# Subsequence Weighting Hackerrank Solution

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence.

You are given a sequence `A` in which every element is a pair of integers  i.e  `A` = [(a1, w1), (a2, w2),…, (aN, wN)].

For a subseqence `B` = [(b1, v1), (b2, v2), …., (bM, vM)] of the given sequence :

• We call it increasing if for every i (1 <= i < M ) , bi < bi+1.
• Weight(B) = v1 + v2 + … + vM.

Given a sequence, output the maximum weight formed by an increasing subsequence.

Input:
The first line of input contains a single integer TT test-cases follow. The first line of each test-case contains an integer N. The next line contains a1, a2 ,… , aN separated by a single space. The next line contains w1, w2, …, wN separated by a single space.

Output:
For each test-case output a single integer: The maximum weight of increasing subsequences of the given sequence.

Constraints:
1 <= T <= 5
1 <= N <= 150000
1 <= ai <= 109, where i ∈ [1..N]
1 <= wi <= 109, where i ∈ [1..N]

Sample Input:

```2
4
1 2 3 4
10 20 30 40
8
1 2 3 4 1 2 3 4
10 20 30 40 15 15 15 50
```

Sample Output:

```100
110
```

Explanation:
In the first sequence, the maximum size increasing subsequence is 4, and there’s only one of them. We choose `B = [(1, 10), (2, 20), (3, 30), (4, 40)]`, and we have `Weight(B) = 100`.

In the second sequence, the maximum size increasing subsequence is still 4, but there are now 5 possible subsequences:

```1 2 3 4
10 20 30 40

1 2 3 4
10 20 30 50

1 2 3 4
10 20 15 50

1 2 3 4
10 15 15 50

1 2 3 4
15 15 15 50
```

Of those, the one with the greatest weight is `B = [(1, 10), (2, 20), (3, 30), (4, 50)]`, with `Weight(B) = 110`.

Please note that this is not the maximum weight generated from picking the highest value element of each index. That value, 115, comes from [(1, 15), (2, 20), (3, 30), (4, 50)], which is not a valid subsequence because it cannot be created by only deleting elements in the original sequence.

Contents

## Solution in Python

```#!/bin/python3

import os
import sys
import bisect

# Complete the solve function below.
def solve(a, w):
b = [[0,0],[10000000000,10000000000]]
for i in range(len(a)):
g = [a[i],w[i]]
bisect.insort(b,g)
ind = b.index(g)
if b[ind+1] != b[ind] and b[ind-1] != b[ind]:
b[ind]+=b[ind-1]
for j in range(ind+1,len(b)):
if b[j] >b[ind]:
break
b = b[:ind+1] + b[j:]
elif b[ind+1] == b[ind]:
b[ind]+=b[ind-1]
if b[ind+1]>=b[ind]:
b.remove(b[ind])
else:
b.remove(b[ind+1])
for j in range(ind+1,len(b)):
if b[j]>b[ind]:
break
b = b[: ind+1] + b[j: ]
elif b[ind-1] ==b[ind]:
b[ind] += b[ind-2]
if b[ind-1] >= b[ind]:
b.remove(b[ind])
else:
for j in range(ind+1,len(b)):
if b[j]>b[ind]:
break
b = b[: ind+1] + b[j: ]
b.remove(b[ind-1])
return b[-2]

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

t = int(input())

for t_itr in range(t):
n = int(input())

a = list(map(int, input().rstrip().split()))

w = list(map(int, input().rstrip().split()))

result = solve(a, w)

fptr.write(str(result) + '\n')

fptr.close()```