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Query #2
a = 2
b = 5
k = 10

Array
10 10 -10 0 0

Query #3
a = 3

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// For example, to find the 3rd element in in arr_2[ ] i.e. i = 2

actual_value(arr_2[2]) = actual_value(arr_2[0]) + actual_value(arr_2[1])

arr_2[0] = 10 //already given

actual_value(arr_2[1]) = arr_2[0] + arr_2[1] 
// 10 + 0 = 10

actual_value(arr_2[2]) = arr_2[1] + arr_2[2] 

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n = 5
m = 3 //queries

Query #1
a = 1
b = 2
k = 10

Array after adding k = 10 from index a to b (1 to 2)
arr_1 = 10 10 0 0 0

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Sample Input
5 3
1 2 10
2 5 10
3 4 10

Sample Output
20

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    /*
    * Problem : https://www.hackerrank.com/challenges/crush/problem
    */
    static long arrayManipulation(int n, int[][] queries) {
        long[] arr = new long[n];
        long max = -1;
        int a,b,k;
            
            for(int i = 0; i < queries.length; i++) {
                a = queries[i][0];