CF 102697076 - Jar of Candies

The jar starts with n candies. After that, we are given the number of candies still present in the jar on each observed day, beginning with the first day on which candies were stolen. For every day, we need to determine how many candies disappeared since the previous observation.

codeforcescompetitive-programming

CF 102697076 - Jar of Candies

Rating: -
Tags: -
Solve time: 51s
Verified: yes

Solution

Problem Understanding

The jar starts with n candies. After that, we are given the number of candies still present in the jar on each observed day, beginning with the first day on which candies were stolen. For every day, we need to determine how many candies disappeared since the previous observation.

The first observation is compared with the original amount n. Every later observation is compared with the previous day's remaining amount. If the remaining amounts are

n = 100, followed by 97 92 85,

then the thefts are 100 - 97 = 3, 97 - 92 = 5, and 92 - 85 = 7.

The output contains one integer per observed day, representing the number of candies stolen on that day.

The published problem does not specify a useful upper bound on the number of daily observations, so the safest complexity target is linear in the amount of input. Since every remaining-candy value must be examined at least once to produce its corresponding answer, O(m) time for m observations is optimal. A quadratic approach would repeatedly revisit previous observations and would become unnecessarily expensive as the input grows. Python integers also avoid any concern about overflow when performing the subtraction.

There are a few boundary cases that are easy to mishandle. If there is only one observation, there is no previous daily value, so it must be compared directly with the initial amount. For example,

10
7

produces

3

A careless implementation that starts by subtracting adjacent elements of the daily list would produce no answer at all.

The first day's theft can also be larger than the theft on later days. For example,

20
13 12 5

produces

7
1
7

The calculation must always use the immediately preceding amount, rather than assuming the theft amount is constant.

Finally, the arithmetic should be performed even when two consecutive observations happen to be equal. For example,

10
7 7 4

produces

3
0
3

Although the original problem describes the amount as decreasing, treating the input as consecutive observations makes the subtraction rule explicit and prevents a careless implementation from skipping an equal value.

Approaches

The direct approach is already the optimal one here. A brute-force implementation could, for every day, search backward through all previous observations to reconstruct how many candies had been present immediately before that day. That would still be correct if the previous amount were eventually found, but in the worst case it could perform about m(m-1)/2 comparisons for m observations, which is O(m²) work.

The structure of the input makes all of that searching unnecessary. The only value needed to calculate today's theft is yesterday's remaining number of candies. We do not need the entire history. Starting with the original n, read each daily value x, compute previous - x, output that difference, and then replace previous with x.

The brute-force approach works because the previous amount can be recovered from the input history, but it fails to exploit the fact that the history can be processed sequentially. The observation that each answer depends only on the immediately preceding remaining amount reduces the problem to a single pass.

Approach Time Complexity Space Complexity Verdict
Brute Force O(m²) O(m) Too slow for large input
Optimal O(m) O(1) auxiliary space Accepted

Here m is the number of daily observations.

Algorithm Walkthrough

  1. Read the initial number of candies in the jar and store it as previous. This is the amount against which the first day's observation must be compared.
  2. Read all remaining daily observations. Each value represents the number of candies left after that day's theft.
  3. For every observed value current, calculate previous - current. This is exactly the number of candies removed between the two observations.
  4. Append that difference to the output and set previous = current. The current day's remaining amount becomes the previous amount needed for the next day.
  5. Print all calculated differences, one per line. Processing the observations in their given order preserves the chronological relationship required by the subtraction.

Why it works

The invariant is that immediately before processing a daily observation current, previous contains exactly the number of candies that were in the jar before that day's theft. For the first observation, this is true because previous starts as the initial amount n. After processing a day, assigning previous = current makes the invariant true for the next day. Since the number stolen on a day is precisely the amount before the theft minus the amount after it, every produced difference is correct.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    remaining = list(map(int, sys.stdin.read().split()))

    previous = n
    answer = []

    for current in remaining:
        answer.append(str(previous - current))
        previous = current

    sys.stdout.write("\n".join(answer))

if __name__ == "__main__":
    solve()

The first line initializes previous with the original number of candies. The rest of the input is read with sys.stdin.read() rather than assuming a particular number of values on the second line. This follows the problem's description of the second line as a series of observations while also making the solution robust to ordinary whitespace formatting.

