Write a function that takes in an array of positive integers and returns the maximum sum of non-adjacent elements in the array.
If the input array is empty, the function should return 0.
Sample input
[75, 105, 120, 75, 90, 135]
Sample output
330 = 75 + 120 + 135
Solution
Time Complexity: O(n)
Space Complexity: O(1)
static int maxSubsetSumNoAdjacent(int[] array) {
int answer = 0;
if (array.length == 0) {
return 0;
} else if (array.length == 1) {
return array[0];
}
int first = Math.max(array[0], array[1]);
int second = array[0];
int current = 0;
for (int i = 2; i < array.length; i++) {
current = Math.max(first, second + array[i]);
second = first;
first = current;
}
answer = first;
return answer;
}
You’re given a non-empty array of positive integers representing the amounts of time that specific queries take to execute. Only one query can be executed at a time, but the queries can be executed in any order.
A query’s waiting time is defined as the amount of time that it must wait before its execution starts. In other words, if a query is executed second, then its waiting time is the duration of the first query; if a query is executed third, then its waiting time is the sum of the durations of the first two queries.
Write a function that returns the minimum amount of total waiting time for all of the queries. For example, if you’re given the queries of durations
[1, 4, 5], then the total waiting time if the queries were executed in the order of [5, 1, 4] would be (0) + (5) + (5 + 1) = 11. The first query of duration 5 would be executed immediately, so its waiting time would be 0, the second query of duration 1 would have to wait 5 seconds (the duration of the first query) to be executed, and the last query would have to wait the duration of the first two queries before being executed.
Note: you’re allowed to mutate the input array.
Solution
Use a map to store previous total waiting time per index.
Time Complexity: O(n log n)
Space Complexity: O(1)
Algorithm used: Greedy Algorithm
static int getMinimumWaitingTime(int[] queries) {
/**
* sort the array to help getting the minimum total waiting time
*/
Arrays.sort(queries);
int size = queries.length;
if (size <= 1) {
return 0;
}
Map<Integer, Integer> map = new HashMap<>();
int totalWaitingTime = 0;
for (int i = 0; i < queries.length; i++) {
int currentWaitingTime = 0;
int previousTotalWaitingTime = 0;
if (i != 0) {
currentWaitingTime = queries[i - 1];
previousTotalWaitingTime = currentWaitingTime + map.get(i - 1);
}
map.put(i, previousTotalWaitingTime);
totalWaitingTime += previousTotalWaitingTime;
}
return totalWaitingTime;
}
The Fibonacci sequence is defined as follows: the first number of the sequence is 0, the second number is 1, and the nth number is the sum of the (n – 1)th and (n – 2)th numbers. Write a function that takes in an integer n and returns the nth Fibonacci number.
I have a solution here.
Write a function that takes in a non-empty string and that returns a boolean representing whether the string is a palindrome.
A palindrome is defined as a string that’s written the same forward and backward. Note that single-character strings are palindromes.
Solution
Loop through half of the string. Compare each character from one half to the each character from the other half.
static boolean isPalindrome(String str) {
for (int i = 0, x = str.length() - 1; i < str.length() / 2; i++, x--) {
if (str.charAt(x) != str.charAt(i)) {
return false;
}
}
return true;
}
Write a function that takes in a sorted array of integers as well as a target integer. The function should use the Binary Search algorithm to determine if the target integer is contained in the array and should return its index if it is, otherwise -1.
I have a solution here.