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Commit 29ac15a

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Merge pull request #324 from cheehwatang/add-923-3SumWithMultiplicity
Add 923. 3Sum With Multiplicity (Hash Table)
2 parents a4c50bb + 333ee50 commit 29ac15a

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‎README.md‎

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<th>Solution</th>
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<th>Topics</th>
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</tr>
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<tr>
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<td align="center">September 3rd</td>
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<td>923. <a href="https://leetcode.com/problems/3sum-with-multiplicity/">3Sum With Multiplicity</a></td>
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<td align="center">$\text{\color{Dandelion}Medium}$</td>
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<td align="center">
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<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/923.%203Sum%20With%20Multiplicity/ThreeSumWithMultiplicity_HashTable.java">Hash Table</a>
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</td>
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<td align="center">
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<a href="#array">Array</a>,
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<a href="#hash-table">Hash Table</a>
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</td>
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</tr>
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<tr>
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<td align="center">September 2nd</td>
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<td>923. <a href="https://leetcode.com/problems/3sum-with-multiplicity/">3Sum With Multiplicity</a></td>
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<tr>
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<td align="center">923</td>
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<td><a href="https://leetcode.com/problems/3sum-with-multiplicity/">3Sum With Multiplicity</a></td>
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<td align="center">
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<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/923.%203Sum%20With%20Multiplicity/ThreeSumWithMultiplicity_Counting.java">Java</a>
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<td align="center">Java with
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<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/923.%203Sum%20With%20Multiplicity/ThreeSumWithMultiplicity_Counting.java">Counting</a> or
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<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/923.%203Sum%20With%20Multiplicity/ThreeSumWithMultiplicity_HashTable.java">Hash Table</a>
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<td align="center">$\text{\color{Dandelion}Medium}$</td>
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<td align="center">
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<a href="#array">Array</a>,
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<a href="#counting">Counting</a>
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<a href="#counting">Counting</a>,
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<a href="#hash-table">Hash Table</a>
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</td>
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<td></td>
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<td align="center">$\text{\color{Dandelion}Medium}$</td>
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<a href="#array">Array</a>,
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<a href="#counting">Counting</a>
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<a href="#counting">Counting</a>,
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<a href="#hash-table">Hash Table</a>
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</td>
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<td>Solution Using
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<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/923.%203Sum%20With%20Multiplicity/ThreeSumWithMultiplicity_HashTable.java"><em>Hash Table</em></a>
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<td></td>
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<td align="center">1497</td>
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</td>
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<td></td>
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</tr>
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<tr>
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<td align="center">923</td>
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<td><a href="https://leetcode.com/problems/3sum-with-multiplicity/">3Sum With Multiplicity</a></td>
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<td align="center">
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<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/923.%203Sum%20With%20Multiplicity/ThreeSumWithMultiplicity_HashTable.java">Java</a>
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</td>
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<td align="center">$\text{\color{Dandelion}Medium}$</td>
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<td align="center">
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<a href="#array">Array</a>,
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<a href="#counting">Counting</a>,
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<a href="#hash-table">Hash Table</a>
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</td>
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<td>Solution Using
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<a href="https://github.com/cheehwatang/leetcode-java/blob/main/solutions/923.%203Sum%20With%20Multiplicity/ThreeSumWithMultiplicity_Counting.java"><em>Counting</em></a>
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</td>
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</tr>
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<td align="center">929</td>
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<td><a href="https://leetcode.com/problems/unique-email-addresses/">Unique Email Addresses</a></td>
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package com.cheehwatang.leetcode;
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import java.util.HashMap;
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import java.util.Map;
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// Time Complexity : O(n^2),
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// where 'n' is the length of 'arr'.
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// We traverse 'arr' in a nested for-loop, checking each triplet for each 'arr[i]'.
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//
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// Space Complexity : O(n^2),
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// where 'n' is the length of 'arr'.
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// For each 'arr[i]', there are 'n' number of possible differences to store in the HashTable.
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public class ThreeSumWithMultiplicity_HashTable {
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// Approach:
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// Using a HashTable to record "target - arr[k]" of the equation, then traverse both 'i' and 'j' to check if equal.
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// Note that we rearranged the equation as arr[i] + arr[j] == target - arr[k].
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// Each loop recording the frequency of "target - arr[k]", add the frequency if equal.
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public int threeSumMulti(int[] arr, int target) {
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int n = arr.length;
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// Store the first value of 'arr[k]' which is arr[n - 1], starting from right to left each iteration.
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Map<Integer, Integer> map = new HashMap<>();
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map.put(target - arr[n - 1], 1);
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long count = 0;
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for (int j = n - 2; j >= 0; j--) {
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// Add the frequency of "target - arr[k]" to the count if found.
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for (int i = j - 1; i >= 0; i--) {
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count += map.getOrDefault(arr[i] + arr[j], 0);
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}
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// Once done with all the 'j' for this loop, increase the frequency for "target - arr[j]".
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int difference = target - arr[j];
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map.put(difference, map.getOrDefault(difference, 0) + 1);
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}
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return (int) (count % (1e9 + 7));
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}
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}

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