Welcome to Subscribe On Youtube

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.

  • 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;
    }
}

############

class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int threeSumMulti(int[] arr, int target) {
        int[] cnt = new int[101];
        for (int v : arr) {
            ++cnt[v];
        }
        long ans = 0;
        for (int j = 0; j < arr.length; ++j) {
            int b = arr[j];
            --cnt[b];
            for (int i = 0; i < j; ++i) {
                int a = arr[i];
                int c = target - a - b;
                if (c >= 0 && c <= 100) {
                    ans = (ans + cnt[c]) % MOD;
                }
            }
        }
        return (int) ans;
    }
}

  • // 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:
    def threeSumMulti(self, arr: List[int], target: int) -> int:
        cnt = Counter(arr)
        ans = 0
        mod = 10**9 + 7
        for j, b in enumerate(arr):
            cnt[b] -= 1
            for i in range(j):
                a = arr[i]
                c = target - a - b
                ans = (ans + cnt[c]) % mod
        return ans

############

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

  • func threeSumMulti(arr []int, target int) int {
	const mod int = 1e9 + 7
	cnt := [101]int{}
	for _, v := range arr {
		cnt[v]++
	}
	ans := 0
	for j, b := range arr {
		cnt[b]--
		for i := 0; i < j; i++ {
			a := arr[i]
			c := target - a - b
			if c >= 0 && c <= 100 {
				ans += cnt[c]
				ans %= mod
			}
		}
	}
	return ans
}

All Problems

All Solutions