Originally published byDev.to
One way is to traverse the array and keep updating the min and max elements .
- initialize min = arr[0] max = arr[o]
traverse the array from index 1
For each element:
If element < min β update min
If element > max β update max
4.return max and min array
5.Time Complexity: O(n)
We traverse the array once
Space Complexity: O(1)
πΊπΈ
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