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-rw-r--r-- | scratch/deepmind/part_two/find-duplicate-optimize-for-space.ts | 70 |
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}`); + } +} |