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The time complexity of this program is O(n^2), where "n" represents the number of elements in the `nums`array. Here's the breakdown:
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1. Sorting: The program sorts the input array `nums`. Sorting typically takes O(n * log(n)) time complexity, where 'n' is the length of the array.
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1.The program starts by sorting the `nums` array, which has a time complexity of O(n * log(n)), where "n" is the length of the array.
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2. Main Loop: The program uses nested loops. The outer loop runs from 0 to `len - 2`, where `len` is the length of the sorted array `nums`. The inner while loop has two pointers (`l` and `r`) that move towards each other. In the worst case, the inner loop can go through the entire array, so its time complexity is O(n).
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2. After sorting, the program uses nested loops:
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- The outer loop runs for `i` from 0 to `len - 2`, where "len" is the length of the array. This loop iterates through each element of the array once, so it has a time complexity of O(n).
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3. The code inside the inner loop contains constant-time operations, such as addition, comparison, and list creation.
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- The inner loop uses two pointers (`l` and `r`) and runs until `r`is greater than `l`. Within the inner loop, constant-time operations are performed, such as calculating the sum of three elements and adjusting the pointers.
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Overall, the dominant time complexity is determined by the sorting step, which is O(n * log(n)). The loop inside the sorting step has a complexity of O(n), but it doesn't dominate the overall complexity.
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3. Inside the inner loop, there are additional while loops that skip duplicate elements. These while loops also perform constant-time operations.
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Since the sorting step has a time complexity of O(n * log(n)), and the nested loops contribute O(n) iterations, the overall time complexity is dominated by the sorting step, resulting in O(n * log(n)).
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So, the total time complexity of the program is O(n * log(n)) due to the sorting step.
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**Space Complexity (Big O Space):**
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The space complexity of this program is O(1) because it uses a constant amount of extra space. The program creates a few integer variables (`l`, `r`, `sum`, and `len`), but the space used by these variables is independent of the input size. Additionally, the `result` list stores the output, but its space is not considered part of the space complexity analysis, as it's required to store the program's output.
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The space complexity of the program is determined by the space used for the output list `result`. In the worst case, when there are many unique triplets that sum to zero, the size of `result` can be significant.
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1. The input array `nums` and the variables `len`, `l`, and `r` all use constant space.
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2. The `result` list stores the triplets, and its size can be at most O(n^2) when there are many unique triplets.
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3. Other variables and temporary lists used inside the loops use constant space.
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So, the space complexity of the program is O(n^2) for the `result` list when considering the worst case.
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Therefore, the overall space complexity is O(1).
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In summary, the provided program has a time complexity of O(n * log(n)) due to sorting and a space complexity of O(n^2) for the `result` list when considering the worst-case scenario.
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