Program to find number of strictly increasing colorful candle sequences are there in Python

Suppose there are n candles which are aligned from left to right. The i-th candle from the left side has the height h[i] and the color c[i]. We also have an integer k, represents there are colors in range 1 to k. We have to find how many strictly increasing colorful sequences of candies are there? The increasing sequence is checked based on heights, and a sequence is said to be colorful if there are at least one candle of each color in range 1 to K are available. If the answer is too large, then return result mod 10^9 + 7.

So, if the input is like K = 3 h = [1,3,2,4] c = [1,2,2,3], then the output will be 2 because it has sequences [1,2,4] and [1,3,4].

To solve this, we will follow these steps −

• Define a function read() . This will take T, i
• s := 0
• while i > 0, do
  • s := s + T[i]
  • s := s mod 10^9+7
  • i := i -(i AND -i)
• return s
• Define a function update() . This will take T, i, v
• while i <= 50010, do
  • T[i] := T[i] + v
  • T[i] := T[i] mod 10^9+7
  • i := i +(i AND -i)
• return v
• From the main method, do the following −
• L := 2^k, R := 0, N := size of h
• for i in range 0 to L - 1, do
  • T := an array of size 50009 and fill with 0
  • t := 0
  • for j in range 0 to N - 1, do
    • if (i after shifting bits (c[j] - 1) times to the right) is odd, then
      • t := t + update(T, h[j], read(T, h[j] - 1) + 1)
      • t := t mod 10^9+7
  • if (number of bits in i) mod 2 is same as k mod 2, then
    • R := R + t
    • R := R mod 10^9+7
  • otherwise,
    • R := (R + 10^9+7) - t
    • R := R mod 10^9+7
• return R

Example

Let us see the following implementation to get better understanding −

def solve(k, h, c):
   def read(T, i):
      s = 0
      while i > 0:
         s += T[i]
         s %= 1000000007
         i -= (i & -i)
      return s

   def update(T, i, v):
      while i <= 50010:
         T[i] += v
         T[i] %= 1000000007
         i += (i & -i)
      return v

   def number_of_bits(b):
      c = 0
      while b:
         b &= b - 1
         c += 1
      return c

   L = 2 ** k
   R = 0
   N = len(h)

   for i in range(L):
      T = [0 for _ in range(50010)]
      t = 0

      for j in range(N):
         if (i >> (c[j] - 1)) & 1:
            t += update(T, h[j], read(T, h[j] - 1) + 1)
            t %= 1000000007

      if number_of_bits(i) % 2 == k % 2:
         R += t
         R %= 1000000007
      else:
         R += 1000000007 - t
         R %= 1000000007
   return R

k = 3
h = [1,3,2,4]
c = [1,2,2,3]

print(solve(k, h, c))

Input

3, [1,3,2,4], [1,2,2,3]

Output

2

Arnab Chakraborty

Published on 23-Oct-2021 07:19:41

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