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Frogs are sitting on the x-axis of the plane. The ...

Created on: July 12, 2025

Answered using O3 by Chat01

Question

Frogs are sitting on the x-axis of the plane. The x-axis is represented by a given string. represents a frog and represents a stone. The string consists of only the above-mentioned characters. You are given a start index array and end index array, and calculate the number of frogs between the two stones including the endpoint. Note-Array is 1 indexed

6018-48

Sc

Input Format

First line contains a string containing (frogs) and (stones).

Second line contains an integer n denoting the number of test cases.

Third line contains for each test case, two integers start and end (1-indexed) specifying the range on the x-axis to check for the frogs.

Output Format

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x

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Output contains for each test case, the number of frogs (*) between the two stones (1) at the given start and end indices, including the endpoints.
code in python

Thought

Clarifying the frog problem

The user's problem involves counting the number of frogs ('') between two stones ('1') in a string, where indices are provided for each test case. The string consists of '' and '1'. The task is to count how many '*' characters exist between two '1's, inclusive of both stones, within a specified range. I'll need to design a Python solution that processes this input correctly, counting frogs in each interval efficiently. Let's break down how to implement this properly.

Clarifying the frog counting problem

The user's problem is a variation of the "Plates Between Candles" problem, where we are tasked with counting frogs ('') between two stones ('1') within a given range, inclusive of the endpoints. The string consists of '' and stones ('1'). For each test case, we count how many frogs exist between two stones. To make this efficient, I’ll use an approach with prefix sums and precomputing nearest stone indices. I'll write a Python solution that handles input, computes the count, and handles edge cases.

Thought for 11 seconds

Answer

Below is a concise, O(N + Q) Python solution (where N is the length of the x-axis string and Q the number of test cases).

The key idea is the same one used for the classic “plates between candles” problem:

  1. Pre-compute

    • prefix[i] – number of frogs (*) up to and including position i.
    • nearest_left_stone[i] – index of the closest stone (1) at or to the left of i (‒1 if none).
    • nearest_right_stone[i] – index of the closest stone (1) at or to the right of i (‒1 if none).

  2. Answer each query [L, R] (1-indexed):

    • Convert to 0-index: l = L-1, r = R-1.
    • Let left = nearest_right_stone[l] (first stone ≥ l).
    • Let right = nearest_left_stone[r] (last stone ≤ r).
    • If either is ‒1 or left >= right, there is no enclosing pair ⇒ answer 0.
    • Otherwise, the frogs strictly between the two stones equal
      prefix[right] – prefix[left].
python
import sys

def preprocess(s: str):
    n = len(s)
    prefix = [0] * (n + 1)           # prefix[0] = 0
    nearest_left = [-1] * n
    nearest_right = [-1] * n

    # prefix sum of frogs
    for i, ch in enumerate(s):
        prefix[i + 1] = prefix[i] + (ch == '*')

    # nearest stone on the left
    last = -1
    for i in range(n):
        if s[i] == '1':          # stone
            last = i
        nearest_left[i] = last

    # nearest stone on the right
    last = -1
    for i in range(n - 1, -1, -1):
        if s[i] == '1':
            last = i
        nearest_right[i] = last

    return prefix, nearest_left, nearest_right


def frogs_between_stones(s: str, queries):
    prefix, left_stone, right_stone = preprocess(s)
    res = []
    for L, R in queries:               # both are 1-indexed
        l, r = L - 1, R - 1
        left = right_stone[l]          # first stone inside/on left bound
        right = left_stone[r]          # last stone inside/on right bound
        if left == -1 or right == -1 or left >= right:
            res.append(0)
        else:
            res.append(prefix[right] - prefix[left])
    return res


def main():
    data = sys.stdin.read().strip().split()
    s = data[0]
    q = int(data[1])
    nums = list(map(int, data[2:]))
    queries = list(zip(nums[::2], nums[1::2]))
    answers = frogs_between_stones(s, queries)
    print("\n".join(map(str, answers)))

if __name__ == "__main__":
    main()

How it works

  • Pre-processing is linear in the string length.
  • Each query is answered in O(1) time using the pre-computed arrays.
  • Memory usage is linear in N for the three helper arrays.

You can run this script, provide the string, the number of test cases, and each start end pair, and it will print one answer per line.

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