POTD_12_OCT_2024_TwoSmallestsInEverySubarray

[Avnee29]

By maximizing consecutive element sum – O(n) time and O(1) auxiliary space

An efficient solution is based on the observation that this problem reduces to finding a maximum sum of two consecutive elements in array. If (x,y) is the pair such that (x+y) is the answer, then x and y must be consecutive elements in the array.

For a subarray with 2 elements, 1st and 2nd smallest elements are those 2 elements. Now, x and y are present in some subarray such that they are the endpoints. Now, x, y must be the smallest 2 elements of that subarray. If there are other elements Z1 , Z2, ……., ZK  between x and y, they are greater than or equal to x and y,

Case1 : 

If there is one element z between x and y, then the smaller subarray with the elements max(x,y) and z , should be the answer, because max(x,y) + z >= x + y

Case2:

If there are more than one elements between x and y, then the subarray within x and y will have all consecutive elements  (Zi + Zi+1) >= (x+y),  so (x,y) pair can’t be the answer. 

So, by contradictions, x and y must be consecutive elements in the array. Time Complexity: O(n), where n is the length of given array Auxiliary Space: O(1)

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