最美情侣中文字幕电影,在线麻豆精品传媒,在线网站高清黄,久久黄色视频

歡迎光臨散文網(wǎng) 會(huì)員登陸 & 注冊(cè)

LeetCode 1814. Count Nice Pairs in an Array

2023-04-11 12:24 作者:您是打尖兒還是住店呢  | 我要投稿

You are given an array?nums?that consists of non-negative integers. Let us define?rev(x)?as the reverse of the non-negative integer?x. For example,?rev(123) = 321, and?rev(120) = 21. A pair of indices?(i, j)?is?nice?if it satisfies all of the following conditions:

  • 0 <= i < j < nums.length

  • nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])

Return?the number of nice pairs of indices. Since that number can be too large, return it?modulo?109?+ 7.

?

Example 1:

Input: nums = [42,11,1,97]

Output: 2

Explanation:?

The two pairs are:?

- (0,3) : 42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121.?

- (1,2) : 11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12.

Example 2:

Input: nums = [13,10,35,24,76]

Output: 4

?

Constraints:

  • 1 <= nums.length <= 105

  • 0 <= nums[i] <= 109

hint 1:

The condition can be rearranged to (nums[i] - rev(nums[i])) == (nums[j] - rev(nums[j])).

hint 2:

Transform each nums[i] into (nums[i] - rev(nums[i])). Then, count the number of (i, j) pairs that have equal values.

hint 3:

Keep a map storing the frequencies of values that you have seen so far. For each i, check if nums[i] is in the map. If it is, then add that count to the overall count. Then, increment the frequency of nums[i].


Runtime:?98 ms, faster than?10.67%?of?Java?online submissions for?Count Nice Pairs in an Array.

Memory Usage:?55.7 MB, less than?59.67%?of?Java?online submissions for?Count Nice Pairs in an Array.


LeetCode 1814. Count Nice Pairs in an Array的評(píng)論 (共 條)

分享到微博請(qǐng)遵守國(guó)家法律
浙江省| 五莲县| 西充县| 湟源县| 常熟市| 马龙县| 桃园县| 图们市| 莱西市| 建水县| 焉耆| 红桥区| 泰和县| 巫山县| 剑阁县| 沙坪坝区| 盐边县| 桑植县| 天峨县| 全州县| 神农架林区| 澄迈县| 建始县| 江阴市| 东台市| 交城县| 牙克石市| 台南市| 博白县| 义马市| 霍山县| 固镇县| SHOW| 当雄县| 兖州市| 拜泉县| 宜川县| 清丰县| 论坛| 阜康市| 乐安县|