# Array Manipulation

### Problem Statement :

```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 the array.

Example:
n=10
queries=[[1,5,3], [4,8,7], [6,9,1]]

Queries are interpreted as follows:

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

Add the values of k between the indices a and b 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 10 after all operations are performed.

Function Description:

Complete the function arrayManipulation in the editor below.

arrayManipulation has the following parameters:

i   1. nt n - the number of elements in the array
2.int queries[q][3] - a two dimensional array of queries where each queries[i] contains three integers, a, b, and k.

Returns:
1. int - the maximum value in the resultant array

Input Format:

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

Constraints:
1.  3<=n<=10^7
2.  1<=m<=2*10^5
3.  1<=a,b<=n
4.  0<=k<=10^9```

### Solution :

```                            ```Solution in C :

In C:

#include <stdio.h>

long long arr[10000005];
long long diff[10000005];

int main(void)
{
int a,b,k,n,m,i;
long long max,val;

scanf("%d%d",&n,&m);

while(m--)
{
scanf("%d%d%d",&a,&b,&k);

if(a==1)
{
arr[1] += k;

if(b<n)
diff[b] -= k;
}
else if(b<n)
{
diff[a-1] += k;
diff[b] -= k;
}
else if(b==n)
diff[a-1] += k;
}

max = arr[1];
val = arr[1];

for(i=1;i<n;i++)
{
val += diff[i];
if(val > max)
max = val;
}

printf("%lld\n",max);

return 0;
}```
```

```                        ```Solution in C++ :

In C++:

#include <iostream>

using namespace std;
const int NMAX = 1e7+2;
long long a[NMAX];
int main()
{
int n, m;
cin >> n >> m;
for(int i=1;i<=m;++i){
int x, y, k;
cin >> x >> y >> k;
a[x] += k;
a[y+1] -= k;
}
long long x = 0,sol=-(1LL<<60);
for(int i=1;i<=n;++i){
x += a[i];
sol = max(sol,x);
}
cout<<sol<<"\n";
return 0;
}```
```

```                        ```Solution in Java :

In Java:

import java.io.StreamTokenizer;

public class Solution {

public static void main(String[] args) throws Exception {
new Solution().run();
}

StreamTokenizer st;

private void run() throws Exception {
int n = nextInt();
int m = nextInt();
long[] a = new long[n + 1];
for (int i = 0; i < m; i++) {
int l = nextInt() - 1;
int r = nextInt();
int v = nextInt();
a[l] += v;
a[r] -= v;
}
long cur = 0;
long max = 0;
for (int i = 0; i < n; i++) {
cur += a[i];
max = Math.max(max, cur);
}
System.out.println(max);
}

private int nextInt() throws Exception {
st.nextToken();
return (int) st.nval;
}
}```
```

```                        ```Solution in Python :

In Python 3:

N,M = [int(_) for _ in input().split(' ')]

Is = {}
for m in range(M) :
a,b,k = [int(_) for _ in input().strip().split(' ')]
if k == 0 :
continue
Is[a] = Is.get(a,0) + k
Is[b+1] = Is.get(b+1,0) - k

m,v = 0,0
for i in sorted(Is) :
v += Is[i]
if v > m :
m = v

print(m)```
```

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