923. 3Sum With Multiplicity (Hash Table)

Author: cheehwatang

solutions/923. 3Sum With Multiplicity/ThreeSumWithMultiplicity_HashTable.java

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+package com.cheehwatang.leetcode;
+
+import java.util.HashMap;
+import java.util.Map;
+
+// Time Complexity  : O(n^2),
+// where 'n' is the length of 'arr'.
+// We traverse 'arr' in a nested for-loop, checking each triplet for each 'arr[i]'.
+//
+// Space Complexity : O(n^2),
+// where 'n' is the length of 'arr'.
+// For each 'arr[i]', there are 'n' number of possible differences to store in the HashTable.
+
+public class ThreeSumWithMultiplicity_HashTable {
+
+    // Approach:
+    // Using a HashTable to record "target - arr[k]" of the equation, then traverse both 'i' and 'j' to check if equal.
+    // Note that we rearranged the equation as arr[i] + arr[j] == target - arr[k].
+    // Each loop recording the frequency of "target - arr[k]", add the frequency if equal.
+
+    public int threeSumMulti(int[] arr, int target) {
+        int n = arr.length;
+        // Store the first value of 'arr[k]' which is arr[n - 1], starting from right to left each iteration.
+        Map<Integer, Integer> map = new HashMap<>();
+        map.put(target - arr[n - 1], 1);
+        long count = 0;
+        for (int j = n - 2; j >= 0; j--) {
+            // Add the frequency of "target - arr[k]" to the count if found.
+            for (int i = j - 1; i >= 0; i--) {
+                count += map.getOrDefault(arr[i] + arr[j], 0);
+            }
+            // Once done with all the 'j' for this loop, increase the frequency for "target - arr[j]".
+            int difference = target - arr[j];
+            map.put(difference, map.getOrDefault(difference, 0) + 1);
+        }
+        return (int) (count % (1e9 + 7));
+    }
+}