leetcode-problemset/leetcode-cn/originData/count-nice-pairs-in-an-array.json

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        "question": {
            "questionId": "1925",
            "questionFrontendId": "1814",
            "categoryTitle": "Algorithms",
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            "title": "Count Nice Pairs in an Array",
            "titleSlug": "count-nice-pairs-in-an-array",
            "content": "<p>You are given an array <code>nums</code> that consists of non-negative integers. Let us define <code>rev(x)</code> as the reverse of the non-negative integer <code>x</code>. For example, <code>rev(123) = 321</code>, and <code>rev(120) = 21</code>. A pair of indices <code>(i, j)</code> is <strong>nice</strong> if it satisfies all of the following conditions:</p>\n\n<ul>\n\t<li><code>0 &lt;= i &lt; j &lt; nums.length</code></li>\n\t<li><code>nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])</code></li>\n</ul>\n\n<p>Return <em>the number of nice pairs of indices</em>. Since that number can be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [42,11,1,97]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The two pairs are:\n - (0,3) : 42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121.\n - (1,2) : 11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [13,10,35,24,76]\n<strong>Output:</strong> 4\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "统计一个数组中好对子的数目",
            "translatedContent": "<p>给你一个数组 <code>nums</code> ，数组中只包含非负整数。定义 <code>rev(x)</code> 的值为将整数 <code>x</code> 各个数字位反转得到的结果。比方说 <code>rev(123) = 321</code> ， <code>rev(120) = 21</code> 。我们称满足下面条件的下标对 <code>(i, j)</code> 是 <strong>好的</strong> ：</p>\n\n<ul>\n\t<li><code>0 &lt;= i &lt; j &lt; nums.length</code></li>\n\t<li><code>nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])</code></li>\n</ul>\n\n<p>请你返回好下标对的数目。由于结果可能会很大，请将结果对 <code>10<sup>9</sup> + 7</code> <b>取余</b> 后返回。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums = [42,11,1,97]\n<b>输出：</b>2\n<b>解释：</b>两个坐标对为：\n - (0,3)：42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121 。\n - (1,2)：11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums = [13,10,35,24,76]\n<b>输出：</b>4\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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            "hints": [
                "The condition can be rearranged to (nums[i] - rev(nums[i])) == (nums[j] - rev(nums[j])).",
                "Transform each nums[i] into (nums[i] - rev(nums[i])). Then, count the number of (i, j) pairs that have equal values.",
                "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]."
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