Taking a O(n⁴) into a O(n²)
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 -228 to 228–1 and the result is guaranteed to be at most 231–1.
Example:
Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]Output:
2Explanation:
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
Results:
Runtime: 67 ms, faster than 34.54% of Java online submissions for 4Sum II.Memory Usage: 60 MB, less than 22.36% of Java online submissions for 4Sum II.
Solution:
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
//A + B = -C - D
int count=0;
HashMap<Integer, Integer> AB=new HashMap<Integer, Integer>();
for(int a:A){
for(int b: B){
if(AB.get(a+b)==null ){
AB.put(a+b,1);
} else{
AB.put(a+b, AB.get(a+b)+1 );
}
}
}
for(int c:C){
for(int d:D){
if( AB.get(-c-d)!=null ){
count+=AB.get(-c-d);
}
}
}
return count;
}
}
