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I think about using a binary tree (maybe a balanced binary tree).
- gf September 19, 2016At first it's necessary to build the tree with the k elements, this takes O(k*log(k)).
Then to find the min just check the leaf at the left, this will be O(log(k)).
When finding the min in the next subarray, at first it's neccesary to remove the element that's leaving O(1) and add the new element O(log(k)), this has to be done n-k+1 times, so the complexity will be O(n*log(k)).
If k = 1, then it will be O(n).