# Difference between revisions of "TADM2E 4.2"

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− | (a) | + | (a) Iterate over the array once, keeping track of the max and min values respectively for x and y. This will have <math>O(n)</math> worst-case time. |

(b) The answer is straight forward: <math>S[1]</math> and <math>S[n],</math> since they are the two extreme values. | (b) The answer is straight forward: <math>S[1]</math> and <math>S[n],</math> since they are the two extreme values. |

## Revision as of 22:33, 14 January 2018

(a) Iterate over the array once, keeping track of the max and min values respectively for x and y. This will have $ O(n) $ worst-case time.

(b) The answer is straight forward: $ S[1] $ and $ S[n], $ since they are the two extreme values.

(c) Sort the array with any $ n\log(n) $ method. Then scan through the sorted array to find the smallest gap, thus the desired pair.

(d) Same as above.