4 Sum II

Medium
Given four lists A, B, C, D of integer values, compute how many tuples(i, j, k, l)there are such thatA[i] + B[j] + C[k] + D[l]is zero.
To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -2^28 to 2^28- 1 and the result is guaranteed to be at most 2^31- 1.
Example:
Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]
Output:
2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

Solution

HashMap
虽然是KSum系列,但是因为有4个数组,所以做法可以跟KSum很不一样,这里用HashMap记录两个数组中的数之和的计数,再遍历另两个数组,看HashMap中是否存在-sum
Take the arrays A and B, and compute all the possible sums of two elements. Put the sum in the Hash map, and increase the hash map value if more than 1 pair sums to the same value.
Compute all the possible sums of the arrays C and D. If the hash map contains the opposite value of the current sum, increase the count of four elements sum to 0 by the counter in the map.
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
HashMap<Integer, Integer> map = new HashMap<>();
int count = 0;
for (int i = 0; i < A.length; i++) {
for (int j = 0; j < B.length; j++) {
int sum = A[i] + B[j];
map.put(sum, map.getOrDefault(sum, 0) + 1);
}
}
for (int k = 0; k < C.length; k++) {
for (int l = 0; l < D.length; l++) {
int sum = (C[k] + D[l]);
count += map.getOrDefault(-sum, 0);
}
}
return count;
}
}
// Time complexity: O(n^2)
// Space complexity: O(n^2)

Reference

Last modified 3yr ago