LeetCode 923. 3Sum With Multiplicity

原题链接在这里：https://leetcode.com/problems/3sum-with-multiplicity/

题目：

Given an integer array A, and an integer target, return the number of tuples i, j, k  such that i < j < k and A[i] + A[j] + A[k] == target.

As the answer can be very large, return it modulo 10^9 + 7.

Example 1:

Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (A[i], A[j], A[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: A = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
A[i] = 1, A[j] = A[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.

Note:

1. 3 <= A.length <= 3000
2. 0 <= A[i] <= 100
3. 0 <= target <= 300

题解：

Perform 3Sum and find the target.

Here there are 2 cases:

1. A[l] != A[r], count the frequency of A[l] and A[r], possible number of tuples is leftCount*rightCount.

e.g. 2333, A[l] = 2, A[r] = 3, leftCount = 1, rightCount = 3, number of types is 1*3 = 3.

2. A[l] == A[r], number of tupes is count*(count-1)/2.

e.g. 222, A[l] = 2, A[r] = 2, number of tupes 3*2/2 = 3. There are 3 ways to pick 2 elements from it.

Time Compelxity: O(n^2). n = A.length.

Space: O(1).

AC Java:

class Solution {
    public int threeSumMulti(int[] A, int target) {
        if(A == null || A.length == 0){
            return 0;
        }
        
        int res = 0;
        int M = 1000000000+7;
        Arrays.sort(A);
        for(int i = 0; i<A.length-2; i++){
            int l = i+1;
            int r = A.length-1;
            while(l<r){
                int sum = A[i] + A[l] + A[r];
                if(sum < target){
                    l++;
                }else if(sum > target){
                    r--;
                }else if(A[l] != A[r]){
                    int leftCount = 1;
                    int rightCount = 1;
                    while(l+1<r && A[l]==A[l+1]){
                        leftCount++;
                        l++;
                    }
                    
                    while(l+1<r && A[r]==A[r-1]){
                        rightCount++;
                        r--;
                    }
                    
                    res += leftCount*rightCount;
                    res = res%M;
                    l++;
                    r--;
                }else{
                    // A[l] == A[r]
                    res += (r-l+1)*(r-l)/2;
                    res = res%M;
                    break;
                }
            }
        }
        
        return res;
    }
}