IBM Interview Question for Java Developers


Country: India
Interview Type: Written Test




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I cannot think of something better than O(n^2) here. Brute-force scan every block and evaluate the lowest neighbour. The block with the highest lowest neighbor wins. (Build list of candidates if candidate equals current best). Is there something better than O(n^2)?

- prettyGood August 12, 2017 | Flag Reply
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If we "can" assume that the function will be invoked frequently, then preprocessing will take an advantage, having a complexity down to O(N) where N is a number of elements in a grid.

- Lizzard October 16, 2017 | Flag Reply


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