1445. Apples & Oranges

Explanation

To solve this problem, we need to find the maximum number of apples and oranges that can be collected while maintaining a balance between them. We can achieve this by iterating through the given arrays of apples and oranges, and for each position, we calculate the total weight of apples and oranges on both sides of that position. We then update the count of apples and oranges collected based on the balance condition.

Algorithmic Steps:

Iterate through the given arrays of apples and oranges.
At each position, calculate the total weight of apples and oranges on both sides of that position.
Update the count of apples and oranges collected based on the balance condition.
Return the maximum count of apples and oranges that can be collected.

Time Complexity

The time complexity of this solution is O(n), where n is the total number of elements in the arrays of apples and oranges.

Space Complexity

The space complexity of this solution is O(1) as we are using a constant amount of extra space.

View Leetcode Problem

class Solution {
    public int maxApplesOranges(int[] apples, int[] oranges) {
        int leftApples = 0, rightApples = 0;
        int leftOranges = 0, rightOranges = 0;
        
        int maxApples = 0, maxOranges = 0;
        
        for (int i = 0; i < apples.length; i++) {
            rightApples += apples[i];
        }
        
        for (int i = 0; i < oranges.length; i++) {
            rightOranges += oranges[i];
        }
        
        for (int i = 0; i < apples.length; i++) {
            leftApples += apples[i];
            rightApples -= apples[i];
            
            leftOranges += oranges[i];
            rightOranges -= oranges[i];
            
            maxApples = Math.max(maxApples, Math.min(leftApples, rightOranges));
            maxOranges = Math.max(maxOranges, Math.min(leftOranges, rightApples));
        }
        
        return Math.max(maxApples, maxOranges);
    }
}

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