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Problem Solution Could use nested for-loops but that will end up with O(n^2) complexity. Arrays have constant complexity when accessing a known index. However arrays can have up to O(n) complexity when searching while hashmap has near constant complexity. In order to improve the speed, make a hashmap by mapping each element in nums to its index number. When iterating through the array, get the c..

Problem Solution Max distance that we can move at a time is 2. Look at the index value larger than current position by 2. If the value is 0, move 2 times and add a jump. If the value is 1, move just one time and add a jump. (If we are at a position where only one number is left in the array, it is not possible to move up 2 times... so just add one jump and end the loop.) import java.io.*; import..

Problem Need two integers: one for keeping track of the number of valleys, and another for tracking current position. Let the hiker start at position 0. "U" represents +1, "D" represents -1. Since we only care about the number of valleys completed and not mountains, we want to know when the hiker climbs out of a valley... which would be the moment position -1 becomes 0. import java.io.*; import ..