Array Manipulation HackerRank Solution

Array Manipulation HackerRank Solution
Source : https://www.hackerrank.com/challenges/crush

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in your array.

For example, the length of your array of zeros . Your list of queries is as follows:

    a b k    1 5 3    4 8 7    6 9 1

Add the values of between the indices and inclusive:

index->  1 2 3  4  5 6 7 8 9 10 [0,0,0, 0, 0,0,0,0,0, 0] [3,3,3, 3, 3,0,0,0,0, 0] [3,3,3,10,10,7,7,7,0, 0] [3,3,3,10,10,8,8,8,1, 0]

The largest value is after all operations are performed.

Function Description

Complete the function arrayManipulation in the editor below. It must return an integer, the maximum value in the resulting array.

arrayManipulation has the following parameters:

• n - the number of elements in your array
• queries - a two dimensional array of queries where each queries[i] contains three integers, a, b, and k.

Input Format

The first line contains two space-separated integers and , the size of the array and the number of operations.
Each of the next lines contains three space-separated integers , and , the left index, right index and summand.

Constraints

Output Format

Return the integer maximum value in the finished array.

Sample Input

5 31 2 1002 5 1003 4 100

Sample Output

200

Explanation

After the first update list will be 100 100 0 0 0.
After the second update list will be 100 200 100 100 100.
After the third update list will be 100 200 200 200 100.
The required answer will be .

Source : https://www.hackerrank.com/challenges/crush

Solution

	// Karthikalapati.blogspot.com
	import java.util.Scanner;
	// For each of the "m" operations, we do not want to take O(n) time to process it. That's because our runtime will end up being O(nm). To get a O(n+m) runtime, we have to process each operation in O(1) time. To do so, we keep track of just the endpoints, which are just 2 numbers, instead of the O(n) numbers in between the endpoints. This is the main idea to decrease our runtime.
	// For a value "k", we can mark its left endpoint "a" in the original array, along with its right endpoint "b" in the original array. To distinguish between "a" and "b" we can store "+k" for "a" and "-k" for "b". Now, we technically have stored all information that the "m" operations represent, into an array. Most importantly, we stored it in O(m) time.
	// The next step is to actually find the maximum value based off of our unique representation of the data. We traverse our array from left to right. Whenever we reach a left endpoint "a" (which we represented with a positive number), we just add that to our sum. Whenever we reach a right endpoint "b" (which we represented with a negative number), we subtract that from our sum (well, technically we add a negative value). By doing so, the value "sum" represents the value that array[i] would have if we had applied all "m" operations to it. The maximum value of "sum" that we get while traversing the array is the value we return. If this algorithm is still unclear to you, try walking through HackerRank's sample input (Testcase 0) with the code below.
	// Time Complexity: O(n + m)
	// Space Complexity: O(n)
	public class Solution {
	public static void main(String[] args) {
	Scanner scan = new Scanner(System.in);
	int N = scan.nextInt();
	int M = scan.nextInt();
	/* Save interval endpoints' "k" values in array */
	long [] array = new long[N + 1];
	while (M-- > 0) {
	int a = scan.nextInt();
	int b = scan.nextInt();
	int k = scan.nextInt();
	array[a-1] += k;
	array[b] -= k; // see comment at end of code
	}
	scan.close();
	/* Find max value */
	long sum = 0;
	long max = 0;
	for (int i = 0; i < N; i++) {
	sum += array[i];
	max = Math.max(max, sum);
	}
	System.out.println(max);
	}
	}
	// Regarding the exact values of "a" and "b", it's a little tricky since "a" and "b" are 1-indexed but our array is 0-indexed. So we want to subtract 1 from both "a" and "b". However, for "b" we re-add 1 because we want to change values from "a" to "b" inclusive as stated in the problem, so we want the end of the interval to be 1 to the right of "b" which is why we re-add the 1.
	// Discuss on HackerRank: https://www.hackerrank.com/challenges/crush/forum/comments/255515