Coder's Cat

LeetCode: 4Sum

2019-11-05

Solution for LeetCode 4sum

Challenge Description

https://leetcode.com/problems/4sum/

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:

[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]

Solution

For time optimization, we need to sort the given array. Then this problem transfered into the 3sum problem,

Time complexity is: $$O(N^3)$$

class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
vector<vector<int> > ans;
sort(num.begin(), num.end());

if(num.size() < 4) return ans;
for(size_t first =0; first < num.size() - 3; first++) {
for(size_t s = first+1; s < num.size()-2; s++) {
int value = target-num[s]-num[first];
int head = s + 1;
int tail = num.size() - 1;
while(head < tail) {
int v = num[head] + num[tail];
if(v == value) {
ans.push_back(vector<int> { num[first], num[s], num[head], num[tail] });
head++, tail--;
while(head < tail && num[head] == num[head-1])
head++;
while(tail > head && num[tail] == num[tail+1])
tail--;

}
else if(v > value)
tail--;
else
head++;
}
while(s < num.size()-2 && num[s+1] == num[s])
s++;
}
while(first < num.size()-3 && num[first] == num[first+1])
first++;
}
return ans;
}
};

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