def find_pairs(arr, target):
n = len(arr)
result = []
for i in range(n):
for j in range(i+1, n):
if arr[i] + arr[j] == target:
result.append((i, j))
return result
Time Complexity: O(n^2)
- Outer loop: n iterations
- Inner loop: (n-1) + (n-2) + ... + 1 = n(n-1)/2 iterations
- Total: O(n^2)
Space Complexity: O(k) where k = number of pairs found
- result array grows with matches
- Worst case: O(n^2) if all pairs match
Optimization Suggestion:
- Use hash table for O(n) time complexity
- Trade space for time: O(n) space
{
"analysis": {
"function": "string",
"language": "string",
"timeComplexity": {
"notation": "O(n^2)",
"bestCase": "O(1)",
"averageCase": "O(n^2)",
"worstCase": "O(n^2)",
"derivation": [
"Step 1: Outer loop runs n times",
"Step 2: Inner loop runs (n-1), (n-2), ..., 1 times",
"Step 3: Total = sum from 1 to n-1 = n(n-1)/2",
"Step 4: Simplify to O(n^2)"
]
},
"spaceComplexity": {
"notation": "O(n)",
"auxiliary": "O(n)",
"total": "O(n)",
"breakdown": {
"input": "O(n) - input array",
"result": "O(k) - output pairs",
"variables": "O(1) - loop counters"
}
},
"recommendations": [
{
"type": "optimization",
"description": "Use hash table approach",
"newComplexity": "O(n) time, O(n) space",
"tradeoff": "Space for time"
}
]
},
"metadata": {
"analyzedAt": "ISO8601 timestamp",
"confidence": "high|medium|low"
}
}