LeetCode 2407: Longest Increasing Subsequence II
LeetCode 2407 Solution Explanation
Explanation:
To solve this problem, we can use dynamic programming to find the longest increasing subsequence with a maximum difference of k
between adjacent elements. The idea is to maintain a list of the longest increasing subsequences ending at each index of the input array. For each index, we can iterate through all previous elements to find the longest increasing subsequence that ends with that element and also satisfies the maximum difference constraint.
Algorithm:
- Initialize a dynamic programming array
dp
of length equal to the input arraynums
. - For each index
i
in the arraynums
, iterate from indexj=0
toi-1
:- If
nums[i] - nums[j] <= k
, updatedp[i]
to the maximum ofdp[i]
anddp[j] + 1
.
- If
- Find the maximum value in the
dp
array which represents the length of the longest increasing subsequence that satisfies the maximum difference constraint. - Return the maximum value as the result.
Time Complexity:
The time complexity of this approach is O(n^2), where n is the length of the input array nums
.
Space Complexity:
The space complexity is O(n) to store the dynamic programming array.
:
LeetCode 2407 Solutions in Java, C++, Python
class Solution {
public int lengthOfLIS(int[] nums, int k) {
int n = nums.length;
int[] dp = new int[n];
Arrays.fill(dp, 1);
int maxLen = 1;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (nums[i] - nums[j] <= k) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
maxLen = Math.max(maxLen, dp[i]);
}
return maxLen;
}
}
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