Leetcode 1995. Count Special Quadruplets
Given a 0-indexed integer array nums
, return the number of distinct quadruplets (a, b, c, d)
such that:
nums[a] + nums[b] + nums[c] == nums[d]
0 <= a, b, c, d < nums.length
a
, b
, c
, and d
are distinct.nums
array?
nums
can go up to 50 as per the constraints given on LeetCode.nums
be negative?
nums
can be negative, zero, or positive.a
, b
, c
, and d
in any order?
a < b < c < d
as specified by the constraints of indices being distinct and having a specific order.To solve the problem, we need an effective way to identify quadruplets that meet the given conditions. Given the maximum size of the array (50), we can use a brute force approach to check all possible quadruplets.
a
, b
, c
, and d
.nums[a] + nums[b] + nums[c] == nums[d]
is met for every combination.#include <vector>
using namespace std;
class Solution {
public:
int countQuadruplets(vector<int>& nums) {
int n = nums.size();
int count = 0;
// Brute-force solution with four nested loops
for (int a = 0; a < n - 3; ++a) {
for (int b = a + 1; b < n - 2; ++b) {
for (int c = b + 1; c < n - 1; ++c) {
for (int d = c + 1; d < n; ++d) {
if (nums[a] + nums[b] + nums[c] == nums[d]) {
count++;
}
}
}
}
}
return count;
}
};
The time complexity of the brute-force solution is O(n^4)
where n
is the length of the array. Given the constraints (n <= 50
), this approach is feasible as it will have a maximum of 50^4 = 6,250,000
iterations which is manageable in most scenarios.
While this approach works given the constraints, in the event of larger input sizes, the solution would need optimization to reduce the time complexity. For instance, using hash maps to store previously computed sums and improving lookup times could be a next step for optimization.
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