I have used in Prefix method

long [] arr= new long[n+2];

 for (List<Integer> q: queries){
    int a= q.get(0);
    int b=q.get(1);
    int k=q.get(2);

     arr [a] +=k;
     arr [b+1] -=k;
     }
     long max=0,current=0;
     for(int i =1; i<=n;i++){
        current += arr[i];
        if(current > max){
            max = current;
        }
     }
     return max;
     }

}

import sys queries=sys.stdin.read().strip().split() it=iter(queries) n=int(next(it)) q=int(next(it)) arr = [0] * (n+2)

for i in range(q): a = int(next(it)) b = int(next(it)) k = int(next(it)) arr[ a ] += k arr[ b +1] -= k

max_value = 0 current = 0 for val in arr: current += val max_value = max(max_value, current)

print(max_value)

import java.io.; import java.util.;

class Result {

/*
 * Complete the 'arrayManipulation' function below.
 *
 * The function is expected to return a LONG_INTEGER.
 * The function accepts following parameters:
 *  1. INTEGER n
 *  2. 2D_INTEGER_ARRAY queries
 */

public static long arrayManipulation(int n, List<List<Integer>> queries) {
    // We use the difference array technique
    long[] arr = new long[n + 2];  // n+2 to avoid index issues

    for (List<Integer> query : queries) {
        int a = query.get(0);
        int b = query.get(1);
        int k = query.get(2);

        arr[a] += k;      // add k at start index
        arr[b + 1] -= k;  // subtract k after end index
    }

    long max = 0;
    long current = 0;

    for (int i = 1; i <= n; i++) {
        current += arr[i];
        if (current > max) {
            max = current;
        }
    }

    return max;
}

}

public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

    String[] firstMultipleInput = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

    int n = Integer.parseInt(firstMultipleInput[0]);
    int m = Integer.parseInt(firstMultipleInput[1]);

    List<List<Integer>> queries = new ArrayList<>();

    for (int i = 0; i < m; i++) {
        String[] queriesRowTempItems = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

        List<Integer> queriesRowItems = new ArrayList<>();

        for (int j = 0; j < 3; j++) {
            int queriesItem = Integer.parseInt(queriesRowTempItems[j]);
            queriesRowItems.add(queriesItem);
        }

        queries.add(queriesRowItems);
    }

    long result = Result.arrayManipulation(n, queries);

    bufferedWriter.write(String.valueOf(result));
    bufferedWriter.newLine();

    bufferedReader.close();
    bufferedWriter.close();
}

}

You cannot directly sum each item in range because you will get a Time Exceeded Exception.

Instead, use prefix sum technique:

// Swift
func arrayManipulation(n: Int, queries: [[Int]]) -> Int {
    // 1-indexed
    var operations: [Int] = Array(repeating: 0, count: n + 1)

    for query in queries {
        let a = query[0]
        let b = query[1]
        let k = query[2]

        operations[a - 1] += k
        operations[b] -= k
    }
    
    var maxValue = 0
    var currentSum = 0
    for value in operations {
        currentSum += value
        maxValue = max(maxValue, currentSum)
    }
    
    return maxValue
}

tricky. Not explained very well.. KEY : prefix sum algorthim

    arr = [0] * (n+2)

    for i in range(len(queries)):
        a = queries[i][0]
        b = queries[i][1]
        k = queries[i][2]
        arr[ a ]   += k
        arr[ b +1] -= k

    max_value = 0
    current = 0
    for val in arr:
        current += val
        max_value = max(max_value, current)

    return (max_value)