Add #0015 3Sum problem and tests

Author: o-p

difficulty/medium/0015 3Sum

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+../../problems/0015-3sum

list/Blind Curated 75/05. 3Sum

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+../../problems/0015-3sum

problems/0015-3sum/3sum.test.ts

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+import threeSum from './3sum';
+
+test.each`
+  nums                     | expected
+  ${[-1, 0, 1, 2, -1, -4]} | ${[[-1, -1, 2], [-1, 0, 1]]} }
+  ${[]}                    | ${[]}
+  ${[0]}                   | ${[]}
+  ${[1, -1]}               | ${[]}
+`('3Sum: ($nums)', ({ nums, expected }) =>
+  expect(threeSum(nums)).toEqual(expected)
+);

problems/0015-3sum/3sum.ts

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+/**
+ * 簡單想法：
+ *
+ *   1. 總和為 0 一定是:
+ *      - 0 + 0 + 0
+ *      - +N + -X + -Y (其中 N = X + Y，並且 X 與 Y 中必定有一項 <= N/2)
+ *      - -N + +X + +Y (其中 N = X + Y，並且 X 與 Y 中必定有一項 <= N/2)
+ */
+function threeSum(nums: number[]): number[][] {
+  /** 各數字出現量 */
+  const counts = nums.reduce(
+    (all, val) => Object.assign(all, {
+      [val]: (all[val] ?? 0) + 1,
+    }), {}
+  );
+
+  const numbers = Object.keys(counts).map(n => parseInt(n, 10));
+  const postives = numbers.filter(n => n > 0);
+  const negatives = numbers.filter(n => n < 0);
+
+  const matched = postives.reduce(
+    (all: number[][], pos: number) => negatives
+      .filter(neg => pos + 2 * neg <= 0 && pos + neg >= 0) // 這邊接受 -N 0 N 的 case
+      .map(neg => [neg, -pos - neg, pos])
+      .filter(([a, b, c]) => a === b ? counts[b] >= 2 : counts[b] >= 1)
+      .concat(all),
+    counts[0] >= 3
+      ? [[0, 0, 0]]
+      : []
+  );
+
+  return negatives.reduce(
+    (all: number[][], neg: number) => postives
+      .filter(pos => pos * 2 + neg >= 0 && pos + neg < 0) // 這邊不收 N 0 -N 的 case
+      .map(pos => [neg, -pos - neg, pos])
+      .filter(([a, b, c]) => b === c ? counts[b] >= 2 : counts[b] >= 1)
+      .concat(all),
+    matched,
+  );
+}
+
+export default threeSum