Problem
Given an integer array arr, and an integer target, return the number of tuples i, j, k such that i < j < k and arr[i] + arr[j] + arr[k] == target.
As the answer can be very large, return it modulo 10^9 + 7.
Example 1:
Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation:
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.
Example 2:
Input: arr = [1,1,2,2,2,2], target = 5
Output: 12
Explanation:
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.
Example 3:
Input: arr = [2,1,3], target = 6
Output: 1
Explanation: (1, 2, 3) occured one time in the array so we return 1.
Constraints:
3 <= arr.length <= 30000 <= arr[i] <= 1000 <= target <= 300
Solution (Java)
class Solution {
private static final int MOD = (int) 1e9 + 7;
private static final int MAX = 100;
public int threeSumMulti(int[] arr, int target) {
int answer = 0;
int[] countRight = new int[MAX + 1];
for (int num : arr) {
++countRight[num];
}
int[] countLeft = new int[MAX + 1];
for (int j = 0; j < arr.length - 1; j++) {
--countRight[arr[j]];
int remains = target - arr[j];
if (remains <= 2 * MAX) {
for (int v = 0; v <= Math.min(remains, MAX); v++) {
if (remains - v <= MAX) {
int count = countRight[v] * countLeft[remains - v];
if (count > 0) {
answer = (answer + count) % MOD;
}
}
}
}
++countLeft[arr[j]];
}
return answer;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).