**Longest Increasing Subsequence - Amazon Top Interview Questions**

### Problem Statement :

Given an unsorted list of integers nums, return the longest strictly increasing subsequence of the array. Bonus: Can you solve it in \mathcal{O}(n \log n)O(nlogn) time? Constraints n ≤ 1,000 where n is the length of nums Example 1 Input nums = [6, 1, 7, 2, 8, 3, 4, 5] Output 5 Explanation Longest increasing subsequence would be [1, 2, 3, 4, 5] Example 2 Input nums = [12, 5, 6, 25, 8, 11, 10] Output 4 Explanation One longest increasing subsequence would be [5, 6, 8, 11]

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
int n = nums.size();
vector<int> dp; // dp[i] = last digit in the list of size i+1, dp is increasing
for (int& i : nums) {
if (dp.empty()) {
dp.push_back(i);
continue;
}
auto it = lower_bound(dp.begin(), dp.end(), i);
if (it == dp.end())
dp.push_back(i);
else {
int idx = it - dp.begin();
dp[idx] = i;
}
}
return dp.size();
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
if not nums:
return 0
def get_index(arr, target):
l, r = 0, len(arr) - 1
res = r + 1
while l <= r:
mid = (l + r) // 2
if target < arr[mid]:
res = mid
r = mid - 1
elif target > arr[mid]:
l = mid + 1
else:
return mid
return res
dp = []
res = 0
for num in nums:
idx = get_index(dp, num)
if idx == res:
dp.append(num)
res += 1
else:
dp[idx] = num
return res
```

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