[JavaScript 刷题] Code Signal - 相似数组(Are Similar?)

Posted 2021-09-29 GoldenaArcher

[javascript 刷题] Code Signal - 相似数组(Are Similar?)

题目地址：Are Similar?

题目

如下：

Two arrays are called similar if one can be obtained from another by swapping at most one pair of elements in one of the arrays.

Given two arrays a and b, check whether they are similar.

Example:

• For a = [1, 2, 3] and b = [1, 2, 3], the output should be areSimilar(a, b) = true.

  The arrays are equal, no need to swap any elements.

• For a = [1, 2, 3] and b = [2, 1, 3], the output should be areSimilar(a, b) = true.

  We can obtain b from a by swapping 2 and 1 in b.

• For a = [1, 2, 2] and b = [2, 1, 1], the output should be areSimilar(a, b) = false.

  Any swap of any two elements either in a or in b won’t make a and b equal.

Input/Output:

• [execution time limit] 4 seconds (js)

• [input] array.integer a

  Array of integers.

  Guaranteed constraints:

  3 ≤ a.length ≤ 105,

  1 ≤ a[i] ≤ 1000.

• [input] array.integer b

  Array of integers of the same length as a.

  Guaranteed constraints:

  b.length = a.length,

  1 ≤ b[i] ≤ 1000.

• [output] boolean

  true if a and b are similar, false otherwise.

解题思路

解题思路和 近乎增长序列(almostIncreasingSequence) 很相似，不过有点稍微变种。

1. 因为先决条件最多只能调换 一个数组中 的 一对元素，所以二者之间的不同只能为 0(不需要调换)，或者 2(调换一对)

2. 当最大不同为 0 时，代表数组完全相等，返回 true 即可

3. 当最大不同为 2 时，因为只需要交换一个数组中的两个元素，所以只要满足下列条件即可：

   $a[y]=b[x],a[x]=b[y]$

这样跑下来的时间复杂度是 $O(n)$，需要遍历一次所有的数组，空间复杂度为 $O(n)$，需要保存所有不同数。当然，后者也可以被优化到 $O(1)$，只需要当 包含所有不同的数组 长度超过 2 时直接返回即可，这样最大也就需要 3 个额外空间去进行存储。

使用 JavaScript 解题

// swapping at most one pair of elements in one of the arrays.
function areSimilar(a, b) {
  const maxDiff = maxDiffs(a, b);

  if (maxDiff.length > 2 || maxDiff.length === 1) return false;

  if (maxDiff.length === 0) return true; // identical

  const [diff1, diff2] = maxDiff;

  return a[diff1] === b[diff2] && a[diff2] === b[diff1];
}

const maxDiffs = (a, b) => {
  let maxDiff = [];
  for (let i = 0; i < a.length; i++) {
    if (a[i] !== b[i]) {
      maxDiff.push(i);
    }
  }

  return maxDiff;
};

console.log(areSimilar([1, 2, 3], [1, 2, 3]));