WIP: Partially solve InterviewCake's find duplicate number

Write a function that finds one duplicate number from a list of numbers 1..n.

The function should satisfy the following performance objectives:
Runtime complexity: O(n*log(n))
Space complexity:   O(1)

1 files changed, 70 insertions, 0 deletions

diff --git a/scratch/deepmind/part_two/find-duplicate-optimize-for-space.ts b/scratch/deepmind/part_two/find-duplicate-optimize-for-space.ts
new file mode 100644
index 000000000000..98f5bb144e76
--- /dev/null
+++ b/scratch/deepmind/part_two/find-duplicate-optimize-for-space.ts
@@ -0,0 +1,70 @@
+function findRepeatBruteForce(xs: Array<number>): number {
+  // InterviewCake asks us to write a function that optimizes for space. Using
+  // brute force, we can write a function that returns an answer using constant
+  // (i.e. O(1)) space at the cost of a quadratic (i.e. O(n^2)) runtime.
+  //
+  // I did not think of this myself; InterviewCake's "Tell me more" hints
+  // did. Since I think this idea is clever, I wrote a solution from memory to
+  // help me internalize the solution.
+  for (let i = 0; i < xs.length; i += 1) {
+    let seeking = xs[i];
+    for (let j = i + 1; j < xs.length; j += 1) {
+      if (xs[j] === seeking) {
+        return seeking;
+      }
+    }
+  }
+}
+
+function findRepeatSort(xs: Array<number>): number {
+  // This version first sorts xs, which gives the function a time-complexity of
+  // O(n*log(n)), which is better than the quadratic complexity of the
+  // brute-force solution. The space requirement here is constant.
+  //
+  // Since we need to sort xs in-place to avoid paying a O(n) space cost for
+  // storing the newly sorted xs, we're mutating our input. InterviewCake
+  // advises us to not mutate our input.
+  xs.sort();
+  let i = 0;
+  let j = 1;
+  for (; j < xs.length; ) {
+    if (xs[i] === xs[j]) {
+      return xs[i];
+    }
+    i += 1;
+    j += 1;
+  }
+}
+
+function findRepeat(xs: Array<number>): number {
+  return 0;
+}
+
+// Tests
+let desc = "just the repeated number";
+let actual = findRepeat([1, 1]);
+let expected = 1;
+assertEqual(actual, expected, desc);
+
+desc = "short array";
+actual = findRepeat([1, 2, 3, 2]);
+expected = 2;
+assertEqual(actual, expected, desc);
+
+desc = "medium array";
+actual = findRepeat([1, 2, 5, 5, 5, 5]);
+expected = 5;
+assertEqual(actual, expected, desc);
+
+desc = "long array";
+actual = findRepeat([4, 1, 4, 8, 3, 2, 7, 6, 5]);
+expected = 4;
+assertEqual(actual, expected, desc);
+
+function assertEqual(a, b, desc) {
+  if (a === b) {
+    console.log(`${desc} ... PASS`);
+  } else {
+    console.log(`${desc} ... FAIL: ${a} != ${b}`);
+  }
+}