LeetCode 3202: Find the Maximum Length of Valid Subsequence II Solution
Master LeetCode problem 3202 (Find the Maximum Length of Valid Subsequence II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
3202. Find the Maximum Length of Valid Subsequence II
Problem Explanation
Explanation
To solve this problem, we can use dynamic programming with a memoization approach. We can define a recursive function that takes the current index and the previous remainder as parameters. We can then iterate over all possible remainders from 0 to k-1 and recursively find the maximum length of valid subsequence starting from the current index with each remainder.
At each index, we have two choices: either we include the current element in the subsequence or we skip it. If we include the element, we update the remainder based on the current element's value. If the remainder matches the expected value, we can increment the length of the subsequence.
Finally, we return the maximum length of valid subsequence starting from any index with a remainder of 0.
Time Complexity
The time complexity of this approach is O(n * k) where n is the number of elements in the array and k is the value of k.
Space Complexity
The space complexity of this approach is O(n * k) for the memoization table.
Solution Code
class Solution {
public int maxSubArray(int[] nums, int k) {
Integer[][] memo = new Integer[nums.length][k];
return helper(nums, 0, 0, k, memo);
}
private int helper(int[] nums, int index, int prevRemainder, int k, Integer[][] memo) {
if (index == nums.length) {
return 0;
}
if (memo[index][prevRemainder] != null) {
return memo[index][prevRemainder];
}
int maxLength = 0;
for (int remainder = 0; remainder < k; remainder++) {
int currentRemainder = (prevRemainder + nums[index]) % k;
int length = (currentRemainder == 0) ? 1 + helper(nums, index + 1, currentRemainder, k, memo) : helper(nums, index + 1, currentRemainder, k, memo);
maxLength = Math.max(maxLength, length);
}
memo[index][prevRemainder] = maxLength;
return maxLength;
}
}Try It Yourself
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