Codilityの勧め ~JavaScriptで解くアルゴリズム~の実践編です。
問題
A non-empty array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.
For example, in array A such that:
A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
the elements at indexes 0 and 2 have value 9,
the elements at indexes 1 and 3 have value 3,
the elements at indexes 4 and 6 have value 9,
the element at index 5 has value 7 and is unpaired.
Write a function:
function solution(A);
that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.
For example, given array A such that:
A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
the function should return 7, as explained in the example above.
Write an efficient algorithm for the following assumptions:
N is an odd integer within the range [1..1,000,000];
each element of array A is an integer within the range [1..1,000,000,000];
all but one of the values in A occur an even number of times.
語彙メモ
-
be paired with~と対になって -
unpairedペアになっていない -
indexes索引 ※indexは「索引」の意味の時の複数形はindexes、「指標」の意味の時はindicesになります
解法
必要知識
- アロー関数
- 高階関数
- Mapオブジェクト
- map()
- find()
- Object.keys()
- +演算子
- reduce()
太字にしているものは初見でした。
特にreduce()は最初は難しかったです。
function solution(A) {
const makeIntegerOcurrencesMap = (map, nextInteger) => {
const isMissingKey = (key) => map[key] === undefined;
if (isMissingKey(nextInteger)) {
map[nextInteger] = 0;
}
++map[nextInteger];
return map;
};
const isOdd = (number) => number % 2 !== 0;
const integerOccurrencesMap = A.reduce(makeIntegerOcurrencesMap, {});
return +Object.keys(integerOccurrencesMap).find((integer) =>
isOdd(integerOccurrencesMap[integer])
);
}
