【Codility Lesson3】FrogJmp

Codilityの勧め ~JavaScriptで解くアルゴリズム~の実践編です。

問題

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

Write a function:

function solution(X, Y, D);

that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:

X = 10
Y = 85
D = 30
the function should return 3, because the frog will be positioned as follows:

after the first jump, at position 10 + 30 = 40
after the second jump, at position 10 + 30 + 30 = 70
after the third jump, at position 10 + 30 + 30 + 30 = 100
Write an efficient algorithm for the following assumptions:

X, Y and D are integers within the range [1..1,000,000,000];
X ≤ Y.

語彙メモ

• located at position X 位置Xにあります
• greater than or equal to Y Y以上
• must perform to reach its target. ターゲットに到達するために実行しなければならない
• Write an efficient algorithm for the following assumptions: (:以下の)仮定のための効率的なアルゴリズムを書け

解法

必要知識

• parseInt()
• 三項演算子

太字にしているものは初見でした。

function solution(X, Y, D) {
  // write your code in JavaScript (Node.js 8.9.4)
  if (X >= Y) {
    return 0;
  } else if (D >= X + Y) {
    return 1;
  } else {
    let minimalJumps = parseInt((Y - X) / D);
    minimalJumps += (Y - X) % D > 0 ? 1 : 0;
    return minimalJumps;
  }
}

比較的簡単でした。

参考

アルゴリズム図鑑 絵で見てわかる26のアルゴリズム

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