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feat: add LeetCode problem 540 (#1217)
* feat: add LeetCode problem 540 * feat: Added a description to the LeetCode problem 540 * feat: Added details in the description of the LeetCode problem 540 * feat: Changed a word in @details of the LeetCode problem 540
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| 485 | [Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones) | [C](./src/485.c) | Easy |
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| 485 | [Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones) | [C](./src/485.c) | Easy |
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| 509 | [Fibonacci Number](https://leetcode.com/problems/fibonacci-number) | [C](./src/509.c) | Easy |
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| 509 | [Fibonacci Number](https://leetcode.com/problems/fibonacci-number) | [C](./src/509.c) | Easy |
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| 520 | [Detect Capital](https://leetcode.com/problems/detect-capital) | [C](./src/520.c) | Easy |
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| 520 | [Detect Capital](https://leetcode.com/problems/detect-capital) | [C](./src/520.c) | Easy |
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| 540 | [Single Element in a Sorted Array](https://leetcode.com/problems/single-element-in-a-sorted-array/) | [C](./src/540.c) | Medium |
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| 561 | [Array Partition](https://leetcode.com/problems/array-partition) | [C](./src/561.c) | Easy |
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| 561 | [Array Partition](https://leetcode.com/problems/array-partition) | [C](./src/561.c) | Easy |
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| 567 | [Permutation in String](https://leetcode.com/problems/permutation-in-string) | [C](./src/567.c) | Medium |
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| 567 | [Permutation in String](https://leetcode.com/problems/permutation-in-string) | [C](./src/567.c) | Medium |
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| 617 | [Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees) | [C](./src/617.c) | Easy |
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| 617 | [Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees) | [C](./src/617.c) | Easy |
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leetcode/src/540.c
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leetcode/src/540.c
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/**
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* Time complexity: O(log n).
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* Space complexity: O(1).
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* @details The array has a pattern that consists in of the existing sub-array to
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* the left of the non-repeating number will satisfy the condition that
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* each pair of repeated elements have their first occurrence at the even index
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* and their second occurrence at the odd index, and that the sub-array to
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* the right of the non-repeating number will satisfy the condition that
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* each pair of repeated elements have their first occurrence at the odd index
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* and their second occurrence at the even index. With this pattern in mind,
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* we can solve the problem using binary search.
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*/
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int singleNonDuplicate(int* nums, int numsSize) {
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int left = 0, right = numsSize - 1;
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while (left < right) {
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int mid = (right + left) / 2;
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if (mid % 2 == 0) {
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if (nums[mid] == nums[mid + 1])
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left = mid + 2;
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else
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right = mid;
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}
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else {
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if (nums[mid] == nums[mid - 1])
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left = mid + 1;
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else
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right = mid - 1;
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}
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}
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return nums[left];
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}
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