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Leetcode 3074. Apple Redistribution into Boxes

Problem Statement

You are given an integer array apples where apples[i] represents the number of apples in the i-th box. Your task is to redistribute the apples in the boxes such that no box has more than a certain number of apples (maxApples). Each redistributable move consists of moving one apple from one box to another.

Given apples and maxApples, return the minimum number of moves required to achieve the goal.

Clarifying Questions

  1. Can we assume apples and maxApples are both non-negative integers?
    • Yes, both apples and maxApples are non-negative.
  2. Is it possible for maxApples to be smaller than the count of apples in the smallest box?
    • Yes, the algorithm should handle this scenario as well.
  3. Are there any constraints on the range of values for the length of apples (n) and the values within apples?
    • Assume reasonable constraints typical in coding competitions (e.g., 1 ≤ n ≤ 10^5 and 0 ≤ apples[i] ≤ 10^9).
  4. Should we consider cases where it is impossible to redistribute apples such that no box exceeds maxApples?
    • We can assume inputs are given such that a valid redistribution is always possible.

Strategy

  1. Identify Boxes Exceeding maxApples: First, we identify all the boxes that have more apples than maxApples.
  2. Calculate Excess Apples: For each box that exceeds maxApples, calculate the excess number of apples.
  3. Redistribute Apples: Redistribute the excess apples to other boxes. We can iterate over the boxes and adjust the counts by moving apples from overloaded boxes to underloaded boxes until all boxes fit within the maxApples limit.
  4. Calculate Moves: For each apple moved, count that as one move. The output is the total number of moves required.

Code

public class AppleRedistribution {
    public int minMoves(int[] apples, int maxApples) {
        int excessApples = 0;
        int moves = 0;
        
        for (int apple : apples) {
            if (apple > maxApples) {
                excessApples += apple - maxApples;
            }
        }
        
        for (int apple : apples) {
            if (apple < maxApples) {
                int need = maxApples - apple;
                int canTake = Math.min(need, excessApples);
                moves += canTake;
                excessApples -= canTake;
            }
        }
        
        return moves;
    }
    
    public static void main(String[] args) {
        AppleRedistribution redist = new AppleRedistribution();
        int[] apples = {5, 8, 6, 7, 3, 2}; // Example input
        int maxApples = 6;
        int result = redist.minMoves(apples, maxApples);
        System.out.println("Minimum moves required: " + result);
    }
}

Time Complexity

This solution efficiently handles the problem within linear time complexity, making it suitable for large arrays.

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