Dare2Solve
To solve the problem of finding the longest valid subsequence where consecutive pairs have equal sums modulo k
, we can utilize a dynamic programming approach. The key insight is to maintain a mapping of remainder values modulo k
to the length of subsequences that end with each remainder.
Initialization: Initialize a dictionary or array remainder_to_length
to keep track of the maximum length of subsequences that end with each remainder modulo k
.
Iterate through nums
: For each number num
in nums
, calculate its remainder modulo k
.
Update lengths: For each remainder, update remainder_to_length
by considering adding num
to subsequences ending with the previous remainder (curr_remainder + num) % k
.
Track maximum length: Maintain a variable max_length
to store the maximum length found during the iteration.
Return result: After iterating through all numbers in nums
, max_length
will contain the length of the longest valid subsequence.
nums
. We iterate through nums
once and perform constant-time operations for each element.k
. The space is primarily used to store mappings of remainders to lengths of subsequences, which can have up to k
entries.The solution efficiently computes the length of the longest valid subsequence using dynamic programming and hashing, ensuring optimal performance.
var maximumLength = function (nums, k) {
maximumLength
that takes two parameters: nums
, an array of integers, and k
, a positive integer. The function aims to find the length of the longest valid subsequence where consecutive pairs have equal sums modulo k
.let dp = new Array(k * k).fill(0);
dp
of size k * k
filled with zeros. This array will store the maximum lengths of subsequences that end with specific remainders modulo k
.var helper = (x, y) => x * k + y;
helper
that calculates indices in the dp
array based on two remainders x
and y
using the formula x * k + y
. This function simplifies accessing and updating elements in the dp
array based on remainders.nums.forEach((num) => {
num %= k;
for (let i = 0; i < k; i++) {
if (dp[helper(num, i)] + 1 > dp[helper(i, num)]) {
dp[helper(i, num)] = dp[helper(num, i)] + 1;
}
}
});
num
in the nums
array. For each num
, computes its remainder modulo k
.num % k
, iterates through all possible remainders i
from 0 to k-1
.num
to the subsequence ending with remainder i
produces a longer valid subsequence than using i
as the current remainder.dp
array with the new maximum length of subsequences if a longer valid subsequence is found. return Math.max(...dp);
};
dp
array using Math.max(...dp)
, which represents the length of the longest valid subsequence found based on the conditions specified.class Solution {
public:
int maximumLength(vector<int>& nums, int k) {
vector<int> dp(k * k, 0);
auto helper = [&](int x, int y, int k) {
return x * k + y;
};
for (int num : nums) {
num %= k;
for (int i = 0; i < k; i++) {
if (dp[helper(num, i, k)] + 1 > dp[helper(i, num, k)]) {
dp[helper(i, num, k)] = dp[helper(num, i, k)] + 1;
}
}
}
return *max_element(dp.begin(), dp.end());
}
};
class Solution:
def maximumLength(self, nums: List[int], k: int) -> int:
dp = [0] * (k * k)
def helper(x, y, k):
return x * k + y
for num in nums:
num %= k
for i in range(k):
if dp[helper(num, i, k)] + 1 > dp[helper(i, num, k)]:
dp[helper(i, num, k)] = dp[helper(num, i, k)] + 1
return max(dp)
class Solution {
public int maximumLength(int[] nums, int k) {
int[] dp = new int[k * k];
for (int num : nums) {
num %= k;
for (int i = 0; i < k; i++) {
if (dp[helper(num, i, k)] + 1 > dp[helper(i, num, k)]) {
dp[helper(i, num, k)] = dp[helper(num, i, k)] + 1;
}
}
}
int max = 0;
for (int value : dp) {
max = Math.max(max, value);
}
return max;
}
private int helper(int x, int y, int k) {
return x * k + y;
}
}
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var maximumLength = function (nums, k) {
let dp = new Array(k * k).fill(0);
var helper = (x, y) => x * k + y;
nums.forEach((num) => {
num %= k;
for (let i = 0; i < k; i++) {
if (dp[helper(num, i)] + 1 > dp[helper(i, num)]) {
dp[helper(i, num)] = dp[helper(num, i)] + 1;
}
}
});
return Math.max(...dp);
};