LeetCode-in-All
300. Longest Increasing Subsequence
Medium
Given an integer array nums, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
1 <= nums.length <= 2500-104 <= nums[i] <= 104
Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?
Solution
#include <stdio.h>
#include <limits.h>
int lengthOfLIS(int* nums, int numsSize) {
if (nums == NULL || numsSize == 0) {
return 0;
}
int* dp = (int*)malloc((numsSize + 1) * sizeof(int));
// Prefill the dp array with INT_MAX
for (int i = 1; i <= numsSize; i++) {
dp[i] = INT_MAX;
}
int left = 1;
int right = 1;
for (int i = 0; i < numsSize; i++) {
int curr = nums[i];
int start = left;
int end = right;
// Binary search within dp to find the correct position for curr
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (dp[mid] > curr) {
end = mid;
} else {
start = mid;
}
}
// Update dp array based on binary search result
if (dp[start] > curr) {
dp[start] = curr;
} else if (curr > dp[start] && curr < dp[end]) {
dp[end] = curr;
} else if (curr > dp[end]) {
dp[++end] = curr;
right++;
}
}
int result = right;
free(dp); // Free the allocated memory for dp array
return result;
}