Given a number N(1<=N<=2000)., The task is to find the number strings of size N that can be obtained after using characters from 'a' to 'z' and by processing the given q(1<=q<=200000) queries.
For each query given two integers L, R (0<=L<=R<=N) such that substring [L, R] of the generated string of size N must be a palindrome. The task is to process all queries and generate a string of size N such that the substrings of this string defined by all queries are palindrome.
The answer can be very large. So, output answer modulo 1000000007.
Note: 1-based indexing is considered for the string.
Input : N = 3 query 1: (1, 2) query 2: (2, 3) Output : 26 Explanation : Substring 1 to 2 should be palindrome and substring 2 to 3 should be palindrome. so, all three characters should be same. so, we can obtain 26 such strings. Input : N = 4 query 1: (1, 3) query 2: (2, 4) Output : 676 Explanation : substring 1 to 3 should be palindrome and substring 2 to 4 should be a palindrome. So, a first and third character should be the same and second and the fourth should be the same. So, we can obtain 26*26 such strings.
Approach : An efficient solution is to use union-find algorithm.
- Find the mid-point of each range (query) and if there are many queries having the same mid-point then only retain that query whose length is max, i.e (where r – l is max).
- This would have reduced the number of queries to 2*N at max since there is a 2*N number of mid-points in a string of length N.
- Now for each query do union of element l with r, (l + 1) with (r – 1), (l + 2) with (r – 2) and so on. We do this because the character which would be put on the index l would be the same as the one we put on index r. Extending this logic to all queries we need to maintain disjoint-set data structure. So basically all the elements of one component of disjoint-set should have the same letter on them.
- After processing all the queries, let the number of disjoint-set components be x, then the answer is 26^x
Below is the implementation of the above approach :
# Python3 program to implement above approach
N = 2005
mod = 10**9 + 7
# To store the size of string and
# number of queries
n, q = 0, 0
# To store parent and rank of ith place
par = [0 for i in range(N)]
Rank = [0 for i in range(N)]
# To store maximum interval
m = dict()
# Function for initialization
for i in range(n + 1):
Rank[i], par[i] = 0, i
# Function to find parent
if (par[x] != x):
par[x] = find(par[x])
# Function to make union
def Union(x, y):
xpar = find(x)
ypar = find(y)
if (Rank[xpar] < Rank[ypar]):
par[xpar] = ypar
elif (Rank[xpar] > Rank[ypar]):
par[ypar] = xpar
par[ypar] = xpar
Rank[xpar] += 1
# Power function to calculate a raised to m1
# under modulo 10000007
def power(a, m1):
if (m1 == 0):
elif (m1 == 1):
elif (m1 == 2):
return (a * a) % mod
elif (m1 & 1):
return (a * power(power(a, m1 // 2), 2)) % mod
return power(power(a, m1 // 2), 2) % mod
# Function to take maxmium interval
def query(l, r):
if l + r in m.keys():
m[l + r] = max(m[l + r], r)
m[l + r] = max(0, r)
# Function to find different possible strings
for i in m:
x = i – m[i]
y = m[i]
# make union of all chracters which
# are meant to be same
while (x < y): Union(x, y) x += 1 y -= 1 # find number of different sets formed ans = 0 for i in range(1, n + 1): if (par[i] == i): ans += 1 # return the required answer return power(26, ans) % mod # Driver Code n = 4 # queries query(1, 3) query(2, 4) print(possiblestrings()) # This code is contributed by Mohit Kumar [tabbyending]
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Improved By : mohit kumar 29