Formatted question description: https://leetcode.ca/all/923.html

923. 3Sum With Multiplicity (Medium)

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

Companies:
Quora

Related Topics:
Two Pointers

Solution 1.


Java

  • class Solution {
        public int threeSumMulti(int[] A, int target) {
            final int MODULO = 1000000007;
            long count = 0;
            Arrays.sort(A);
            int length = A.length;
            int leftStart = 0, leftEnd = length - 3;
            for (int left = leftStart; left <= leftEnd; left++) {
                if (A[left] > target)
                    break;
                int remainTarget = target - A[left];
                int mid = left + 1, right = length - 1;
                while (mid < right) {
                    int sum = A[mid] + A[right];
                    if (sum == remainTarget) {
                        if (A[mid] != A[right]) {
                            int midCount = 1, rightCount = 1;
                            while (mid + 1 < right && A[mid] == A[mid + 1]) {
                                mid++;
                                midCount++;
                            }
                            while (right - 1 > mid && A[right] == A[right - 1]) {
                                right--;
                                rightCount++;
                            }
                            count += midCount * rightCount;
                            count %= MODULO;
                            mid++;
                            right--;
                        } else {
                            int curCount = right - mid + 1;
                            count += curCount * (curCount - 1) / 2;
                            count %= MODULO;
                            break;
                        }
                    } else if (sum < remainTarget)
                        mid++;
                    else
                        right--;
                }
            }
            return (int) count;
        }
    }
    
  • // OJ: https://leetcode.com/problems/3sum-with-multiplicity/solution/
    // Time: O(N^2)
    // Space: O(N)
    class Solution {
    public:
        int threeSumMulti(vector<int>& A, int target) {
            unordered_map<int, long> m;
            long ans = 0, mod = 1e9 + 7;
            for (int n : A) m[n]++;
            for (auto &[a, ca] : m) {
                for (auto &[b, cb] : m) {
                    int c = target - a - b;
                    if (m.count(c) == 0) continue;
                    int cc = m[c];
                    if (a == b && b == c) ans += ca * (ca - 1) * (ca - 2) / 6; // all three are equal
                    else if (a == b && b != c) ans += ca * (ca - 1) / 2 * cc; // first two are equal and the 3rd is not
                    else if (a < b && b < c) ans += ca * cb * cc; // all three are not equal
                }
            }
            return ans % mod;
        }
    };
    
  • class Solution(object):
        def threeSumMulti(self, A, target):
            """
            :type A: List[int]
            :type target: int
            :rtype: int
            """
            count = collections.Counter(A)
            Aset = set(A)
            Alist = list(Aset)
            Alist.sort()
            _lenA = len(Alist)
            res = 0
            for i in range(_lenA):
                for j in range(i, _lenA):
                    c = target - Alist[i] - Alist[j]
                    if c >= Alist[j] and c in Aset:
                        if Alist[i] != Alist[j] != c:
                            res += count[Alist[i]] * count[Alist[j]] * count[c]
                        elif Alist[i] == Alist[j] and Alist[j] != c:
                            res += count[c] * self.caculate(count[Alist[i]], 2)
                        elif Alist[j] == c and Alist[i] != Alist[j]:
                            res += count[Alist[i]] * self.caculate(count[Alist[j]], 2)
                        elif Alist[i] == c and Alist[j] != c:
                            res += count[Alist[j]] * self.caculate(count[c], 2)
                        else:
                            res += self.caculate(count[Alist[i]], 3)
            return res % (10 ** 9 + 7)
        
        def caculate(self, x, i):
            if i == 2:
                return x * (x - 1) / 2
            elif i == 3:
                return x * (x - 1) * (x - 2) / 6
    

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