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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

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]))

[[0, 4, 4], [0, 5, 12], [2, 6, 9], [4, 8, 10]], [2, 5, 4, 6, 8]

4

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