Question

use c ++
Description
Find the maximum value of an array with length N after X operations. The initial value of all elements of the array is 0

Constraint
10 <= N <= 10 ^ 6

10 <= X <= 10 ^ 6

10 <= array [N] <= 10 ^ 6

Example
Input:

11 4
1 4 5
3 10 2
7 9 10
2 5 6
Output:

13
How to count

Line length (n) = 11
Many operations (x) = 4

| Operation to | start (s) | end (e) | value (v) |
| 1 | 1 | 4 | 5 | ... (1)
| 2 | 3 | 10 | 2 | ... (2)
| 3 | 7 | 9 | 10 | ... (3)
| 4 | 2 | 5 | 6 | ... (4)

Index -> 1- 2- 3- 4- 5- 6- 7- 8- 9- 10 11
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] <- Initial value of the row
        [5, 5, 5, 5, 0, 0, 0, 0, 0, 0] <- Value after 1st operation
        [5, 5, 7, 7, 2, 2, 2, 2, 2, 2, 0] <- Value after the 2nd operation
        [5, 5, 7, 7, 2, 2, 12, 12, 12, 2, 0] <- Value after the 3rd operation
        [5, 11, 13, 13, 8, 2, 12, 12, 12, 2, 0] <- Value after the 4th operation

Max-> 13

Answer 1

The answer to the question is given below.

The code in c++ is given below.

#include<bits/stdc++.h>
using namespace std;
int main()
{
   int n,k;
   cin>>n>>k;
   int arr[n]={0}; /*creating array of n elements with value 0*/
   /* while loop for k operation */
   while(k--)
   {
       int start,end,value;
       cin>>start>>end>>value;
       /*appling the operation on element from start to end*/
       for(int i=start;i<=end;i++)
       {
           arr[i-1]=arr[i-1]+value;
       }
   }
   /* displaying the maximum element in the array*/
   cout<<"Max Element after these operation: "<<*max_element(arr,arr+n);
   return 0;
}

The screenshot of the running code is given below.






The screenshot of the output is given below.






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