diff --git a/README.md b/README.md
index 83c3c50..b6a2e0e 100644
--- a/README.md
+++ b/README.md
@@ -27,6 +27,18 @@
| Solution |
Topics |
+
+ | September 2nd |
+ 923. 3Sum With Multiplicity |
+ $\text{\color{Dandelion}Medium}$ |
+
+ Counting
+ |
+
+ Array,
+ Counting
+ |
+
| September 1st |
16. 3Sum Closest |
@@ -76,19 +88,6 @@
Array
-
- | August 28th |
- 3. Longest Substring Without Repeating Characters |
- $\text{\color{Dandelion}Medium}$ |
-
- Hash Table & Sliding Window
- |
-
- Hash Table,
- Sliding Window,
- String
- |
-
@@ -952,6 +951,19 @@
|
+
+ | 923 |
+ 3Sum With Multiplicity |
+
+ Java
+ |
+ $\text{\color{Dandelion}Medium}$ |
+
+ Array,
+ Counting
+ |
+ |
+
| 929 |
Unique Email Addresses |
@@ -3270,6 +3282,19 @@
Hash Table
+
+ | 923 |
+ 3Sum With Multiplicity |
+
+ Java
+ |
+ $\text{\color{Dandelion}Medium}$ |
+
+ Array,
+ Counting
+ |
+ |
+
| 1497 |
Check If Array Pairs Are Divisible by k |
diff --git a/solutions/923. 3Sum With Multiplicity/ThreeSumWithMultiplicity_Counting.java b/solutions/923. 3Sum With Multiplicity/ThreeSumWithMultiplicity_Counting.java
new file mode 100644
index 0000000..dd8157e
--- /dev/null
+++ b/solutions/923. 3Sum With Multiplicity/ThreeSumWithMultiplicity_Counting.java
@@ -0,0 +1,53 @@
+package com.cheehwatang.leetcode;
+
+// Time Complexity : O(n),
+// where 'n' is the length of 'arr'.
+// We traverse 'arr' once to count the frequency of the elements.
+//
+// Space Complexity : O(1),
+// as the auxiliary space used is independent of the input 'arr',
+// with the counting array with size of 101 regardless of the input 'arr'.
+
+public class ThreeSumWithMultiplicity_Counting {
+
+ // Approach:
+ // Using a counting array, get the frequency of all the integers,
+ // then check all the possible combinations to get the sum.
+ // Note that, i < j < k and arr[i] + arr[j] + arr[k] == target,
+ // do not require the index as the sum is not dependent on the index.
+
+ public int threeSumMulti(int[] arr, int target) {
+
+ // With the constraint of 0 <= arr[i] <= 100, count the frequency of all the integers/
+ int[] counting = new int[101];
+ for (int integer : arr) counting[integer]++;
+
+ // Keep the count as long type, so as only need to perform the modulo once at the end.
+ long count = 0;
+
+ // Traverse the counting array in two for-loops to get the value for 'i' and 'j'.
+ for (int i = 0; i <= 100; i++) {
+ for (int j = i; j <= 100; j++) {
+ // Get the k integer to check if it is in the counting array.
+ int k = target - i - j;
+ if (k < 0 || k > 100) continue;
+ if (counting[k] > 0) {
+ // If the triplet is of the same integers (eg. [1,1,1], target = 3),
+ // then perform geometric sum for 3 elements.
+ if (i == k && j == k)
+ count += (long) counting[i] * (counting[i] - 1) * (counting[i] - 2) / 6;
+ // If i and j is identical, but different from k (eg. [1,1,2], target = 4),
+ // then perform geometric sum for i & j, then multiply with k.
+ else if (i == j)
+ count += (long) counting[i] * (counting[i] - 1) / 2 * counting[k];
+ // If both i and j is smaller than k, then get the multiples of all the frequency.
+ // This is to make sure no duplicates of results
+ // (eg, [1,2,3], target = 6), only [1,2,3] is counted, but not [2,1,3] or [3,2,1], etc.)
+ else if (i < k && j < k)
+ count += (long) counting[i] * counting[j] * counting[k];
+ }
+ }
+ }
+ return (int) (count % (1e9 + 7));
+ }
+}