**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;
}
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;
}
In Java:
import java.io.BufferedReader;
import java.io.InputStreamReader;
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 {
st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
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;
}
}
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)
```

## View More Similar Problems

## Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

View Solution →## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

View Solution →## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

View Solution →## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

View Solution →## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

View Solution →