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apcsaArraysCountIncrements


Given an array of ints, return the number of times that one element is directly followed by another element with a value of one greater. Thus, if the element at index 3 has the value 40 and the element at index 4 has the value 41, that counts once toward the number you are building.

HINT
You will need to compare two side-by-side elements using your loop control variable and either the variable plus one or minus one. Be careful about the edge cases at the beginning and end!

apcsaArraysCountIncrements([3, 4, 5]) → 2
apcsaArraysCountIncrements([5, 4, 3]) → 0
apcsaArraysCountIncrements([3, 5, 6]) → 1

...Save, Compile, Run (ctrl-enter)

public int apcsaArraysCountIncrements(int[] nums) { }

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