Solution 1 for Scaler Topics Fortnightly Contest - 11

This article is part of the Scaler Topics Fortnightly Contest - 11

Solution Approach

• We can find the position of each of the values from 0 to N-1
• Then we can traverse i from 0 to N-2 and check the leftmost and rightmost boundary of the position of the values considered so far. If the length equals to the number of elements considered, then it is a good subarray. We will just have to find the good subarray of maxmium length.

Time complexity: O(N)
Space complexity: O(N)

C++ Implementation

Java Implementation

Python Implementation