Suppose, we have N items, which are marked as 0, 1, 2, …, N-1. Then, we are given a 2D list of size S called sets. Here, we can purchase the i-th set for the price sets[i,2], and we receive every item between sets[i, 0] to sets[i, 1]. Also, we have a list of size N called removals, where we can throw away 1 instance of the i-th element for the price removals[i]. So, we have to find out the minimum cost to purchase precisely one of every element from 0 to N-1 inclusive, or the result is -1 if this can't be done.
So, if the input is like sets = [ [0, 4, 4], [0, 5, 12], [2, 6, 9], [4, 8, 10] ]
removals = [2, 5, 4, 6, 8], then the output will be 4.
To solve this, we will follow these steps −
N := size of removals
graph := a new matrix for size (N + 1) x (N + 1)
for each s, e, w in sets, do
add [e+1, w] to graph[s]
for each i, r in index i and item r in removals, do
add [i, r] to graph[i + 1]
pq := a new heap
dist := a new map
dist[0] := 0
while pq is not empty, do
d, e := remove the smallest item from heap pq
if dist[e] < d is non-zero, then
continue the next iteration
if e is same as N, then
return d
for each nei, w in graph[e], do
d2 := d + w
if d2 < dist[nei] is non-zero, then
dist[nei] := d2
add [d2, nei] to pq
return -1
Example
Let us see the following implementation to get better understanding −
import heapq
from collections import defaultdict
class Solution:
def solve(self, sets, removals):
N = len(removals)
graph = [[] for _ in range(N + 1)]
for s, e, w in sets:
graph[s].append([e + 1, w])
for i, r in enumerate(removals):
graph[i + 1].append([i, r])
pq = [[0, 0]]
dist = defaultdict(lambda: float("inf"))
dist[0] = 0
while pq:
d, e = heapq.heappop(pq)
if dist[e] < d:
continue
if e == N:
return d
for nei, w in graph[e]:
d2 = d + w
if d2 < dist[nei]:
dist[nei] = d2
heapq.heappush(pq, [d2, nei])
return -1
ob = Solution()
print (ob.solve([
[0, 4, 4],
[0, 5, 12],
[2, 6, 9],
[4, 8, 10]
], [2, 5, 4, 6, 8]))Input
[[0, 4, 4], [0, 5, 12], [2, 6, 9], [4, 8, 10]], [2, 5, 4, 6, 8]
Output
4