CF 1399B - Gifts Fixing

CF 1399B - Gifts Fixing

Rating: 800
Tags: greedy
Solve time: 1m 29s
Verified: yes

Solution

Problem Understanding

We are given n gifts, where each gift consists of a certain number of candies and a certain number of oranges. Each gift i has a_i candies and b_i oranges. Our goal is to make all gifts identical in terms of both candies and oranges, using the fewest possible moves. In one move, we can eat one candy, one orange, or both from a single gift, but we cannot remove more than what a gift contains.

The input consists of multiple test cases. For each test case, we are given the number of gifts and two arrays: one for candies and one for oranges. We need to output the minimum number of moves required to make all gifts equal.

Constraints are moderate: n can go up to 50 and each a_i or b_i can be as large as 10^9. This means that we cannot simulate every single move individually if the numbers are huge, because that would be too slow. Instead, we need an approach that computes the total number of moves directly from differences.

Edge cases that can trip up a naive solution include: when all gifts already have the same number of candies or oranges, or when one gift has a very large number of candies or oranges compared to the rest. For example, if a = [1, 1000000000] and b = [1, 1], a naive simulation of removing one candy at a time would be infeasible.

Approaches

A brute-force approach would try to repeatedly choose gifts and decrement candies and oranges until all gifts match. This would be correct but inefficient, as each operation could be repeated up to 10^9 times for some gifts. With n=50, that could result in 50 * 10^9 operations, which is impossible within 1 second.

The key observation that leads to the optimal solution is that the target number of candies for all gifts must be at least the minimum among all a_i, and similarly, the target number of oranges must be at least the minimum among all b_i. Reducing every gift to its respective minimum is always feasible and requires the fewest moves per gift. For a single gift, we can perform "both at once" moves optimally. Specifically, for a gift with a_i candies and b_i oranges, if we want to reduce it to min_a and min_b, we can reduce both simultaneously as many times as min(a_i - min_a, b_i - min_b), then finish the rest individually. This greedy approach ensures minimal moves.

The brute-force and optimal approach comparison is:

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n * max(a_i, b_i))	O(n)	Too slow
Optimal	O(n) per test case	O(n)	Accepted

Algorithm Walkthrough

1. For each test case, read n and the arrays a and b representing candies and oranges.
2. Compute min_a = min(a) and min_b = min(b). These represent the target number of candies and oranges for all gifts.
3. Initialize moves = 0. This will accumulate the total moves required.
4. Iterate over each gift i from 0 to n-1.
5. For gift i, compute the excess candies delta_a = a[i] - min_a and excess oranges delta_b = b[i] - min_b.
6. The number of moves for this gift is max(delta_a, delta_b). We use the maximum because we can reduce both candies and oranges simultaneously whenever possible. For example, if delta_a=3 and delta_b=1, one move can reduce both, leaving 2 moves for candies alone, totaling max(3,1)=3.
7. Add the moves for this gift to moves.
8. After processing all gifts, output the total moves.

Why it works: the invariant is that we always reduce each gift exactly to the minimum candies and oranges. By performing "both at once" moves greedily, we never perform unnecessary single-item reductions. Each gift’s required moves are independent, so summing them gives the global minimum.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    
    min_a = min(a)
    min_b = min(b)
    
    moves = 0
    for i in range(n):
        delta_a = a[i] - min_a
        delta_b = b[i] - min_b
        moves += max(delta_a, delta_b)
    
    print(moves)

The code first reads the number of test cases. For each test case, it reads the arrays, computes the minimum candies and oranges, and then iterates through each gift to compute the minimal moves using the maximum of the differences. This approach directly implements the optimal algorithm and handles very large numbers efficiently. The choice of max(delta_a, delta_b) ensures we optimally apply combined reductions first.

Worked Examples

Example 1

Input:

3
3 5 6
3 2 3

Step trace:

Gift	a_i	b_i	min_a	min_b	delta_a	delta_b	moves_i
1	3	3	3	2	0	1	1
2	5	2	3	2	2	0	2
3	6	3	3	2	3	1	3

Total moves = 1 + 2 + 3 = 6

This matches the sample output.

Example 2

Input:

5
1 2 3 4 5
5 4 3 2 1

Step trace:

Gift	a_i	b_i	min_a	min_b	delta_a	delta_b	moves_i
1	1	5	1	1	0	4	4
2	2	4	1	1	1	3	3
3	3	3	1	1	2	2	2
4	4	2	1	1	3	1	3
5	5	1	1	1	4	0	4

Total moves = 4 + 3 + 2 + 3 + 4 = 16

Matches sample output.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Computing min and iterating once over n gifts
Space	O(n)	Storing the arrays a and b

With t <= 1000 and n <= 50, the total number of operations is at most 1000*50 = 50,000, which easily fits within the time limit. Memory usage is minimal.

Test Cases

# helper function
import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    # solution
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
        b = list(map(int, input().split()))
        
        min_a = min(a)
        min_b = min(b)
        
        moves = 0
        for i in range(n):
            delta_a = a[i] - min_a
            delta_b = b[i] - min_b
            moves += max(delta_a, delta_b)
        print(moves)
    return output.getvalue().strip()

# provided samples
assert run("5\n3\n3 5 6\n3 2 3\n5\n1 2 3 4 5\n5 4 3 2 1\n3\n1 1 1\n2 2 2\n6\n1 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\n1 1 1 1 1 1\n3\n10 12 8\n7 5 4") == "6\n16\n0\n4999999995\n7"

# custom tests
assert run("1\n1\n100\n100") == "0", "single gift"
assert run("1\n2\n1 100\n1 1") == "99", "one gift large"
assert run("1\n3\n5 5 5\n2 2 2") == "0",