Not getting any approach to solve

#1

Please help me find the approach to this question and I have also read the article on geeksforgeeks but still I not getting logic

#2

this problem can be solved by using 2 pointer approach and maintaining a frequency array/map.
first pointer: start of sub-array
second pointer: end of sub-array
if current sub-array has distinct numbers, you can add more numbers, if it does not have all distinct numbers, then you can remove numbers
time complexity: O(N)

#3

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