**Sum of Four Numbers Less Than Target - Amazon Top Interview Questions**

### Problem Statement :

You are given four lists of integers A, B, C, D and an integer target. Return the number of different unique indices i, j, k, l such that A[i] + B[j] + C[k] + D[l] ≤ target. Constraints a ≤ 1,000 where a is the length of A b ≤ 1,000 where b is the length of B c ≤ 1,000 where c is the length of C d ≤ 1,000 where d is the length of D Example 1 Input A = [2, 3] B = [5, 2] C = [0] D = [1, 2] target = 6 Output 3 Explanation We can pick the following combinations: [2, 2, 0, 1] [2, 2, 0, 2] [3, 2, 0, 1] Example 2 Input A = [1, 1] B = [0] C = [0] D = [0] target = 1 Output 2

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D, int target) {
vector<int> v1;
vector<int> v2;
int res = 0;
for (int i : A) {
for (int j : B) {
v1.push_back(i + j);
}
}
for (int i : C) {
for (int j : D) {
v2.push_back(i + j);
cout << (i + j) << endl;
}
}
sort(v2.begin(), v2.end());
for (int i : v1) {
int l = 0, r = v2.size() - 1;
int pos = -1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (v2[mid] + i <= target) {
pos = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
res += pos + 1;
}
return res;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
private int ans = 0;
public int solve(int[] A, int[] B, int[] C, int[] D, int target) {
int[] prefixsumArr = new int[1005];
Map<Integer, Integer> map = new HashMap();
int idx = 0;
for (int i = 0; i < A.length; i++) {
for (int j = 0; j < B.length; j++) {
int val = A[i] + B[j];
map.put(val, map.getOrDefault(val, 0) + 1);
}
}
for (int key : map.keySet()) {
prefixsumArr[key] = map.get(key);
}
for (int i = 1; i < prefixsumArr.length; i++)
prefixsumArr[i] = prefixsumArr[i] + prefixsumArr[i - 1];
for (int i = 0; i < C.length; i++) {
for (int j = 0; j < D.length; j++) {
int val = C[i] + D[j];
int remain = target - val;
if (remain >= 0)
ans += prefixsumArr[remain];
}
}
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, A, B, C, D, target):
ab = sorted(a + b for a in A for b in B)
return sum(bisect_right(ab, target - c - d) for c in C for d in D)
```

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