}
}
}
return target.length - lis.size();
}
}Try It Yourself
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