The loop performs exactly one subtraction for each observation. The assignment previous = current must happen after calculating the difference. Reversing those two operations would make every subtraction equal to zero.

The output is accumulated as strings and written once at the end. This avoids repeatedly calling print, which is useful when there are many observations. There is no integer-overflow issue in Python.

Worked Examples

Sample 1

For the published sample, the jar initially contains 100 candies and the observations are 97 92 85 80 72 65.

Step Previous Current Stolen
1 100 97 3
2 97 92 5
3 92 85 7
4 85 80 5
5 80 72 8
6 72 65 7

The resulting output is:

3
5
7
5
8
7

The trace shows why only one state variable is needed. After each subtraction, the current observation becomes the reference value for the next calculation.

Example 2

Consider:

20
13 12 5
Step Previous Current Stolen
1 20 13 7
2 13 12 1
3 12 5 7

The output is:

7
1
7

This example demonstrates that the theft amount does not need to be constant. Each answer is determined independently from the two consecutive jar levels.

Complexity Analysis

Measure Complexity Explanation
Time O(m) Every daily observation is processed exactly once
Space O(m) The implementation stores the observations and output strings before writing them

The computation itself uses only O(1) auxiliary state, namely previous. The presented implementation stores the input and output because it reads the observations as a complete sequence. Even so, it is linear in the input size. Since the problem provides no explicit large bound for the number of observations, linear processing is the appropriate target and avoids the quadratic behavior of repeatedly searching the history.

Test Cases

import sys
import io

def solve():
    n = int(input())
    remaining = list(map(int, sys.stdin.read().split()))

    previous = n
    answer = []

    for current in remaining:
        answer.append(str(previous - current))
        previous = current

    sys.stdout.write("\n".join(answer))

def run(inp: str) -> str:
    global input
    old_stdin = sys.stdin
    old_input = input

    try:
        sys.stdin = io.StringIO(inp)
        input = sys.stdin.readline

        output = io.StringIO()
        old_stdout = sys.stdout
        sys.stdout = output

        try:
            solve()
        finally:
            sys.stdout = old_stdout

        return output.getvalue()
    finally:
        sys.stdin = old_stdin
        input = old_input

# Provided sample
assert run(
    "100\n"
    "97 92 85 80 72 65\n"
) == "3\n5\n7\n5\n8\n7", "sample 1"

# Minimum-size observation sequence
assert run(
    "10\n"
    "7\n"
) == "3", "single observation"

# All values equal, exercising a zero difference
assert run(
    "10\n"
    "7 7 4\n"
) == "3\n0\n3", "equal consecutive observations"

# Boundary-style decreasing sequence
assert run(
    "5\n"
    "4 3 2 1\n"
) == "1\n1\n1\n1", "consecutive one-candy thefts"

# Large input size
large = "100000\n" + " ".join(
    str(100000 - i) for i in range(1, 100000)
) + "\n"
expected = "\n".join(["1"] * 99999)
assert run(large) == expected, "large linear input"
Test input Expected output What it validates
10 / 7 3 A single daily observation
10 / 7 7 4 3, 0, 3 Consecutive equal observations and state updates
5 / 4 3 2 1 1, 1, 1, 1 Repeated boundary-sized differences
100000 / 99999 ... 1 99999 lines containing 1 Linear performance on a large input

Edge Cases

For a single observation, such as

10
7

the algorithm starts with previous = 10, reads current = 7, and computes 10 - 7 = 3. It then updates previous to 7 and finishes. The output is 3. There is no attempt to access an earlier daily observation, so there is no off-by-one problem.

For varying theft amounts, consider

20
13 12 5

The first iteration computes 20 - 13 = 7, the second computes 13 - 12 = 1, and the third computes 12 - 5 = 7. The output is 7, 1, 7. The algorithm does not assume that the same number of candies is stolen every day.

For equal consecutive observations, consider

10
7 7 4

The first difference is 10 - 7 = 3. The state then becomes 7, so the second difference is 7 - 7 = 0. Finally, the state becomes 7 again and the last difference is 7 - 4 = 3. The output is 3, 0, 3. The sequential subtraction handles this naturally without needing a special case.