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C-Plus-Plus/search/Longest_Increasing_Subsequence_using_binary_search.cpp
Naman Jain 05d784d844 Longest Increasing subsequence using binary search most optimal approach (Modified) (#2776)
* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem (Modified)

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem(done)

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Floyd warshall

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem

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Co-authored-by: realstealthninja <68815218+realstealthninja@users.noreply.github.com>
2024-10-18 21:59:49 +05:30

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/**
* @file
* @brief find the length of the Longest Increasing Subsequence (LIS)
* using [Binary Search](https://en.wikipedia.org/wiki/Longest_increasing_subsequence)
* @details
* Given an integer array nums, return the length of the longest strictly
* increasing subsequence.
* The longest increasing subsequence is described as a subsequence of an array
* where: All elements of the subsequence are in increasing order. This subsequence
* itself is of the longest length possible.
* For solving this problem we have Three Approaches :-
* Approach 1 :- Using Brute Force
* The first approach that came to your mind is the Brute Force approach where we
* generate all subsequences and then manually filter the subsequences whose
* elements come in increasing order and then return the longest such subsequence.
* Time Complexity :- O(2^n)
* It's time complexity is exponential. Therefore we will try some other
* approaches.
* Approach 2 :- Using Dynamic Programming
* To generate all subsequences we will use recursion and in the recursive logic we
* will figure out a way to solve this problem. Recursive Logic to solve this
* problem:-
* 1. We only consider the element in the subsequence if the element is grater then
* the last element present in the subsequence
* 2. When we consider the element we will increase the length of subsequence by 1
* Time Complexity: O(N*N)
* Space Complexity: O(N*N) + O(N)
* This approach is better then the previous Brute Force approach so, we can
* consider this approach.
* But when the Constraints for the problem is very larger then this approach fails
* Approach 3 :- Using Binary Search
* Other approaches use additional space to create a new subsequence Array.
* Instead, this solution uses the existing nums Array to build the subsequence
* array. We can do this because the length of the subsequence array will never be
* longer than the current index.
* Time complexity: O(nlog(n))
* Space complexity: O(1)
* This approach consider Most optimal Approach for solving this problem
* @author [Naman Jain](https://github.com/namanmodi65)
*/
#include <cassert> /// for std::assert
#include <iostream> /// for IO operations
#include <vector> /// for std::vector
#include <algorithm> /// for std::lower_bound
#include <cstdint> /// for std::uint32_t
/**
* @brief Function to find the length of the Longest Increasing Subsequence (LIS)
* using Binary Search
* @tparam T The type of the elements in the input vector
* @param nums The input vector of elements of type T
* @return The length of the longest increasing subsequence
*/
template <typename T>
std::uint32_t longest_increasing_subsequence_using_binary_search(std::vector<T>& nums) {
if (nums.empty()) return 0;
std::vector<T> ans;
ans.push_back(nums[0]);
for (std::size_t i = 1; i < nums.size(); i++) {
if (nums[i] > ans.back()) {
ans.push_back(nums[i]);
} else {
auto idx = std::lower_bound(ans.begin(), ans.end(), nums[i]) - ans.begin();
ans[idx] = nums[i];
}
}
return static_cast<std::uint32_t>(ans.size());
}
/**
* @brief Test cases for Longest Increasing Subsequence function
* @returns void
*/
static void tests() {
std::vector<int> arr = {10, 9, 2, 5, 3, 7, 101, 18};
assert(longest_increasing_subsequence_using_binary_search(arr) == 4);
std::vector<int> arr2 = {0, 1, 0, 3, 2, 3};
assert(longest_increasing_subsequence_using_binary_search(arr2) == 4);
std::vector<int> arr3 = {7, 7, 7, 7, 7, 7, 7};
assert(longest_increasing_subsequence_using_binary_search(arr3) == 1);
std::vector<int> arr4 = {-10, -1, -5, 0, 5, 1, 2};
assert(longest_increasing_subsequence_using_binary_search(arr4) == 5);
std::vector<double> arr5 = {3.5, 1.2, 2.8, 3.1, 4.0};
assert(longest_increasing_subsequence_using_binary_search(arr5) == 4);
std::vector<char> arr6 = {'a', 'b', 'c', 'a', 'd'};
assert(longest_increasing_subsequence_using_binary_search(arr6) == 4);
std::vector<int> arr7 = {};
assert(longest_increasing_subsequence_using_binary_search(arr7) == 0);
std::cout << "All tests have successfully passed!\n";
}
/**
* @brief Main function to run tests
* @returns 0 on exit
*/
int main() {
tests(); // run self test implementation
return 0;
}