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923. 3Sum With Multiplicity.md

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923. 3Sum With Multiplicity

leetcode Daily Challenge on March 23th, 2021.

Difficulty : Medium

Related Topics : Two Pointers

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 109 + 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.

Constraints:

  • 3 <= arr.length <= 3000
  • 0 <= arr[i] <= 100
  • 0 <= target <= 300

Solution

  • mine
    • Java

      • Runtime: 2 ms, faster than 97.14%, Memory Usage: 38.2 MB, less than 97.50% of Java online submissions

        // O(N + W^2)time
// O(W) space
// W = 101
public int threeSumMulti(int[] A, int target) {
    long[] c = new long[101];
    for (int a : A) c[a]++;
    long res = 0;
    for (int i = 0; i <= 100; i++)
        for (int j = i; j <= 100; j++) {
            int k = target - i - j;
            if (k > 100 || k < 0) continue;
            if (i == j && j == k)
                //  3
                //C a
                res += c[i] * (c[i] - 1) * (c[i] - 2) / 6;
            else if (i == j && j != k)
                //  1     2
                //C a * C b
                res += c[i] * (c[i] - 1) / 2 * c[k];
            else if (j < k)
                //  1     1     1
                //C a * C b * C c
                res += c[i] * c[j] * c[k];
        }
    return (int)(res % (1e9 + 7));
}
  • the most votes
  • Runtime: 2 ms, faster than 97.14%, Memory Usage: 38.2 MB, less than 97.50% of Java online submissions 
    public int threeSumMulti(int[] A, int target) {
    long[] c = new long[101];
    for (int a : A) c[a]++;
    long res = 0;
    for (int i = 0; i <= 100; i++)
        for (int j = i; j <= 100; j++) {
            int k = target - i - j;
            if (k > 100 || k < 0) continue;
            if (i == j && j == k)
                res += c[i] * (c[i] - 1) * (c[i] - 2) / 6;
            else if (i == j && j != k)
                res += c[i] * (c[i] - 1) / 2 * c[k];
            else if (j < k)
                res += c[i] * c[j] * c[k];
        }
    return (int)(res % (1e9 + 7));
}