Problem
Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array.
Input
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 m lines contains three space-separated integers a, b and k, the left index, right index and summand.
Output
int: the maximum value in the resultant array
Constraints
- 1 <= n <= 107
- 1 <= m <= 2 x 105
- 1 <= a <= b <= n
- 1 <= k <= 107
Sample Input
5 3 1 2 100 2 5 100 3 4 100
Sample Output
200 Explanation After the first update the list is 100 100 0 0 0. After the second update list is 100 200 100 100 100. After the third update list is 100 200 200 200 100. The maximum value is 200.
Solution
Approach: Optimal Brute Force and Sorting
Code Implementation
//
// main.cpp
// Array Manipulation
//
// Created by Himanshu on 30/03/22.
//
#include <iostream>
#include <vector>
#include <tuple>
#include <algorithm>
#include <climits>
using namespace std;
typedef long long ll;
//Custom comparator to sort operations (a, b, k)
//in ascending order based on 'a'
bool cmp (const tuple<ll, ll, ll> &e, const tuple<ll, ll, ll> &f) {
if (get<0>(e) == get<0>(f)) {
return (get<1>(e) <= get<1>(f));
} else {
return (get<0>(e) <= get<0>(f));
}
}
//Custom comparator to sort operations (a, b, k)
//in ascending order based on 'b'
bool cmp2 (const tuple<ll, ll, ll> &e, const tuple<ll, ll, ll> &f) {
if (get<1>(e) == get<1>(f)) {
return (get<0>(e) <= get<0>(f));
} else {
return (get<1>(e) <= get<1>(f));
}
}
int main() {
ll n, m, a, b, k;
ll inc = 0, leftIdx = 0, rightIdx = 0, ans = INT_MIN;
cin>>n>>m;
vector<tuple<ll, ll, ll>> inp1, inp2;
for (int i=0; i<m; i++) {
cin>>a>>b>>k;
inp1.push_back(make_tuple(a, b, k));
inp2.push_back(make_tuple(a, b, k));
}
sort(inp1.begin(), inp1.end(), &cmp);
sort(inp2.begin(), inp2.end(), &cmp2);
for (int i=0; i<=n; i++) {
//Add the value 'k' to the ans,
//if i is in range, or equal to 'a'
while (i == get<0>(inp1[leftIdx])) {
inc += get<2>(inp1[leftIdx]);
leftIdx++;
}
//Subtract the value 'k' from the ans,
//if i is not in range, or greater than 'b'
while (i == get<1>(inp2[rightIdx]) + 1) {
inc -= get<2>(inp2[rightIdx]);
rightIdx++;
}
//Select the max values as ans
ans = max(ans, inc);
}
cout<<ans;
return 0;
}
Time complexity: O(n + mlogm)