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Removed fence mark in explanations.
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en/1-1000/1-two-sum.md

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@@ -20,9 +20,7 @@ You can return the answer in any order.
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**Explanation**:
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```
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Because nums[0] + nums[1] == 9, we return [0, 1].
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```
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### [Example 2]
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en/1-1000/15-3sum.md

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**Explanation**:
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```
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nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
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nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
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nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
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The distinct triplets are [-1,0,1] and [-1,-1,2].
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Notice that the order of the output and the order of the triplets does not matter.
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```
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### [Example 2]
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**Explanation**:
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```
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The only possible triplet does not sum up to 0.
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```
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### [Example 3]
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en/1-1000/202-happy-number.md

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**Explanation**:
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```
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1^2 + 9^2 = 82
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8^2 + 2^2 = 68
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6^2 + 8^2 = 100
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1^2 + 0^2 + 0^2 = 1
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```
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### [Example 2]
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en/1-1000/209-minimum-size-subarray-sum.md

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**Explanation**:
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```
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The subarray [4,3] has the minimal length under the problem constraint.
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```
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### [Example 2]
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en/1-1000/27-remove-element.md

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**Explanation**:
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```
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Your function should return k = 2, with the first two elements of nums being 2.
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It does not matter what you leave beyond the returned k (hence they are underscores).
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```
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### [Example 2]
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**Explanation**:
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```
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Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
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Note that the five elements can be returned in any order.
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It does not matter what you leave beyond the returned k (hence they are underscores).
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```
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### [Constraints]
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en/1-1000/28-find-the-index-of-the-first-occurrence-in-a-string.md

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**Explanation**:
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```
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"sad" occurs at index 0 and 6.
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The first occurrence is at index 0, so we return 0.
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```
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### [Example 2]
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**Explanation**:
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```
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"leeto" did not occur in "leetcode", so we return -1.
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```
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### [Constraints]
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en/1-1000/303-range-sum-query-immutable.md

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**Explanation**:
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```
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NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
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numArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1
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numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1
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numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3
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```
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### [Constraints]
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en/1-1000/454-4sum-ii.md

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**Explanation**:
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```
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The two tuples are:
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1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
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2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
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```
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### [Example 2]
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en/1-1000/707-design-linked-list.md

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**Explanation**:
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```
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MyLinkedList myLinkedList = new MyLinkedList();
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myLinkedList.addAtHead(1);
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myLinkedList.addAtTail(3);
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myLinkedList.addAtIndex(1, 2); // linked list becomes 1->2->3
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myLinkedList.get(1); // return 2
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myLinkedList.deleteAtIndex(1); // now the linked list is 1->3
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myLinkedList.get(1); // return 3
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```
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### [Constraints]
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en/1-1000/833-find-and-replace-in-string.md

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**Explanation**:
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```
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"a" occurs at index 0 in s, so we replace it with "eee".
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"cd" occurs at index 2 in s, so we replace it with "ffff".
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```
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### [Example 2]
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**Explanation**:
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```
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"ab" occurs at index 0 in s, so we replace it with "eee".
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"ec" does not occur at index 2 in s, so we do nothing.
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```
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### [Constraints]
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