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search/Longest_Increasing_Subsequence_using_binary_search.cpp
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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. | ||
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* For solving this problem we have Three Approaches :- | ||
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* 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. | ||
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* 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) | ||
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* This approach is better then the previous Brute Force approach so, we can | ||
* consider this approach. | ||
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* But when the Constraints for the problem is very larger then this approach fails | ||
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* 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. | ||
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* Time complexity: O(n∗log(n)) | ||
* Space complexity: O(1) | ||
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* This approach consider Most optimal Approach for solving this problem | ||
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* @author [Naman Jain](https://github.com/namanmodi65) | ||
*/ | ||
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#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 | ||
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/** | ||
* @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; | ||
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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()); | ||
} | ||
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/** | ||
* @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); | ||
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std::vector<int> arr2 = {0, 1, 0, 3, 2, 3}; | ||
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assert(longest_increasing_subsequence_using_binary_search(arr2) == 4); | ||
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std::vector<int> arr3 = {7, 7, 7, 7, 7, 7, 7}; | ||
assert(longest_increasing_subsequence_using_binary_search(arr3) == 1); | ||
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std::vector<int> arr4 = {-10, -1, -5, 0, 5, 1, 2}; | ||
assert(longest_increasing_subsequence_using_binary_search(arr4) == 5); | ||
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std::vector<double> arr5 = {3.5, 1.2, 2.8, 3.1, 4.0}; | ||
assert(longest_increasing_subsequence_using_binary_search(arr5) == 4); | ||
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std::vector<char> arr6 = {'a', 'b', 'c', 'a', 'd'}; | ||
assert(longest_increasing_subsequence_using_binary_search(arr6) == 4); | ||
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std::vector<int> arr7 = {}; | ||
assert(longest_increasing_subsequence_using_binary_search(arr7) == 0); | ||
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std::cout << "All tests have successfully passed!\n"; | ||
} | ||
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/** | ||
* @brief Main function to run tests | ||
* @returns 0 on exit | ||
*/ | ||
int main() { | ||
tests(); // run self test implementation | ||
return 0; | ||
} |
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