4Sum II
Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l)
there are such that A[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
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
int n = A.length;
HashMap<Integer, Integer> map = new HashMap<>();
for (int a : A) {
for (int b : B) {
int sum = a + b;
if (!map.containsKey(sum))
map.put(sum, 1);
else
map.put(sum, map.get(sum) + 1);
}
}
int count = 0;
for (int c : C) {
for (int d : D) {
int sum = c + d;
if (map.containsKey(-sum))
count += map.get(-sum);
}
}
return count;
}
}
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